{"id":9596,"date":"2021-04-07T13:00:03","date_gmt":"2021-04-07T07:30:03","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=9596"},"modified":"2021-04-07T15:05:42","modified_gmt":"2021-04-07T09:35:42","slug":"samacheer-kalvi-12th-chemistry-solutions-chapter-7","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-12th-chemistry-solutions-chapter-7\/","title":{"rendered":"Samacheer Kalvi 12th Chemistry Solutions Chapter 7 Chemical Kinetics"},"content":{"rendered":"

Are you searching for the Samacheer Kalvi 12th Chemistry Chapter Wise Solutions PDF? Then, get your Samacheer Kalvi 12th Chapter Wise Solutions PDF for free on our website. Students can Download Chemistry Chapter 7 Chemical Kinetics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 7 Chemical Kinetics<\/h2>\n

All concepts are explained in an easy way on our website. So, students can easily learn and practice Tamilnadu State Board 12th Chemistry Chapter 7 Chemical Kinetics Question and Answers. You can enjoy the digitized learning with the help of the Tamilnadu State Board Chemistry Online Material.<\/p>\n

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Text Book Evalution<\/h3>\n

I. Choose the correct answer.<\/p>\n

12th Chemistry Chapter 7 Book Back Answers Question 1.<\/strong>
\nFor a first order reaction A \u2192 B the rate constant is x min-1<\/sup>. If the initial concentration of A is 0.01 M, the concentration of A after one hour is given by the expression.
\n(a) 0.01 e-x<\/sup>
\n(b) 1 x 10-2<\/sup> (1 – e-60x<\/sup>)
\n(c) (1 x 10-2<\/sup>) e-60x<\/sup>
\n(d) none of these
\nAnswer:
\n(c) (1 x 10-2<\/sup>) e-60x<\/sup>
\nAnswer:
\nSolutions:
\n\"12th
\nIn this case
\nk = x min-1<\/sup> and [A0<\/sub>] = 0.01 M
\n= 1 x 10-2<\/sup> M
\nt = 1 hour = 60 min
\n[A] = 1 x 10-2<\/sup>(e-60x<\/sup>)<\/p>\n

12th Chemistry 7th Lesson Book Back Answers Question 2.<\/strong>
\nA zero order reaction X \u2192 Product, with an initial concentration 0.02M has a half life of 10 min. If one starts with concentration 0.04M, then the half life is …………….
\n(a) 10 s
\n(b) 5 min
\n(c) 20 min
\n(d) cannot be predicted using the given information
\nAnswer:
\n(c) 20 min
\nSolutions:
\n\"12th
\nGiven,
\n[A0<\/sub>] = 0.02 M ; t1\/2<\/sub> = 10 min
\n[A0<\/sub>] = 0.04 M ; t1\/2<\/sub> = ?
\nSubstitute in (1)
\n10 min ? 0.02 M ……………………..(2)
\nt1\/2<\/sub> \u221d 0.04 M ……………………..(3)
\nDividing Eq.(3) by Eq. (2) we get,
\n\\(\\frac { { t }^{ 1\/2 } }{ 10min }\\) = \\(\\frac { 0.04M }{ 0.02M }\\)
\nt1\/2<\/sub> = 2 x 10 min = 20 min<\/p>\n

Chemical Kinetics Book Back Answers Question 3.<\/strong>
\nAmong the following graphs showing variation of rate constant with temperature (T) for a reaction, the one that exhibits Arrhenius behavior over the entire temperature range is ……………
\n\"Chemical
\nAnswer:
\n\"Samacheer
\nSolution:
\n\"Samacheer
\nIn k = In A – \\(\\left( \\frac { { E }_{ a } }{ R } \\right)\\) \\((\\frac { 1 }{ T })\\)
\nthis equation is in the form of a straight
\nline equation y = c + m x
\na plot of ink vs \\((\\frac { 1 }{ T })\\) is a straight line with negative slope.<\/p>\n

Samacheer Kalvi Guru 12th Chemistry Question 4.<\/strong>
\nFor a first order react ion A \u2192 product with initial concentration x mol L-1<\/sup>, has a half life period of 2.5 hours. For the same reaction with initial concentration mol L-1<\/sup> the half life is
\n(a) (2.5 x 2) hours
\n(b) \\((\\frac { 2.5 }{ 2 })\\) hours
\n(c) 2.5 hours
\n(d) Without knowing the rate constant, t1\/2<\/sub> cannot be determined from the given data
\nAnswer:
\n(d) Without knowing the rate constant, t1\/2<\/sub> cannot be determined from the given data.
\nSolutions:
\nFor a first order reaction
\nt1\/2<\/sub> = \\(\\frac { 0.693 }{ k }\\) t1\/2<\/sub> does not depend on the initial concentration and it remains constant (whatever may be the initial concentration)
\nt1\/2<\/sub> = 2.5 hrs .<\/p>\n

Samacheer 12 Chemistry Solutions Question 5.<\/strong>
\nFor the reaction, 2NH3<\/sub> \u2192 N2<\/sub> + 3H2<\/sub>, if
\n\"Samacheer
\nthen the relation between
\nk1<\/sub>, k2<\/sub> and k3<\/sub> is
\n(a) k1<\/sub> = k2<\/sub> = k3<\/sub>
\n(b) k1<\/sub> = 3 k2<\/sub> = 2 k3<\/sub>
\n(c) 1.5k1<\/sub> = 3 k2<\/sub> = k3<\/sub>
\n(d) 2k1<\/sub> = k2<\/sub> = 3 k3<\/sub>
\nAnswer:
\n(c) 1.5k1<\/sub> = 3 k2<\/sub> = k3<\/sub>
\nSolution:
\n\"Chemical<\/p>\n

Samacheer Kalvi 12th Chemistry Question 6.<\/strong>
\nThe decomposition of phosphine (PH3<\/sub>) on tungsten at low pressure is a first order reaction. It is because the …………….
\n(a) rate is proportional to the surface coverage
\n(b) rate is inversely proportional to the surface coverage
\n(c) rate is independent of the surface coverage
\n(d) rate of decomposition is slow
\nAnswer:
\n(c) rate is independent of the surface coverage
\nSolution:
\nGiven:
\nAt low pressure the reaction follows first order, therefore Rate \u221d [reactant]1<\/sup> Rate \u221d (surface area) At high pressure due to the complete coverage of surface area, the reaction follows zero order. Rate \u221d [reactant]\u00b0. Therefore the rate is independent of surface area.<\/p>\n

Chemical Kinetics Solutions Question 7.<\/strong>
\nFor a reaction Rate = k [acetone]3\/2<\/sup> then unit of rate constant and rate of reaction respectively is …………..
\n(a) (mol L-1<\/sup> s-1<\/sup>), (mol-1\/2<\/sup> L1\/2<\/sup> s-1<\/sup>)
\n(b) (mol-1\/2<\/sup> L1\/2<\/sup> s-1<\/sup>), (mol L-1<\/sup> s-1<\/sup>)
\n(c) (mol1\/2<\/sup> L1\/2<\/sup> s-1<\/sup>), (mol L-1<\/sup> s-1<\/sup>)
\n(d) (mol L s-1<\/sup>), (mol1\/2<\/sup> L1\/2<\/sup> s)
\nAnswer:
\n(b) (mol1\/2<\/sup> L1\/2<\/sup> s-1<\/sup>), (mol L-1<\/sup> s-1<\/sup>)
\nSolution:
\nRate = k [A]n<\/sup>
\nRate = \\(\\frac { -d[A] }{ dt } \\)
\nunit of rate = \\(\\frac { mol{ L }^{ -1 } }{ s }\\) = mol L-1<\/sup> s-1<\/sup>
\nunit of rate constant = \\(\\frac { (mol{ L }^{ -1 }{ S }^{ -1 }) }{ { (mol{ L }^{ -1 }) }^{ n } }\\)
\n= mol1-n<\/sup> Ln-1<\/sup> S-1<\/sup>
\nin this case, rate k [Acetone]3\/2<\/sup>
\nn = 3\/2
\nmol1-(3\/2)<\/sup> L(3\/2)-1<\/sup> s-1<\/sup>
\nmol-(1\/2)<\/sup> L(1\/2)<\/sup> s-1<\/sup><\/p>\n

Samacheer Kalvi Class 12 Chemistry Solutions Question 8.<\/strong>
\nThe addition of a catalyst during a chemical reaction alters which of the following quantities?
\n(a) Enthalpy
\n(b) Activation energy
\n(c) Entropy
\n(d) Internal energy
\nAnswer:
\n(b) Activation energy
\nSolution:
\nA catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.<\/p>\n

Chemical Kinetics In Tamil Question 9.<\/strong>
\nConsider the following statements:
\n(i) increase in concentration of the reactant increases the rate of a zero order reaction.
\n(ii) rate constant k is equal to collision frequency A if Ea<\/sub> = o
\n(iii) rate constant k is equal to collision frequency A if Ea<\/sub> = o
\n(iv) a plot of ln (k) vs T is a straight line.
\n(v) a plot of In (k) vs \\((\\frac { 1 }{ T })\\) is a straight line with a positive slope.<\/p>\n

Correct statements are
\n(a) (ii) only
\n(b) (ii) and (iv)
\n(c) (ii) and (v)
\n(d) (i), (ii) and (v)
\nAnswer:
\n(a) (ii) only
\nSolutions:
\nIn zero order reactions, increase in the concentration of reactant does not alter the rate, So statement (i) is wrong.
\n\"Samacheer
\nif Ea<\/sub> = O (so, statement (ii) is correct, and statement (iii) is wrong)
\nk = A e\u00b0
\nk = A
\nin k = A – \\(\\left( \\frac { { E }_{ a } }{ R } \\right)\\) \\(\\frac { 1 }{ T }\\)
\nthis equation is in the form of a straight line equation yc + m x. a plot of Ink vs \\(\\frac { 1 }{ T }\\) is a straight line with negative slope so statements (iv) and (v) are wrong.<\/p>\n

Samacheer Kalvi Guru Class 12 Chemistry Question 10.<\/strong>
\nIn a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively – x kJ mol-1<\/sup> and y kJ mol-1<\/sup>. Therefore, the energy of activation in the backward direction is ………..
\n(a) (v – x)kJ mol-1<\/sup>
\n(b) (x + y) J mol-1<\/sup>
\n(c) (x – y) kJ mol-1<\/sup>
\n(d) (x + y) x 103<\/sup> J mol-1<\/sup>
\nAnswer:
\n(d) (x + y) x 103<\/sup> J mol-1<\/sup>
\nSolution:
\n\"Chemical<\/p>\n

12th Chemistry Samacheer Kalvi Question 11.<\/strong>
\nWhat is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400K? (R 8.314 JK-1<\/sup> mol-1<\/sup>)
\n(a) 234.65 kJ mol-1<\/sup> K-1<\/sup>
\n(b) 434.65 kJ mol-1<\/sup> K-1<\/sup>
\n(c) 434.65 J mol-1<\/sup> K-1<\/sup>
\n(d) 334.65 J mol-1<\/sup> K-1<\/sup>
\nAnswer:
\n(c)434.65 J mol-1<\/sup> K-1<\/sup>
\nSolutions:
\n\"Samacheer<\/p>\n

Question 12.
\n\"12th
\nThis reaction follows first order kinetics. The rate constant at particular temperature is 2.303 x 102<\/sup> hourd. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? (Log 2 = 0.30 10)
\n(a) 0.125 M
\n(b) 0.215 M
\n(c) 0.25 x 2.303 M
\n(d) 0.05 M
\nAnswer:
\n(b) 0.2 15 M
\nSolution:
\n\"Samacheer<\/p>\n

Question 13.
\nFor a first order reaction, the rate constant is 6.909 min-1<\/sup>.The time taken for 75% conversion in minutes is …………
\n(a) \\((\\frac { 3 }{ 2 })\\) log 2
\n(b) \\((\\frac { 3 }{ 2 })\\) log 2
\n(c) \\((\\frac { 3 }{ 2 })\\) log \\((\\frac { 3 }{ 4 })\\)
\n(d) \\((\\frac { 2 }{ 3 })\\) log \\((\\frac { 4 }{ 3 })\\)
\nAnswer:
\n(b) \\((\\frac { 3 }{ 2 })\\) log 2
\nSolution:
\nk = \\((\\frac { 2.303 }{ t })\\) log \\(\\left( \\frac { \\left[ { A }_{ 0 } \\right] }{ \\left[ A \\right] } \\right)\\)
\n[A0<\/sub>] = 100
\n[A] = 25
\n[A0<\/sub>]= 100; [A]=25
\n6.909 = \\((\\frac { 2.303 }{ t })\\) log \\((\\frac { 100 }{ 25 })\\)
\nt = \\((\\frac { 2.303 }{ 6.909 })\\) log (4) \u21d2 t = \\((\\frac { 1 }{ 3 })\\) log 22<\/sup>
\nt = \\((\\frac { 2 }{ 3 })\\) log 2<\/p>\n

Question 14.
\nIn a first order reaction x \u2192 y; if k is the rate constant and the initial concentration of the reactant x is 0.1 M, then, the half life is ……..
\n(a) \\((\\frac { log2 }{ k })\\)
\n(b) \\((\\frac { 0.693 }{ (0.1)k })\\)
\n(c) \\((\\frac { In2 }{ k })\\)
\n(d) none of these
\nAnswer:
\n(c) \\((\\frac { In2 }{ k })\\)
\nSolution:
\nk = \\((\\frac { 1 }{ t })\\) In \\(\\left( \\frac { \\left[ { A }_{ 0 } \\right] }{ \\left[ A \\right] } \\right)\\)
\n[A0<\/sub>] = 0.1
\n[A] = 0.05
\nk = \\(\\left( \\frac { 1 }{ { t }_{ 1\/2 } } \\right)\\) In \\((\\frac { 0.1 }{ 0.05 })\\)
\nk = \\(\\left( \\frac { 1 }{ { t }_{ 1\/2 } } \\right)\\) In (2) \u21d2 t1\/2<\/sub> = \\((\\frac { In(2) }{ k })\\)<\/p>\n

Question 15.
\nPredict the rate law of the following reaction based on the data given below:
\n2A + B \u2192 C + 3D
\n\"Samacheer
\n(a) rate = k [A]2<\/sup> [B]
\n(b) rate = k [A][B]2<\/sup>
\n(c) rate = k [A][B]
\n(d) rate = k [A]1\/2<\/sup> [B]3\/2<\/sup>
\nAnswer:
\n(b) rate = k [A][B]2<\/sup>
\nSolution:
\nrate1<\/sub> = k [0.1]n<\/sup> [0.1]m<\/sup> ……………(1)
\nrate2<\/sub> = k [0.2]n<\/sup> [0.1]m<\/sup> …………(2)
\nDividing Eq.(2) by Eq.(1)
\n\"Samacheer
\n\\(\\frac { 2x }{ x }\\) = 2n<\/sup>
\n\u2234 n = 1
\nrate3<\/sub> = k [0.1]n<\/sup> [0.2]m<\/sup> …………..(3)
\nrate4<\/sub> = k [0.2]n<\/sup> [0.2]m<\/sup> …………..(4)
\nDividing Eq.(4) by Eq.(2)
\n\"Samacheer
\n\\(\\frac { 8 }{ 2 } \\) = 2m<\/sup>
\n\u2234m = 1
\n\u2234 rate = k [A]1<\/sup> [B]2<\/sup><\/p>\n

Question 16.
\nAssertion: rate of reaction doubles when the concentration of the reactant is doubles if it is a first order reaction.
\nReason: rate constant also doubles
\n(a) Both assertion and reason are true and reason is the correct explanation of assertion.
\n(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
\n(c) Assertion is true but reason is false.
\n(d) Both assertion and reason are false.
\nAnswer:
\n(c) Assertion is true but reason is false.
\nSolution:
\nFor a first reaction, when the concentration of reactant is doubled, then the rate of reaction also doubled. Rate constant is independent of concentration and is a constant at a constant temperature, i.e., it depends on the temperature and hence, it will not be doubled and when the concentration of the reactant is doubled.<\/p>\n

Question 17.
\nThe rate constant of a reaction is 5.8 x 102 s1<\/sup>. The order of the reaction is ………….
\n(a) First order
\n(b) zero order
\n(c) Second order
\n(a) Third order
\nAnswer:
\n(a) First order
\nSolution:
\nThe unit of rate constant is s-1<\/sup> and it indicates that the reaction is first order.<\/p>\n

Question 18.
\nFor the reaction N2<\/sub> O5(g)<\/sub> \u2192 2NO2(g)<\/sub> +\\(\\frac { 1 }{ 2 }\\) – O2(g)<\/sub> the value of rate of disappearance of N2<\/sub>O5<\/sub> is given as 6.5 x 10-2<\/sup> mol L-1<\/sup>s-1<\/sup> The rate of formation of NO2<\/sub> and O2<\/sub> is given respectively as …………….
\n(a) (3.25 x 10-2<\/sup> mol L-1<\/sup>s-1<\/sup>) and (1.3 x 10-2<\/sup> mol L-1<\/sup>s-1<\/sup>)
\n(b) (1.3 x 10-2<\/sup> mol L-1<\/sup>s-1<\/sup>) and (3.25 x 102<\/sup> mol L-1<\/sup>s-1<\/sup>)
\n(c) (1.3 x 10-1<\/sup> mol L-1<\/sup>s-1<\/sup>) and (3.25 x 10-2<\/sup> mol L-1<\/sup>s-1<\/sup>)
\n(d) None of these
\nAnswer:
\n(c) (1.3 x 10-1<\/sup> mol L-1<\/sup>s-1<\/sup>) and (3.25 x 10-2<\/sup> mol L-1<\/sup>s-1<\/sup>)
\n\"Samacheer<\/p>\n

Question 19.
\nDuring the decomposition of H2<\/sub>O2<\/sub> to give dioxygen, 48g O2<\/sub> is formed per minute at certain point of time. The rate of formation of water at this point is …………….
\n(a) 0.75 mol min-1<\/sup>
\n(b) 1.5 mol min-1<\/sup>
\n(c) 2.25 mol min-1<\/sup>
\n(d) 3.0 mol min-1<\/sup>
\nAnswer:
\n(d) 3.0 mol min-1<\/sup>
\nSolution:
\nH2<\/sub>O2<\/sub> \u2192 H2<\/sub>O + \\(\\frac { 1 }{ 2 }\\)O2<\/sub>
\nRate = \"Samacheer
\nNo. of moles of oxygen = \\((\\frac { 48 }{ 32 })\\) = 1.5 mol
\nRate of formation of oxygen = 2 x 1.5
\n= 3 mol min-1<\/sup><\/p>\n

Question 20.
\nIf the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is …………
\n(a) Zero
\n(b) one
\n(c) Fraction
\n(d) none
\nAnswer:
\n(a) Zero
\nSolution:
\nFor a first order reaction t1\/2<\/sup> is independent of initial concentration .i.e., n \\(\\neq\\) 1 for such cases
\n\"Samacheer<\/p>\n

Question 21.
\nIn a homogeneous reaction A ? B + C + D, the initial pressure was P0<\/sub> and after time t it was P. Expression for rate constant in terms of P0<\/sub>, P and t will be ……….
\n\"Samacheer
\nAnswer:
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 22.
\nIf 75% of a first order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in ………
\n(a) 20 minutes
\n(b) 30 minutes
\n(c) 35 minutes
\n(d) 75 minutes
\nAnswer:
\n(b) 30 minutes
\nSolution:
\n\"Samacheer<\/p>\n

Question 23.
\nThe half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
\n(a) \\(\\frac { 1 }{ 2 }\\) g
\n(b) \\(\\frac { 1 }{ 4 }\\) g
\n(c) \\(\\frac { 1 }{ 8 }\\) g
\n(d) \\(\\frac { 1 }{ 16 }\\) g
\nAnswer:
\n(d) \\(\\frac { 1 }{ 16 }\\) g
\nSolution:
\nin 140 days \u21d2 initial concentration reduced to \\(\\frac { 1 }{ 2 }\\) g
\nin 280 days \u21d2 initial concentration reduced to \\(\\frac { 1 }{ 4 }\\) g
\nin 420 days \u21d2 initial concentration reduced to \\(\\frac { 1 }{ 8 }\\) g
\nin 560 days \u21d2 initial concentration reduced to \\(\\frac { 1 }{ 8 }\\) g<\/p>\n

Question 24.
\nThe correct difference between first and second order reactions is that …………
\n(a) A first order reaction can be catalysed a second order reaction cannot be catalysed.
\n(b) The half life of a first order reaction does not depend on [A0<\/sub>] the half life of a second order reaction does depend on [A0<\/sub>].
\n(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
\n(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations,
\nAnswer:
\n(b) The half life of a first order reaction does not depend on [A0<\/sub>]; the half life of a second order reaction does depend on [A0<\/sub>].
\nSolution:
\nFor a first order reaction
\nt1\/2<\/sub> = \\(\\frac { 0.6932 }{ k }\\)
\nFor a second order reaction
\n\"Samacheer<\/p>\n

Question 25.
\nAfter 2 hours, a radioactive substance becomes \\((\\frac { 1 }{ 16 })\\)th<\/sup> of original amount. Then the half life (in mm) is ………………
\n(a) 60 minutes
\n(b) 120 minutes
\n(c) 30 minutes
\n(d) 15 minutes
\nAnswer:
\n(c) 30 minutes
\nSolution:
\n\"Samacheer<\/p>\n

II . Answer the following questions:<\/p>\n

Question 1.
\nDefine average rate and instantaneous rate.
\nAnswer:
\n1. Average rate:
\nThe average rate of a reaction is defined as the rate of change of concentration of a reactant (or of a product) over a specified measurable period of time.<\/p>\n

2. insantaneous rate:
\nInstantaneous rate of reaction gives the tendency of the reaction at a particular point of time during its course (or) The time derivative of the concentration of a reactant (or product) converted to a positive number is called the instantaneous rate of reaction.<\/p>\n

Question 2.
\nDefine rate law and rate constant.
\n1. Rate law:
\nThe expression in which reaction rate is given in terms of molar concentration of the reactants with each term raised to some power, which may or may not be same as the Stoichiometric coefficient of the reacting species in a balanced chemical equation.
\nx A + y B \u2192 products
\nRate = k [A]m<\/sup> [B]m<\/sup>
\nk = Rate constant<\/p>\n

2. Rate constant:
\nFor a reaction involving the reactants A and B, Reaction rate = k [A]m<\/sup> [B]m<\/sup> The constant k is called rate constant of the reaction. If [A] = 1 M and [B] = 1 M; Reaction rate = k Thus, the rate constant (k) of a reaction is equal to the rate of reaction when the concentration of each reactant is equal to 1 mol L-1<\/sup>. The change in the concentration of reactant or product per unit time under the condition of unit concentration of all the reactant.<\/p>\n

Question 3.
\nDerive integrated rate law for a zero order reaction A product. A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentrations is called as zero order reactions. Such reactions are rare. Let us consider the following hypothetical zero order reaction.
\nA \u2192 Product
\nThe rate law can be written
\nRate = k [A]\u00b0
\n(\u2234[A]\u00b0 = 1)
\n– d [A] k (l)
\n\\(\\frac { -d[A] }{ dt }\\) = k(1)
\n-d[A] = k dt
\nIntegrate the above equation between the limits of [A0<\/sub>] at zero time and [A] at some later time \u2018t\u2019,
\n\"Samacheer<\/p>\n

Question 4.
\nDefine half life of a reaction. Show that for a first order reaction half life is independent of Initial concentration.
\nAnswer:
\nHalf life of a reaction is defined as the time required for the reactant concentration to reach one half of its initial value. For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration. The rate constant for a first order reaction is given by,
\n\"Samacheer
\nQuestion 5.
\nWhat is an elementary reaction? Give the differences between order and molecularity of a reaction.
\nAnswer:
\nElementary reaction – Each and every single step in a reaction mechanism is called an elementary reaction. Differences between order and molecularity:
\nOrder of a reaction:<\/p>\n

    \n
  1. It is the sum of the powers of concentration terms involved in the experimentally determined rate law.<\/li>\n
  2. It can be zero (or) fractional (or) integer.<\/li>\n
  3. It is assigned for a overall reaction.<\/li>\n<\/ol>\n

    Molecularity of a reaction:<\/p>\n

      \n
    1. It is the total number of reactant species that are involved in an elementary step.<\/li>\n
    2. It is always a whole number, cannot be zero or a fractional number.<\/li>\n
    3. It is assigned for each elementary step of mechanism.<\/li>\n<\/ol>\n

      Question 6.
      \nExplain the rate determining step with an example.
      \nAnswer:
      \n1. Most of the chemical reactions occur by multistep reactions. In the sequence of steps it is found that one of the steps is considerably slower than the others. The overall rate of the reaction cannot be lower in value than the rate of the slowest step.<\/p>\n

      2. Thus in a multistep reaction the experimentally determined rate corresponds to the rate of the slowest step. The step which has the lowest rate value among the other steps of the reaction is called as the rate determining step (or) rate limiting step.<\/p>\n

      3. Consider the reaction,
      \n2A + B \u2192 C + D
      \ngoing by two steps as follows,
      \n\"Samacheer
      \nHere the overall rate of the reaction corresponds to the rate of the first step which is the slow step and thus the first step is called as the rate determining step of the reaction. In the above equation, the rate of the reaction depends upon the rate constant k ( only. The rate of second step dosn’t contribute experimentally determined overall rate of the reaction.
      \nFor example,
      \nNO2(g)<\/sub> + CO2(g)<\/sub> \u2192 NO(g)<\/sub> + CO2(g)<\/sub>
      \nWhich occurs in two elementary steps:<\/p>\n

        \n
      • NO2<\/sub> + NO2<\/sub> \u2192 NO + NO3<\/sub> (Slow)<\/li>\n
      • NO3<\/sub>+ CO \u2192 NO2<\/sub> + CO2<\/sub> (fast)<\/li>\n<\/ul>\n

        Because the first step is the lowest step, the overall reaction cannot proceed any faster than the rate of the first elementary step. The first elementary step in this example is therefore the rate determining step.<\/p>\n

        The rate equation for this reaction is equal to the rate is constant of step-1 multiplied by the reactants of that first step. If the rate constant of step-1 is denoted as k1<\/sub> then the rate of the first step in the reaction (and the total reaction) will be,
        \nRate = k, [NO2<\/sub>] [NO2<\/sub>]
        \n= k1<\/sub> [NO2<\/sub>]2<\/sub><\/p>\n

        Question 7.
        \nDescribe the graphical representation of first order reaction.
        \nAnswer:
        \nRate constant for first order reaction is,
        \nkt = ln\\(\\left( \\frac { \\left[ { A }_{ 0 } \\right] }{ \\left[ A \\right] } \\right)\\)
        \nkt = In [A0<\/sub>] – In [A]
        \nIn[A] = In [A0<\/sub>] – kty = c + mx
        \nIf we follow the reaction by measuring the concentration of the reactants at regular time interval ‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope. From this, the rate constant is calculated.
        \n\"Samacheer<\/p>\n

        Question 8.
        \nWrite the rate law for the following reactions.<\/p>\n

          \n
        1. A reaction that is 3\/2 order in x and zero order in y.<\/li>\n
        2. A reaction that is second order in NO and first order in Br2<\/sub>.<\/li>\n<\/ol>\n

          Answer:
          \n1. \\(\\frac { 3 }{ 2 }\\) x + y (excess) \u2192 products
          \n– \\(\\frac { 3 }{ 2 }\\) \\(\\frac { d[x] }{ dt }\\) = k [x]3\/2<\/sup><\/p>\n

          2. 2NO + Br2<\/sub> \u2192 products
          \n– \\(\\frac { 1 }{ 2 }\\) \\(\\frac { d[NO] }{ dt }\\) = k [NO]2<\/sup> [Br2<\/sub>]<\/p>\n

          Question 9.
          \nExplain the effect of catalyst on reaction rate with an example.
          \nAnswer:<\/p>\n

            \n
          1. Significant changes in the reaction can be brought out by the addition of a substance called catalyst.<\/li>\n
          2. A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change.<\/li>\n
          3. They may participate in the reaction, but again regenerated and the end of the reaction.<\/li>\n
          4. In the presence of a catalyst, the energy of activation is lowered and hence, greater number of molecules can cross the energy barrier and change over to products, thereby increasing the rate of the reaction.<\/li>\n
          5. For example, decomposition of potassium chlorate is enhanced by addition of MnO2<\/sub>.<\/li>\n<\/ol>\n

            2KClO3<\/sub> \\(\\frac{\\mathrm{MnO}_{4}}{\\Delta} 2 \\mathrm{KCl}\\) + 3O2<\/sub> (MnO2<\/sub> – Catalyst)
            \n\"Samacheer<\/p>\n

            Question 10.
            \nThe rate law for a reaction of A, B and C has been found to be rate = k[A]2<\/sup> [B][L]3\/2<\/sup>. How would the rate of reaction change when<\/p>\n

              \n
            1. Concentration of [L] is quadrupled<\/li>\n
            2. Concentration of both [A] and [B] are doubled<\/li>\n
            3. Concentration of [A] is halved<\/li>\n
            4. Concentration of [A] is reduced to(1\/3) and concentration of [L] is quadrupled.<\/li>\n<\/ol>\n

              Solution:
              \nRate = k [A]2<\/sup> [B] [L]3\/2<\/sup> ………….(1)
              \n1. when [L] = [4L]
              \nRate = k [A]2<\/sup> [B] [4L]3\/2<\/sup>
              \nRate = 8 (k[A]2<\/sup> [B] [L]3\/2<\/sup>) …………………..(2)
              \nComparing (1) and (3) rate is increased by 8 times.<\/p>\n

              2. when [A] = [2A] and [B] = [2B]
              \nRate = k[2A]2<\/sup> [2B ] [L]3\/2<\/sup>
              \nRate = 8 (k[A]2<\/sup> [B] [L]3\/2<\/sup> …………….(3)
              \nComparing (1) and (3); rate is increased by 8 times.<\/p>\n

              3. when [A] = \\([\\frac { A }{ 2 }]\\)
              \nRate = k \\([\\frac { A }{ 2 }]\\)2<\/sup> [L]\\(\\frac { 3 }{ 2 }\\)
              \nRate = \\(\\frac { 1 }{ 4 }\\) (k[A]2<\/sup> [B] [L]3\/2<\/sup>) ……………..(4)
              \nComparing (1) and ( 4); rate is reduced to \\(\\frac { 1 }{ 4 }\\) times.<\/p>\n

              4. when [A] = \\([\\frac { A }{ 3 }]\\) and [L] = [4L]
              \nRate k\\(\\frac { A }{ 3 }\\)2<\/sup> [B] [4L]3\/2<\/sup>
              \nRate = \\([\\frac { 8 }{ 9 }]\\) (k[A]2<\/sup> [B] [L]3\/2<\/sup>) ……………….(5)
              \nComparing (1) and (5); rate is reduced to 8\/9 times.<\/p>\n

              Question 11.
              \nThe rate of formation of a dimer in a second order reaction is 7.5 x 10-3<\/sup> mol L-1<\/sup>s-1<\/sup> at 0.05 mol L-1<\/sup> monomer concentration. Calculate the rate constant.
              \nSolution:
              \nLet us consider the dimensation of a monomer M
              \n2M \u2192 (M)2<\/sub>
              \nRate = k [M]n<\/sup>
              \nGiven that n =2 and [M] = 0.05 mol L-1<\/sup>
              \nRate = 7.5 x 10-3<\/sup> mol L-1<\/sup>s-1<\/sup>
              \nRate 7.5 x 103<\/sup> mol L-1<\/sup> s-1<\/sup>
              \nk = \\(\\frac { Rate }{ { \\left[ M \\right] }^{ n } }\\)
              \nk= =\\(\\frac { 7.5\\times { 10 }^{ -3 } }{ { \\left( 0.05 \\right) }^{ 2 } }\\) = 3 mol-1<\/sup> Ls-1<\/sup><\/p>\n

              Question 12.
              \nFor a reaction x +y + z \u2192 products, the rate law is given by rate = k [x]3\/2<\/sup> [y]1\/2<\/sup> what is the overall order of the reaction and what is the order of the reaction with respect to z.
              \nSolution:
              \nRate = k [x]3\/2<\/sup> [y]1\/2<\/sup>
              \noverall order = \\(\\left( \\frac { 3 }{ 2 } +\\frac { 1 }{ 2 } \\right)\\) = 2
              \ni.e., second order reaction.
              \nSince the rate expression does not contain the concentration of Z , the reaction is zero order with respect to Z.<\/p>\n

              Question 13.
              \nExplain briefly the collision theory of bimolecular reactions.
              \nAnswer:
              \nCollision theory is based on the kinetic theory of gases. According to this theory, chemical reactions occur as a result of collisions between the reacting molecules. Let us understand this theory by considering the following reaction.
              \nA2(g)<\/sub> + B2(g)<\/sub> \u2192 2AB(g)<\/sub><\/p>\n

              If we consider that, the reaction between A2<\/sub> and B2<\/sub> molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second. Rate ix Number of molecules colliding per litre per second (or) Rate \u221d Collision rate. The number of collisions is directly proportional to the concentration of both A2<\/sub> and B2<\/sub>.
              \nCollision rate \u221d [A2<\/sub>] [B2<\/sub>] …………………(1)
              \nCollision rate = Z [A2<\/sub>] [B2<\/sub>] ……………………..(2)
              \nWhere, Z is a constant.<\/p>\n

              The collision rate \u00a1n gases can be calculated from kinetic theory of gases. For a gas at room temperature (298K) and 1 atm pressure, each molecule undergoes approximately 109<\/sup> collisions per second, i.e., I collision in 109<\/sup> second. Thus, if every collision resulted in reaction, the reaction would be complete in 109<\/sup> second.<\/p>\n

              In actual practice this does not happen. It implies that all collisions are not effective to lead to the reaction. In order to react, the colliding molecules must possess a minimum energy called activation energy. The molecules that collide with less energy than activation energy will remain intact and no reaction occurs.<\/p>\n

              Fraction of effective collisions (f) is given by the following expression, \\({ e }^{ \\frac { { -E }_{ a } }{ RT } }\\)
              \nFraction of collisions is further reduced due to orientation factor i.e., even if the reactant collide with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state. The fraction of effective collisions (f) having proper orientation is given by the steric factor P.<\/p>\n

              Rate = P x f x collision rate
              \nRate= P x \\({ e }^{ \\frac { { -E }_{ a } }{ RT } }\\) x Z [A2<\/sub>] [B2<\/sub>] ……..(1)
              \nAs per the rate law, Rate = k [A2<\/sub>] [B2<\/sub>] ………………….(2)
              \nWhere k is the rate constant
              \nOn comparing equation (1) and (2), the rate constant k is,
              \nk = p Z \\({ e }^{ \\frac { { -E }_{ a } }{ RT } }\\)<\/p>\n

              Question 14.
              \nWrite Arrhenius equation and explains the terms involved.
              \nAnswer:
              \nArrhenius equation:
              \nk = A\\({ e }^{ \\frac { { -E }_{ a } }{ RT } }\\)
              \nA = Arrhenius factor (frequency factor)
              \nR = Gas constant
              \nk = Rate constant
              \nEa<\/sub> = Activation energy
              \nT = Absolute temperature (in K)<\/p>\n

              Question.15.
              \nThe decomposition of Cl2<\/sub>O7<\/sub> at 500K in the gas phase to Cl2<\/sub> and O2<\/sub> is a first order reaction. After 1 minute at 500K, the pressure of Cl2<\/sub>O7<\/sub> falls from 0.08 to 0.04 atm. Calculate the rate constant in s-1<\/sup>.
              \nAnswer:
              \nSolution:
              \nk = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac{\\left[\\mathrm{A}_{0}\\right]}{[\\mathrm{A}]}\\)
              \nk = \\(\\frac { 2.303 }{ 1 min }\\) log \\(\\frac { [0.08] }{ [0.04] }\\)
              \nk = 2.303 log 2
              \nk = 2.303 x 0.3010
              \nk = 0.693 2 min-1<\/sup>
              \nk = \\((\\frac { 0.6932 }{ 60 })\\) s-1
              \nk = 1.153 x 10-2<\/sup> s-1<\/sup><\/p>\n

              Question 16.
              \nGive the examples for a zero order reaction.
              \nAnswer:
              \nExamples for a zero order reaction:
              \n1. Photochemical reaction between H2<\/sub> and Cl
              \nH2(g)<\/sub> + Cl2(g) <\/sub>\\(\\underrightarrow { h\\nu }\\) 2HCI(g)<\/sub><\/p>\n

              2. Decomposition of N2<\/sub>O on hot platinum surface
              \nN2<\/sub> O(g)<\/sub> \\(\\rightleftharpoons\\) N2(g)<\/sub> + \\(\\frac { 1 }{ 2 }\\) O2(g)<\/sub><\/p>\n

              3. lodination of acetone in acid medium is zero order with respect to iodine.
              \nCH3<\/sub>COCH3<\/sub> + I2<\/sub> \\(\\underrightarrow { { H }^{ + } } \\) ICH2<\/sub>COCH3<\/sub> + HI
              \nRate k [CH3<\/sub>COCI3<\/sub>] [H+<\/sup>]<\/p>\n

              Question 17.
              \nExplain pseudo first order reaction with an example.
              \nAnswer:
              \nA second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,
              \nCH3<\/sub>COOCH3(aq)<\/sub> +H2<\/sub> O(1)<\/sub> \\(\\underrightarrow { { H }^{ + } } \\) CH3<\/sub>COOH(aq) <\/sub>+ CH3<\/sub>OH(aq)<\/sub>
              \nRate = k [CH3<\/sub>COOCH3<\/sub>] [H2<\/sub>O]<\/p>\n

              If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e., concentration of water remains almost a constant. Now we can define k [H2<\/sub>O] = k
              \n\u2234 The above rate equation becomes
              \nRate k [CHCOOCH] Thus it follows first order kinetics.<\/p>\n

              Question 18.
              \nIdentify the order for the following reactions<\/p>\n

                \n
              1. Rusting of Iron<\/li>\n
              2. Radioactive disintegration of 92<\/sub>U23<\/sup><\/li>\n
              3. 2 A+ B \u2192 products; rate = k [A]1\/2<\/sup> [B]2<\/sup><\/li>\n<\/ol>\n

                Answer:
                \n1.\"Samacheer
                \nTheoritically order value may be more than one but practically one.<\/p>\n

                2. All radioactive disintegrations are first order reactions<\/p>\n

                3. 2A + 3B \u2192 products:
                \nrate = k[A]1\/2<\/sup> [B]2<\/sup>
                \nOrder = \\(\\frac { 1 }{ 2 }\\) + 2 = \\(\\frac { 5 }{ 2 }\\) = 2.5<\/p>\n

                Question 19.
                \nA gas phase reaction has energy of activation 200 kJ mol-1<\/sup>. If the frequency factor of the reaction is 1.6 x 1013<\/sup> s-1<\/sup>. Calculate the rate constant at 600 K. (e-40.09<\/sup> = 3.8 x I0-18<\/sup> )
                \nSolution:
                \n\"Samacheer<\/p>\n

                Question 20.
                \nFor the reaction 2x +y \u2192 L find the rate law from the following data.
                \n\"Samacheer
                \nAnswer:
                \nRate = k [x]n<\/sup> [y]m<\/sup>
                \n0.15 = k [0.2]n<\/sup> [0.02]m<\/sup> ……………..(1)
                \n0.30 = k [0.4]n<\/sup> [0.02]m<\/sup> ……………… (2)
                \n1.20 = k [0.4]n<\/sup> [0.08]m<\/sup> ……………… (3)
                \nDividing Eq(3) by Eq (2) we get
                \n\"Samacheer
                \n\"Samacheer<\/p>\n

                Question 21.
                \nHow do concentrations of the reactant influence the rate of reaction?
                \nAnswer:
                \nThe rate of a reaction increases with the increase in the concentration of the reactants. The effect of concentration is explained on the basis of collision theory of reaction rates.<\/p>\n

                According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.<\/p>\n

                Question 22.
                \nHow do nature of the reactant influence rate of reaction?
                \nAnswer:
                \nNature and state of the reactant:
                \nWe know that a chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product.<\/p>\n

                The net energy involved in this process is dependent on the nature of the reactant and hence the rates arc different for different reactants. Let us compare the following two reactions that we carried out in volumetric analysis.<\/p>\n

                  \n
                1. Redox reaction between ferrous ammonium sulphate (FAS) and KMnO4<\/sub><\/li>\n
                2. Redox reaction between oxalic acid and KMnO4<\/sub><\/li>\n<\/ol>\n

                  The oxidation of oxalate ion by KMnO4<\/sub> is relatively slow compared to the reaction between KMnO4<\/sub> and Fe . In fact heating is required for the reaction between KMnO4<\/sub> and Oxalate ion and is carried out at around 60\u00b0C. The physical state of the reactant also plays an important role to influence the rate of reactions. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants.
                  \n\"Samacheer
                  \nFor example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine. Let us consider another example that we carried out in inorganic qualitative analysis of lead salts.<\/p>\n

                  If we mix the aqueous solution of colorless potassium iodide with the colorless solution of lead nitrate, precipitation of yellow lead iodide take place instantancously, whereas if we mix the solid lead nitrate with solid potassium iodide, yellow coloration will appear slowly.<\/p>\n

                  Question 23.
                  \nThe rate constant for a first order reaction is 1.54 x 10 s-1<\/sup>. Calculate its half life time.
                  \nAnswer:
                  \nWe know that, t, 0.693 k
                  \nt1\/2<\/sub> = 0.693\/1.54 x 1o-3<\/sup> = 450 s<\/p>\n

                  Question 24.
                  \nThe half life of the homogeneous gaseous reaction SO2<\/sub>CI2<\/sub> \u2192 SO2<\/sub> + Cl2<\/sub> which obeys first order kinetics Is 8.0 minutes. How long will it take for the concentration of SO2<\/sub>Cl2<\/sub> to be reduced to 1% of the initial value?
                  \nAnswer:
                  \nWe know that, k = 0.693\/ t1\/2<\/sub>
                  \nk = 0.693\/8.0 minutes = 0.087 minutes -1<\/sup>
                  \nFor a first order reaction,
                  \nk = \\(\\frac { 2.303 }{ k }\\) log \\(\\left( \\frac { \\left[ { A }_{ 0 } \\right] }{ \\left[ A \\right] } \\right)\\)
                  \nt = \\(\\frac { 2.303 }{ 0.087{ min }^{ -1 } }\\) log\\(\\frac { 100 }{ 1 }\\)
                  \nt = 52.93 mm<\/p>\n

                  Question 25.
                  \nThe time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?
                  \nAnswer:
                  \n1. Order of a reaction = 1
                  \nt1\/2<\/sub> = 60
                  \nseconds, k = ?
                  \nk = \\(\\frac { 2.303 }{ 60 }\\)
                  \nWe know that, k = \\(\\frac { 2.303 }{ { t }_{ 1\/2 } }\\)
                  \nk = \\(\\frac { 2.303 }{ 60 }\\) = 0.01155 s-1<\/sup><\/p>\n

                  2. [A0<\/sub>] = 100%
                  \nt = 180 s
                  \nk = 0.01155 seconds-1<\/sup>
                  \n[A] = ?
                  \nFor the first order reaction k = \\(\\frac { 2.303 }{ 60 }\\) log \\(\\left( \\frac { \\left[ { A }_{ 0 } \\right] }{ \\left[ A \\right] } \\right)\\)
                  \n0.9207 = log 100 – log [A]
                  \nlog [A] = log 100 – 0.9207
                  \nlog [A] = 2 – 0.9207
                  \nlog[A] = 1.0973
                  \n[A] = antilog of (1.0973)
                  \n[A] = l2.5%<\/p>\n

                  Question 26.
                  \nA zero order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
                  \nAnswer:
                  \n1. A = 100%, x = 20%, Therefore, a – x =100 – 20 = 80
                  \nFor the zero order reaction k= \\((\\frac { x }{ t })\\) \u21d2
                  \nk = \\((\\frac { 20 }{ 20 })\\) = 1
                  \nRate constant for a reaction = 1<\/p>\n

                  2. To calculate the time for 80% of completion
                  \nk = 1, a = l00, x = 80%, t = ?
                  \nTherefore, t = \\((\\frac { x }{ k })\\) = \\((\\frac { 80 }{ 1 })\\) = 80 min<\/p>\n

                  Question 27.
                  \nThe activation energy of a reaction is 225 k cal mol-1<\/sup> and the value of rate constant at 40\u00b0C is 1.8 x 10-5<\/sup> s-1<\/sup>. Calculate the frequency factor, A. Here, we arc given that
                  \nAnswer:
                  \nEa<\/sub> = 22.5 kcal mol-1<\/sup> = 22500 cal mol-1<\/sup>
                  \nT = 40\u00b0C = 40 + 273 = 313 K
                  \nk = 1.8 x 10-5<\/sup> sec-1<\/sup>
                  \nSubstituting the values in the equation
                  \nlog A = log k + \\(\\left( \\frac { { E }_{ a } }{ 2.303RT } \\right)\\)
                  \nlog A = log (l .8 x 10-5<\/sup>) + \\(\\left( \\frac { 22500 }{ 2.303\\times 1.987\\times 313 } \\right)\\)
                  \nlog A = log (l.8) – 5 + (15.7089)
                  \nlog A = (10.9642)
                  \nA = antilog ( 10.9642)
                  \nA = 9.208 x 1010<\/sup> collisions s-1<\/sup><\/p>\n

                  Question 28.
                  \nBenzene diazonium chloride in aqueous solution decomposes according to the equation
                  \nC6<\/sub>H5<\/sub>N2<\/sub>CI C6<\/sub>H5<\/sub>CI + N2<\/sub>. Starting with an initial concentration of 10 g L-1<\/sup> volume of N2<\/sub>. gas obtained at 50\u00b0C at different intervals of time was found to be as under:
                  \n\"Samacheer
                  \nShow that the above reaction follows the first order kinetics. What is the value of the rate constant ?
                  \nSolution:
                  \nFor a first order reaction
                  \nk = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac { a }{ (a-x) }\\)
                  \nk = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac { { V }_{ \\infty } }{ { V }_{ \\infty }-{ V }_{ t } }\\)
                  \nIn this case, V\u221e<\/sub> = 58.3 ml
                  \nThe value of k at different time can be calculated as follows:
                  \n\"Samacheer
                  \nSince the value of k comes out to be nearly constant, the given reaction is of the first order. The mean value of k = 0.0674 min-1<\/sup><\/p>\n

                  Question 29.
                  \nFrom the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order:
                  \n\"Samacheer
                  \nWhere t is the volume of standard KMnO4<\/sub> solution required for titrating the same volume of the reaction mixture.
                  \nSolution:
                  \nVolume of KMnO4<\/sub> solution used Amount of H2<\/sub>O2<\/sub> present. Hence if the given reaction is of the first order, it must obey the equation
                  \nk = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac { a }{ (a-x) }\\)
                  \nk = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac { { V }_{ 0 } }{ { V }_{ t } }\\)
                  \nIn this case,V0<\/sub> = 46.1 ml
                  \nThe value of k at each instant can be calculated as follows:
                  \n\"Samacheer
                  \nThus, the value of k comes out to be nearly constant. Hence it is a reaction of the first order.<\/p>\n

                  Question 30.
                  \nA first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. in what time will the reaction be 80% complete?
                  \nAnswer:
                  \n1. For the first order reaction k = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac { a }{ (a-x) }\\)
                  \nAssume, a = 100 %, x = 40%, t = 50 minutes
                  \nTherefore, a – x = 100 – 40 = 60
                  \nk = (2.303\/50) log (100\/60)
                  \nk = 0.010216 min-1<\/sup>
                  \nHence the value of the rate constant is 0.010216 min-1<\/sup><\/p>\n

                  2. t = ?, when x = 8O%
                  \nTherefore, a – x = 100 – 80 = 20
                  \nFrom above, k = 0.0102 16 min-1<\/sup>
                  \nt = (2.303 \/ 0.010216) log (100 \/ 20)
                  \nt = 157.58 min
                  \nThe time at which the reaction will be 80% complete is 157.58 min.<\/p>\n

                  Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Evaluate yourself<\/strong><\/p>\n

                  Question 1.
                  \nWrite the rate expression for the following reactions, assuming them as elementary reactions.<\/p>\n

                    \n
                  1. 3A + 5B2<\/sub> \u2192 4CD<\/li>\n
                  2. X2<\/sub> + Y2<\/sub> \u2192 2XY<\/li>\n<\/ol>\n

                    Answer:
                    \n1. 3A + 5B2<\/sub> \u2192 4CD
                    \nRate = – \\(\\frac { 1 }{ 3 }\\) \\(\\frac { \\triangle [A] }{ dt }\\)
                    \n= – \\(\\frac { 1 }{ 5 }\\) \\(\\frac { \\triangle [{ B }_{ 2 }] }{ dt }\\)
                    \n= + \\(\\frac { 1 }{ 4 }\\) \\(\\frac { \\triangle [CD] }{ dt }\\)<\/p>\n

                    2. X2<\/sub> + Y2<\/sub> \u2192 2XY
                    \nRate = – \\(\\frac { \\triangle [{ X }_{ 2 }] }{ dt }\\)
                    \n= + \\(\\frac { 1 }{ 2 }\\) \\( [latex]\\frac { \\triangle [{ XY }_{ 2 }] }{ dt }\\)<\/p>\n

                    Question 2.
                    \nConsider the decomposition of N2<\/sub>O5(g)<\/sub> to form NO2(g)<\/sub> and O2(g)<\/sub>. At a particular instant N2<\/sub>O5<\/sub> disappears at a rate of 2.5 x 10-2<\/sup> mol dm-3<\/sup> s-1<\/sup>. At what rates are NO2<\/sub> and O2<\/sub> formed? What is the rate of the reaction?
                    \nSolution:
                    \n2N2<\/sub>O5(g)<\/sub> \u2192 4NO2(g)<\/sub> + O2(g)<\/sub>
                    \nfrom the stoichiometry of the reaction.
                    \n– \\(\\frac { 1 }{ 2 }\\) \\(\\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\\)
                    \n= \\(\\frac { 1 }{ 4 }\\) \\(\\frac { d[{ N }{ O }_{ 2 }] }{ dt }\\)
                    \n= –\\(\\frac { d[{ N }{ O }_{ 2 }] }{ dt }\\)
                    \n= 2 –\\(\\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\\)
                    \nRate of disappearance of N2<\/sub>O5<\/sub> is 2.5 x 10-2<\/sup> mol dm-3<\/sup> s-1<\/sup>
                    \n\u2234 The rate of formation of NO2 at this temperature is 2 x 2.5 x 10-2<\/sup> = 5 x 10-2<\/sup> mol dm-3<\/sup> s-1<\/sup>.
                    \n– \\(\\frac { 1 }{ 2 }\\) \\(\\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\\)
                    \n= – \\(\\frac { d[{ O }_{ 2 }] }{ dt }\\)
                    \n\u2234 \\(\\frac { d[{ O }_{ 2 }] }{ dt }\\) = \\(\\frac { 1 }{ 2 }\\) x 2.5 x 10-2<\/sup> mol dm-3<\/sup> s-1<\/sup>
                    \n= 1.25 x 10-2<\/sup> mol dm-3<\/sup> s-1<\/sup><\/p>\n

                    Question 3.
                    \nFor a reaction, X + Y \u2192 Product quadrupling [x], increases the rate by a factor of 8. Quailrupling both [x] and [y] increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall order of the reaction?
                    \nSolution:
                    \n\"Samacheer
                    \nz = k[x]m<\/sup> [y]n<\/sup> …………(1)
                    \n8z = k[x]m<\/sup> [y]n<\/sup> …………(2)
                    \n16z = k[x]m<\/sup> [y]n<\/sup> ……………(3)
                    \nDividing Eq (2) by Eq (1) we get,
                    \n\"Samacheer
                    \n8 = 4m<\/sup> \u21d223<\/sup> = (22<\/sup>)m \u21d2 2 = 22m<\/sup>
                    \n2m = 3
                    \nm = 3\/2
                    \n1.5 order with respect to x.
                    \nDividing Eq (3) by Eq (1) we get,
                    \n\"Samacheer
                    \n16 = 4m<\/sup>. 4n<\/sup>
                    \n16 = 42<\/sup>. 4n<\/sup>
                    \n\\(\\frac { 16 }{ 16 }\\) = 4n<\/sup>
                    \n1 = 4n
                    \n\u2234 n = 0 [Zero order with respect to y]
                    \nOverall order of the reaction.
                    \nk [x]m<\/sup> [y]n<\/sup>
                    \nk [x]1.5<\/sup> [y]0<\/sup>
                    \nOrder (1.5+0) = 1.5<\/p>\n

                    Question 4.
                    \nFind the individual and overall order of the following reaction using the given data.
                    \n2NO(g) + Cl2<\/sub> (g) \u2192 2NOCI(g)
                    \n\"Samacheer
                    \nSolution:
                    \nRate = k [NO]m<\/sup> [CI2<\/sub>]
                    \nFor experiment 1, the rate law is,
                    \nRate1<\/sub> = k [NO]m<\/sup> [CI2<\/sub>]n<\/sup>
                    \n7.8 x 10-5<\/sup> k[0.1]m<\/sup> [0.1]n<\/sup> ………………(1)<\/p>\n

                    For experiment 2, the rate law is.
                    \nRate2<\/sub> = k [NO]m<\/sup> [CI2<\/sub>]n<\/sup>
                    \n3.12 x 10-4<\/sup> = k[O.2]m<\/sup> [0.1]n<\/sup> ……………….(2)<\/p>\n

                    For experiment 3, the rate law is,
                    \nRate3<\/sub> = k [NO]m<\/sup> [CI2<\/sub>]n<\/sup>
                    \n9.36 x 10-4<\/sup> = k [O.2]m<\/sup> [0.3]m<\/sup> ……………(3)
                    \nDividing Eq (2) by Eq (l) we get,
                    \n\"Samacheer
                    \n4 = \\(\\frac { 0.2 }{ 0.1 }\\)m<\/sup>
                    \n\u21d2 22<\/sup> = 2m<\/sup>
                    \n\u2234m = 2
                    \nTherefore the reaction is secondary order with respect to NO.
                    \nDividing Eq (3) by Eq (2) we get,
                    \n\"Samacheer
                    \nTherefore the reaction is first order with respect to Cl2<\/sub>
                    \nThe rate law is, Rate = k [NO]2<\/sub> [Cl2<\/sub>]1<\/sup>
                    \nThe overall order of the reaction (2 +1) = 3.<\/p>\n

                    Question 5.
                    \nIn a first order reaction A \u2192 products, 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction?
                    \nSolution:
                    \nk = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac{\\left[\\mathrm{A}_{0}\\right]}{[\\mathrm{A}]}\\)
                    \nk = \\(\\frac { 2.303 }{ 40min }\\) log \\(\\frac { 100 }{ (100-60) }\\)
                    \nk = 0.0575 (0.3979) \u21d2 k = 0.02287 min-1<\/sup>
                    \nt1\/2<\/sub> = \\(\\frac { 0.6932 }{ k }\\) log \\(\\frac { 0.6932 }{ 0.02287 }\\)
                    \nt1\/2<\/sub> = 30.31 min.<\/p>\n

                    Question 6.
                    \nThe rate constant for a first order reaction is 2.3 x 10-4<\/sup> s-1<\/sup>. If the initial concentration of the reactant is 0.01 M. what concentration will remain after 1 hour?
                    \nSolution:
                    \nRate constant of a first order reaction k = 2.3 x 10-4<\/sup> s-1<\/sup>
                    \nInitial concentration of the reactant [A0<\/sub>] = 0.01 M
                    \nInitial concentration ot the reactant [A0<\/sub>] = 0.01 M
                    \nConcentration will remain after 1 hour [A] =7
                    \nk = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac{\\left[\\mathrm{A}_{0}\\right]}{[\\mathrm{A}]}\\)
                    \n2.3 x 10-4<\/sup> = \\(\\frac { 2.303 }{ 1 hour }\\) log \\(\\frac { [0.01] }{ [A] }\\)
                    \n\\(\\frac { 2.3\\times { 10 }^{ -4 }\\times 1 }{ 2.303 }\\) = log [0.01] – log[A]
                    \n9.986 x 10-5<\/sup> = – 2 – log [A]
                    \n11.986 x 10-5<\/sup> = – log [A]
                    \n[A] = Antilog (-11.986 x 10-5<\/sup>)
                    \n[A] = 0.997 M<\/p>\n

                    Question 7.
                    \nHydrolysis of an ester in an aqueous solution was studied by titrating the liberated carboxylic acid against sodium hydroxide solution. The concentrations of the ester at different time intervals are given below.
                    \n\"Samacheer
                    \nShow that, the reaction follows first order kinetics.
                    \nSolution:
                    \nThe value of k at different time can be calculated as follows:
                    \n\"Samacheer
                    \nThis value shows that reaction follows first order kinetics.<\/p>\n

                    Question 8.
                    \nFor a first order reaction the rate constant at 500K is 8 x 10-4<\/sup> s-1<\/sup>. Calculate the frequency factor, if the energy of activation for the reaction is 190 kJ mol-1<\/sup>.
                    \nk = 8 x 10-4<\/sup>s
                    \nT = 500K
                    \nEa<\/sub> = 190 kJ mol-1<\/sup> A = ?
                    \nAccording to Arrhenius equation,
                    \n\"Samacheer<\/p>\n

                    Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Text book Example problems<\/strong><\/p>\n

                    Question 1.
                    \nThe oxidation of nitric oxide (NO)
                    \n2NO(g)<\/sub> + O2(g)<\/sub> \u2192 NO2(g)<\/sub>
                    \nSeries of experiments are conducted by keeping the concentration of one of the reactants constant and the changing the concentration of the others.
                    \n\"Samacheer
                    \nFind out the individual overall order of the reaction.
                    \nSolution:
                    \nRate = k [NO]m<\/sup> [O2<\/sub>]n<\/sup>
                    \nFor experiment 1, the rate law is
                    \nRate2<\/sub> = k [NO]m<\/sup> [O2<\/sub>]n<\/sup>
                    \n19.26 x 10-2<\/sup> = k[1.3]m<\/sup> [1.1]n<\/sup> ………………(1)
                    \nSimilarly for experiment 2
                    \nRate2<\/sub> = k [No]m<\/sup> [O2<\/sub>]n<\/sup>
                    \n38.40 x 10-2<\/sup> = k [1.3]m<\/sup> [2.2]n<\/sup> …………………..(2)
                    \nFor experiment 3
                    \nRate3<\/sub> = k [NO]m<\/sup> [O2<\/sub>]n<\/sup>
                    \n76.8 x 10-2<\/sup> = k [2.6]m<\/sup> [1.1]n<\/sup> ……….(3)
                    \nDividing Eq (2) by Eq (1)
                    \n\\(\\frac { 38.40\\times { 10 }^{ -2 } }{ 19.26\\times { 10 }^{ -2 } }\\) = \\(\\frac { k{ \\left[ 1.3 \\right] }^{ m }{ \\left[ 2.2 \\right] }^{ n } }{ k{ \\left[ 1.3 \\right] }^{ m }{ \\left[ 1.1 \\right] }^{ n } }\\)
                    \n2 = \\({ \\left( \\frac { 2.2 }{ 1.1 } \\right) }^{ n }\\)
                    \n2 = 2n<\/sup>
                    \ni.e., n= 1
                    \nTherefore the reaction is first order with respect to O2<\/sub>
                    \nDividing Eq (3) byEq (1)
                    \n\\(\\frac { 76.8\\times { 10 }^{ -2 } }{ 19.26\\times { 10 }^{ -2 } }\\) = \\(\\frac { k{ \\left[ 2.6 \\right] }^{ m }{ \\left[ 1.1 \\right] }^{ n } }{ k{ \\left[ 1.3 \\right] }^{ m }{ \\left[ 1.1 \\right] }^{ n } }\\)
                    \n4 = \\({ \\left( \\frac { 2.6 }{ 1.3 } \\right) }^{ m }\\)
                    \n4 = 2m<\/sup>
                    \ni.e., m = 2
                    \nTherefore the reaction is second order with respect to NO
                    \nThe rate law is Rate1<\/sub> = k [NO]2<\/sup> [O2<\/sub>]1<\/sup>
                    \nThe overall order of the reaction = (2 + 1) = 3<\/p>\n

                    Question 2.
                    \nConsider the oxidation of nitric oxide to form NO2<\/sub>
                    \n2NO(g)<\/sub> + O2(g)<\/sub> \u2192 2NO2(g)<\/sub><\/p>\n

                      \n
                    • Express the rate of the reaction in terms of changes in the concentration of NO, O2<\/sub> and NO2<\/sub>.<\/li>\n
                    • At a particular instant, when [O2<\/sub>] is decreasing at 0.2 mol L-1<\/sup> s-1<\/sup> at what rate is
                      \n[NO2<\/sub>] increasing at that instant?<\/li>\n<\/ul>\n

                      Solution:
                      \n\"Samacheer<\/p>\n

                      Question 3.
                      \nWhat is the order with respect to each of the reactant and overall order of the following reactions?
                      \n1.\u00a0 5Br–<\/sup>(aq)<\/sub> + BrO3<\/sub>–<\/sup> (aq)<\/sub>+\u00a06H+<\/sup>(aq)<\/sub>\u00a0 \u2192\u00a03Br2(1)<\/sub> +3H2<\/sub>O(1)<\/sub>
                      \nThe experimental rate law is
                      \nRate = k [Br–<\/sup>] [BrO3<\/sub>–<\/sup>][H+<\/sup>]2<\/sup><\/p>\n

                      2.\u00a0 CH3<\/sub> CHO(g)<\/sub> \\(\\underrightarrow { \\triangle }\\) CH4(g)<\/sub> + CO(g)<\/sub>
                      \nthe experimental rate law is
                      \nRate = k [CH3<\/sub>CHO]3\/2<\/sup>
                      \nSolution:
                      \n1. First order with respect to Br–<\/sup>, first order with respect to BrO3<\/sub>–<\/sup> and second order with respect to H+<\/sup>. Hence the overall order of the reaction is equal to 1+1+2=4<\/p>\n

                      2. Order of the reaction with respect to acetaldehyde is \\(\\frac { 3 }{ 2 }\\)and overall order is also \\(\\frac { 3 }{ 2 }\\)<\/p>\n

                      Question 4.
                      \nThe rate of the reaction x + 2y \u2192 product is 4 x 10-3<\/sup> mol L-1<\/sup> s-1<\/sup>, if [x] = [y] = 0.2 M and rate constant at 400K is 2 x 10-2<\/sup> s-1<\/sup>. what is the overall order of the reaction.
                      \nSolution:
                      \n\"Samacheer
                      \nComparing the powers on both sides, the overall order of the reaction n + m = 1<\/p>\n

                      Question 5.
                      \nA first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989; log10 = 1)
                      \nSolution:
                      \nFor a first order reaction,
                      \n\"Samacheer
                      \nLet [A0<\/sub>] = 100M
                      \nWhen
                      \nt = t90%<\/sub>; [A] = 10M (given that t90%<\/sub> = 8 hours)
                      \nt = t80%<\/sub>; [A] = 20M
                      \n\"Samacheer
                      \nFind the value of k using the given data
                      \nk = \\(\\frac{2.303}{t_{90 \\%}}\\) log \\((\\frac { 100 }{ 10 })\\)
                      \nk = \\(\\frac { 2.303 }{ 8 hours }\\) log 10
                      \nk = \\(\\frac { 2.303 }{ 8 hours }\\) (1)
                      \nSubstitute the value of k in equation (2)
                      \nt80%<\/sub> = \\(\\frac { 2.303 }{ 2.303\/8 hours }\\) log (5)
                      \nt80%<\/sub> = 8 hours x 0.6989
                      \nt80%<\/sub> = 5.59 hours<\/p>\n

                      Question 6.
                      \nThe half life of a first order reaction x \u2192 products is 6.932 x 104<\/sup> s at 500K . What percentage of x would be decomposed on heating at 500K for 100 min. (e0.06<\/sup> = 1.06)
                      \nSolution:
                      \nGiven t1\/2<\/sub> = 0.6932 x 104<\/sup> s
                      \nTo solve: when t = 100 min
                      \n\"Samacheer
                      \nWe know that for a first order reaction,t1\/2<\/sub> = \\(\\frac { 0.6932 }{ k }\\)
                      \n\"Samacheer<\/p>\n

                      Question 7.
                      \nShow that in case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.
                      \nSolution:
                      \nLet [A0<\/sub>] = 100
                      \nWhen t = t99.9%<\/sub>; [A] = (100 – 99.9) = 0.1
                      \n\"Samacheer<\/p>\n

                      Question 8.
                      \nThe rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s-1<\/sup> respectively. Calculate the value of activation energy.
                      \nAnswer:
                      \nAccording to Arrhenius equation.
                      \n\"Samacheer<\/p>\n

                      Question 9.
                      \nRate constant k of a reaction varies with temperature T according to the following Arrhenius equation. log k = log A \\(\\frac { { E }_{ a } }{ 2.303R }\\) \\(\\frac { 1 }{ T }\\).<\/p>\n

                      Where E is the activation energy. When a graph is plotted for log k Vs \\(\\frac { 1 }{ T }\\) a straight line with a slope of 4000K is obtained. Calculate the activation energy.
                      \nSolution:
                      \n\"Samacheer
                      \nEa<\/sub> = – 2.303 Rm
                      \nEa<\/sub> = – 2.303 x 8.314 JK-1<\/sup> mol-1<\/sup> x (-4000k)
                      \nEa<\/sub> = 76,589 J mol-1<\/sup>
                      \nEa<\/sub> = 76,589 kJ mol-1<\/sup><\/p>\n

                      Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Additional Questions<\/h3>\n

                      Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 1 Mark Questions and Answers<\/strong><\/p>\n

                      I. Choose the correct answer.<\/p>\n

                      Question 1.
                      \nWhich one of the following is a slow reaction?
                      \n(a) Rusting of iron
                      \n(b) Combustion of carbon
                      \n(c) Reaction between BaCl2<\/sub> and dil. H2<\/sub>SO4<\/sub>
                      \n(d) Reaction between acidified K2<\/sub>Cr2<\/sub>O7<\/sub> with NaCl.
                      \nAnswer:
                      \n(a) Rusting of iron<\/p>\n

                      Question 2.
                      \nWhich one of the following is the unit of rate of reaction?
                      \n(a) s-1<\/sup>
                      \n(b) mol s-1<\/sup>
                      \n(c) mol L-1<\/sup> s-1<\/sup>
                      \n(d) mol L s
                      \nAnswer:
                      \n(c) mol L-1<\/sup> s-1<\/sup><\/p>\n

                      Question 3.
                      \nFor a gas phase reaction, the unit of reaction rate is ……………
                      \n(a) s-1<\/sup>
                      \n(b) atm s-1<\/sup>
                      \n(c) mol L-1<\/sup> s-1<\/sup>
                      \n(d) mol-1<\/sup> L-1<\/sup> s-1<\/sup>
                      \nAnswer:
                      \n(b) atm s-1<\/sup><\/p>\n

                      Question 4.
                      \nFor the reaction A \u2192 2B, the rate of the reaction is ………….
                      \n(a) +\\(\\frac { d[B] }{ dt }\\) = 2 – \\(\\frac { d[A] }{ dt }\\)
                      \n(b) +\\(\\frac { d[A] }{ dt }\\) = \\(\\frac { 1 }{ 2 }\\) \\(\\frac { d[B] }{ dt }\\)
                      \n(c) Rate = \\(\\frac { 1 }{ 2 }\\) = \\(\\frac { d[A] }{ dt }\\)
                      \n(d) Rate = 2\\(\\frac { d[B] }{ dt }\\)
                      \nAnswer:
                      \n(a) +\\(\\frac { d[B] }{ dt }\\) = 2 – \\(\\frac { d[A] }{ dt }\\)<\/p>\n

                      Question 5.
                      \nConsider the following statement.
                      \n(i) In ionisation of cyclopropane, if the concentration of cyclopropane is reduced half, the rate increases twice.
                      \n(ii) The rate of the reaction depends upon the concentration of the reactant.
                      \n(iii) Order values must be determined experimentally.<\/p>\n

                      Which of the above statement (s) is \/ are not correct?
                      \n(a) (i) only
                      \n(b) (ii) and (iii)
                      \n(c) (iii) only
                      \n(d) (ii) only
                      \nAnswer:
                      \n(a) (i) only<\/p>\n

                      Question 6.
                      \nIn the reaction 2NO(g)<\/sub> + O2(g)<\/sub> \u2192 2NO2(g)<\/sub> the order of the reaction with respect to NO is…………
                      \n(a) first order
                      \n(b) second order
                      \n(c) third order
                      \n(d) zero order
                      \nAnswer:
                      \n(b) second order<\/p>\n

                      Question 7.
                      \nIn the reaction 2NO(g)<\/sub> + O2(g)<\/sub> \u2192 2NO2(g)<\/sub>. the order of the reaction with respect to O2<\/sub>is …….
                      \n(a) zero order
                      \n(b) first order
                      \n(c) second order
                      \n(d) third order
                      \nAnswer:
                      \n(b) first order<\/p>\n

                      Question 8.
                      \nThe overall order of the reaction 2NO(g)<\/sub> + O2(g)<\/sub> \u2192 2NO2(g)<\/sub> is …………….
                      \n(a) 2
                      \n(b) 1
                      \n(c) 3
                      \n(d) 0
                      \nAnswer:
                      \n(c) 3<\/p>\n

                      Question 9.
                      \nConsider the following statements.
                      \n(i) Rate of the reaction does not depend on the initial concentration of the reactants.
                      \n(ii) Rate constant of the reaction depends on the initial concentration of reactants.
                      \n(iii) Rate constant of the reaction is equal to the rate of the reaction, when the concentration of each of the reactants is unity.
                      \nWhich of the above statement(s) is \/ are not correct?
                      \n(a) (i) only
                      \n(b) (ii) only
                      \n(c) (i) and (ii)
                      \n(d) (iii) only
                      \nAnswer:
                      \n(a) (iii) only<\/p>\n

                      Question 10.
                      \nThe overall molecularity of the reaction 2H2<\/sub> O2(aq)<\/sub> \\(\\underrightarrow { { I }^{ – } }\\) 2H2<\/sub>O1<\/sub> + O2(g)<\/sub> is …………
                      \n(a) unimolecular
                      \n(b) bimolecular
                      \n(c) termolecular
                      \n(d) pentamolecular
                      \nAnswer:
                      \n(b) bimolecular<\/p>\n

                      Question 11.
                      \nWhich of the following is the order of decomposition of hydrogen peroxide catalysed by I–<\/sup> ………….
                      \n(a) First order
                      \n(b) Second order
                      \n(c) Zero order
                      \n(a) Third order
                      \nAnswer:
                      \n(a) First order<\/p>\n

                      Question 12.
                      \nConsider the following statements.
                      \n(i) order cannot be zero.
                      \n(ii) Molecularity can be zero (or) fractional (or) integer.
                      \n(iii) order can be determined only by experiment.<\/p>\n

                      Which of the above statement(s) is \/ are not correct?
                      \n(a) (i) only
                      \n(b) (ii) only
                      \n(c) (iii) only
                      \n(d) (i) and (ii)
                      \nAnswer:
                      \n(c) (iii) only<\/p>\n

                      Question 13.
                      \nThe overall order of the reaction 5Br–<\/sup> + BrO3<\/sub>–<\/sup> + 6H+<\/sup> is ……..
                      \n(a) 4
                      \n(b) 3\/2
                      \n(c) 12
                      \n(d) 1
                      \nAnswer:
                      \n(a) 4<\/p>\n

                      Question 14.
                      \nWhich one of the following reaction is a fractional order reaction?
                      \n(a) 2NO +O2<\/sub> \u2192 2NO2<\/sub>
                      \n(b) CH3<\/sub>CHO(g)<\/sub> \u2192 CH4(g)<\/sub> + CO(g)<\/sub>
                      \n(c) 2H2<\/sub>O2<\/sub> \u2192 2H2<\/sub>)O(1)<\/sub> + O2(g)<\/sub>
                      \n(d) H2<\/sub> + Br2<\/sub> \u2192 2HBr
                      \nAnswer:
                      \n(b) CH3<\/sub>CHO(g)<\/sub> \u2192 CH4(g)<\/sub> + CO(g)<\/sub><\/p>\n

                      Question 15.
                      \nThe order of decomposition of acetaldehyde is …………….
                      \n(a) 1
                      \n(b) 1.5
                      \n(c) 2
                      \n(d) 5\/2
                      \nAnswer:
                      \n(b) 1.5<\/p>\n

                      Question 16.
                      \nWhich one of the following is the unit of rate constant for a first order reaction?
                      \n(a) mol-1<\/sup> L s-1<\/sup>
                      \n(b) mol L-1<\/sup> s-1<\/sup>
                      \n(c) s-1<\/sup>
                      \n(d) mol L S
                      \nAnswer:
                      \n(c) s-1<\/sup><\/p>\n

                      Question 17.
                      \nWhich one of the following is an example for first order reaction?
                      \n(a) 2NO(g)<\/sub>+ O2(g)<\/sub> \u2192 2NO2(g)<\/sub>
                      \n(b) CH3<\/sub>CHO(g)<\/sub> \u2192 CH4(g)<\/sub> + CO(g)<\/sub>
                      \n(c) SO2<\/sub> Cl2(1)<\/sub> \u2192 SO2(g)<\/sub> + Cl2(g)<\/sub>
                      \n(d) 2HBr \u2192 H2<\/sub> + Br2<\/sub>
                      \nAnswer:
                      \n(c) SO2<\/sub> Cl2(1)<\/sub> \u2192 SO2(g)<\/sub> + Cl2(g)<\/sub><\/p>\n

                      Question 18.
                      \nWhich one of the following is not an example for first order reaction?
                      \n(a) N2<\/sub>O5(g)<\/sub> \u2192 2NO2(g)<\/sub> \\(\\frac { 1 }{ 2 }\\) O2(g)<\/sub>
                      \n(b) SO2<\/sub>Cl2(1)<\/sub> \u2192 SO2(g)<\/sub> + Cl2(g)<\/sub>
                      \n(e) H2<\/sub>O2<\/sub>(aq)<\/sub> \u2192 H2<\/sub>O1<\/sub>\\(\\frac { 1 }{ 2 }\\)O2(g)<\/sub>
                      \n(d) CH3<\/sub>CHO(g)<\/sub> \u2192 CH4(g)<\/sub> + CO(g)<\/sub>
                      \nAnswer:
                      \n(d) CH3<\/sub>CHO(g)<\/sub> \u2192 CH4(g)<\/sub> + CO(g)<\/sub><\/p>\n

                      Question 19.
                      \nWhat is the order of isomerisation of cyclopropane to propene?
                      \n(a) 1.5
                      \n(b) 3\/2
                      \n(c) 5\/2
                      \n(d) 1
                      \nAnswer:
                      \n(d) 1<\/p>\n

                      Question 20.
                      \nWhich one of the following is an example of pseudo first order reaction?
                      \n(a) CH3<\/sub>CHO4(g)<\/sub> \u2192 CH4(g)<\/sub> + CO(g)<\/sub>
                      \n(b) 2H2<\/sub>O2(aq)<\/sub> \u2192 H2<\/sub>O(1)<\/sub> +O2(g)<\/sub>
                      \n(c) CH3<\/sub>COOCH3(aq)<\/sub> + H2<\/sub>O(1)<\/sub> \\(\\underrightarrow { { H }^{ + } }\\) CH3<\/sub>COOH(aq)<\/sub> + CH3<\/sub>OH(aq)<\/sub>
                      \n(d) Isomerisation of cyclo propane to propene
                      \nAnswer:
                      \n(c) CH3<\/sub>COOCH3(aq)<\/sub> + H2<\/sub>O(1)<\/sub> \\(\\underrightarrow { { H }^{ + } }\\) CH3<\/sub>COOH(aq)<\/sub> + CH3<\/sub>OH(aq)<\/sub><\/p>\n

                      Question 21.
                      \nWhich one of the following is called pseudo first order reaction?
                      \n(a) Decomposition of acetaldehyde
                      \n(b) Acid hydrolysis of an ester
                      \n(c) Isomerisation of cyclopropane to propene
                      \n(d) Decomposition of hydrogen peroxide
                      \nAnswer:
                      \n(b) Acid hydrolysis of an ester<\/p>\n

                      Question 22.
                      \nWhich of the following is an example of zero order reaction?
                      \n(a) lodination of acetone in acid medium
                      \n(b) Hydrolysis of an ester in acid medium
                      \n(c) Decomposition of acetaldehyde
                      \n(d) Isomerisation of cyclopropane to propene
                      \nAnswer:
                      \n(a) lodination of acetone in acid medium<\/p>\n

                      Question 23.
                      \nWhich one of the follow is not zero order reaction?
                      \n(a) H2(g)<\/sub> + Cl2(g) <\/sub>\\(\\underrightarrow { h\\nu }\\)\u00a0 2HCI(g)<\/sub>
                      \n(b) N2<\/sub>O(g)<\/sub> \\(\\rightleftharpoons\\) N2(g)<\/sub> + \\(\\frac { 1 }{ 2 }\\) O2(g)<\/sub>
                      \n(c) CH3<\/sub>CHO(g)<\/sub> \u2192 CH4(g)<\/sub>+ CO(g)<\/sub>
                      \n(d) CH3<\/sub>COCH3<\/sub> + I2<\/sub> \\(\\underrightarrow { { H }^{ + } }\\) CH2<\/sub>COCH3<\/sub> + HI
                      \nAnswer:
                      \n(c) CH3<\/sub>CHO(g)<\/sub> \u2192 CH4(g)<\/sub>+ CO(g)<\/sub><\/p>\n

                      Question 24.
                      \nConsider the following statements.
                      \n(i) For a first order reaction, half life period is independent of initial concentration.
                      \n(ii) Photo chemical reaction between H2<\/sub> and Cl2<\/sub> is a zero order reaction
                      \n(iii) Acid hydrolysis of an ester is a second order reaction<\/p>\n

                      Which of the above statement is\/are correct?
                      \n(a) (i) only
                      \n(b) (iii) only
                      \n(c) (i) & (ii)
                      \n(d) (ii) & (iii)
                      \nAnswer:
                      \n(c) (i) & (ii)<\/p>\n

                      Question 25.
                      \nThe formula of half life for an nth order reaction involving reactant A and n \\(\\neq\\) 1 is
                      \n\"Samacheer
                      \nAnswer:
                      \n\"Samacheer<\/p>\n

                      Question 26.
                      \nThe half life period of first order reaction is 10 seconds. What is the time required for 99.9% completion of that reaction?
                      \n(a) 20 seconds
                      \n(b) 1000 seconds
                      \n(c) 100 seconds
                      \n(d) 999 seconds
                      \nAnswer:
                      \n(c) loo seconds
                      \nHint:
                      \n10 x t1\/2<\/sub> = t99.9%<\/sub>
                      \n\u2234 t99.9%<\/sub> = 10 x 10 sec = 100 sec<\/p>\n

                      Question 27.
                      \nWhich one of the following is known as arrhenius equation?
                      \n\"Samacheer
                      \nAnswer:
                      \n\"Samacheer<\/p>\n

                      Question 28.
                      \nWhich one of the following does not affect the rate of the reaction?
                      \n(a) Nature of the reactant
                      \n(b) Concentration of the reactants
                      \n(c) Surface area and temperature
                      \n(d) pressure
                      \nAnswer:
                      \n(d) pressure<\/p>\n

                      Question 29.
                      \nConsider the following statements.
                      \n(i) Higher the concentration, slower is the possibility for collision and rate also slower
                      \n(ii) Increase in surface area of reactant leads to more collisions per litre per second
                      \n(iii) Gas phas reactions are slower as compared to solid or liquid reactants<\/p>\n

                      Which of the above statement is\/are not correct?
                      \n(a) (ii)
                      \n(b) (i) & (iii)
                      \n(c) (ii) & (iii)
                      \n(d) (i) & (ii)
                      \nAnswer:
                      \n(b) (i) & (iii)<\/p>\n

                      Question 30.
                      \nWhich of the following reaction take place at a faster rate?
                      \n\"Samacheer
                      \nAnswer:
                      \n\"Samacheer<\/p>\n

                      Question 31.
                      \nWhich one of the following graph is not correct ………..
                      \n\"Samacheer
                      \nAnswer:
                      \n\"Samacheer<\/p>\n

                      Question 32.
                      \nThe half life of paracetamol with in the body is ………
                      \n(a) 2 hours
                      \n(b) 2.5 hours
                      \n(c) 6 hours
                      \n(d) 10 hours
                      \nAnswer:
                      \n(b) 2.5 hours<\/p>\n

                      Question 33.
                      \nWhat is the order of radioactive decay?
                      \n(a) first order
                      \n(b) zero order
                      \n(c) second order
                      \n(d) third order
                      \nAnswer:
                      \n(a) first order<\/p>\n

                      Question 34.
                      \nt1\/2<\/sub> of the reaction increases with increase in initial concentration of the reaction means the order of the reaction will be …………
                      \n(a) first order
                      \n(b) zero order
                      \n(c) second order
                      \n(d) third order
                      \nAnswer:
                      \n(b) zero order<\/p>\n

                      Question 35.
                      \nThe reaction rate that does not decrease with time is …………
                      \n(a) pseudo first order reaction
                      \n(b) first order reaction
                      \n(c) zero order reaction
                      \n(d) second order reaction
                      \nAnswer:
                      \n(c) zero order reaction<\/p>\n

                      Question 36.
                      \nThe rate of the reaction X \u2192 Y becomes 8 times when the concentration of the reactant \u2018X\u2019 is doubled. The rate law of the reaction is ………
                      \n(a) – \\(\\frac { d[x] }{ dt }\\) = k[X]2<\/sup>
                      \n(b) – \\(\\frac { d[x] }{ dt }\\) = k[X]3<\/sup>
                      \n(c) – \\(\\frac { d[x] }{ dt }\\) = k[X]4<\/sup>
                      \n(d) – \\(\\frac { d[x] }{ dt }\\) = k[X]8<\/sup>
                      \nAnswer:
                      \n(b) – \\(\\frac { d[x] }{ dt }\\) = k[X]3<\/sup><\/p>\n

                      Question 37.
                      \nThe decomposition of ammonia gas on platinum surface has a rate constant k = 2.5 x 10-4<\/sup> mol L-1<\/sup> s-1<\/sup> What is the order of the reaction?
                      \n(a) first order
                      \n(b) second order
                      \n(c) third order
                      \n(d) zero order
                      \nAnswer:
                      \n(d) zero order<\/p>\n

                      Question 38.
                      \nA reaction is 50% completed in 2 hours and 75% completed in 4 hours. Then the order of the reaction is ………….
                      \n(a) first order
                      \n(b) zero order
                      \n(c) second order
                      \n(d) third order
                      \nAnswer:
                      \n(a) first order
                      \nAnswer:
                      \n(a) first order<\/p>\n

                      Question 39.
                      \nWhat is the rate equation for the reaction A + B \u2192 C has zero order?
                      \n(a) Rate = k
                      \n(b) Rate = k [A]
                      \n(c) Rate = k [A]. [B]
                      \n(a) Rate = k. \\(\\frac { 1 }{ [c] }\\)
                      \nAnswer:
                      \n(c) Rate = k [A]. [B]<\/p>\n

                      Question 40.
                      \nHow does the value of rate constant vary with reactant concentration?
                      \n\"Samacheer
                      \nAnswer:
                      \n\"Samacheer<\/p>\n

                      Question 41.
                      \nIdentify the reaction order if the unit of rate constant is s-1<\/sup> ……….
                      \n(a) zero order reaction
                      \n(b) second order reaction
                      \n(c) first order reaction
                      \n(d) third order reaction
                      \nAnswer:
                      \n(c) first order reaction<\/p>\n

                      Question 42.
                      \nWhat is unit of zero order reaction?
                      \n(a) s-1<\/sup>
                      \n(b) mol-1<\/sup> L-1<\/sup> s-1<\/sup>
                      \n(c) mol L-1<\/sup> s-1<\/sup>
                      \n(d) mol L s-1<\/sup>
                      \nAnswer:
                      \n(c) mol L-1<\/sup> s-1<\/sup><\/p>\n

                      Question 43.
                      \nWhich of the following factor affect the rate of the reaction\u2019?
                      \n(a) volume
                      \n(b) pressure
                      \n(c) cone
                      \n(d) all the above
                      \nAnswer:
                      \n(c) cone<\/p>\n

                      Question 44.
                      \nAcid hydrolysis of an ester is an example of ………
                      \n(a) zero order reaction
                      \n(b) Pseudo first order reaction
                      \n(c) second order reaction
                      \n(d) first order reaction
                      \nAnswer:
                      \n(b) Pseudo first order reaction<\/p>\n

                      Question 45.
                      \nPolymerisation reactions follows ………………. order kinetics.
                      \n(a) fractional
                      \n(b) first
                      \n(c) zero
                      \n(d) Pseudo first
                      \nAnswer:
                      \n(a) fractional<\/p>\n

                      Question 46.
                      \nActivation energy of a chemical reaction can be determined by ……….
                      \n(a) changing concentration of the reactants
                      \n(b) Evaluating rate constants at standard temperature
                      \n(c) Evaluating rate constants at two different temperature
                      \n(d) Evaluating reIocities of reaction at two different temperature
                      \nAnswer:
                      \n(c) Evaluating rate constants at two different temperature<\/p>\n

                      Question 47.
                      \nWhich of the following is the fastest reaction?
                      \n\"Samacheer
                      \nAnswer:
                      \n\"Samacheer<\/p>\n

                      Question 48.
                      \nHalf life period of a reaction is found to be inversely proportional to the cube of its initial concentration. The order of the reaction is ………
                      \n(a) 2
                      \n(b) 5
                      \n(c) 3
                      \n(d) 4
                      \nAnswer:
                      \n(d) 4<\/p>\n

                      Question 49.
                      \nA large increase in the rate of a reaction for a rise in temperature is due to ………
                      \n(a) the decrease in the number of collisions
                      \n(b) increase in the number of activated molecules
                      \n(c) the shortening of mean free path
                      \n(d) the lowering of activation energy
                      \nAnswer:
                      \n(b) increase in the number of activated molecules<\/p>\n

                      Question 50.
                      \nThe minimum energy of a molecule would possess in order to enter into a fruitful collision is known as …………..
                      \n(a) Reaction energy
                      \n(b) collision energy
                      \n(c) Activation energy
                      \n(d) Threshold energy
                      \nAnswer:
                      \n(a) Threshold energy<\/p>\n

                      II. Fill in the blanks.<\/p>\n

                        \n
                      1. The unit of the rate of a reaction is ………..<\/li>\n
                      2. For a ……….. reaction, the unit of the reaction rate is atm s<\/li>\n
                      3. An elementary step is characterised by its ………..<\/li>\n
                      4. The total number of reactant species involved in an elementary step is called ………..<\/li>\n
                      5. The sum of powers of concentration terms involved in the experimentally determined rate law is called ………..<\/li>\n
                      6. The overall order of decomposition of acetaldehyde to methane and carbon monoxide is ………..<\/li>\n
                      7. A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, it is called ………..<\/li>\n
                      8. A reaction in which rate is independent of the concentration of the reactant over a wide range of concentration is called ………..<\/li>\n
                      9. All radioactive disentegration reactions follow ……….. kinetics.<\/li>\n
                      10. For a first order reaction, half life does not depend on ………..<\/li>\n
                      11. Half life period of zero order reaction is ……….. proportional to initial concentration of the reactant.<\/li>\n
                      12. Half life period ……….. reaction is directly proportional to initial concentration of the reactant.<\/li>\n
                      13. \u00a0……….. was proposed by Max Trautz and William lewis.<\/li>\n
                      14. Collision theory was proposed by ……….. in 1916 and in ……….. 1918.<\/li>\n
                      15. For a gas at room temperature (298 K) and I atm, each molecule undergoes approximately ……….. per second.<\/li>\n
                      16. In order to react, the collidng molecules must possess a minimum energy called ………..<\/li>\n
                      17. Generally the reaction rate tends to double when the temperature is increased by ………..<\/li>\n
                      18. The number of collisions of reactant molecules per second is known as ………..<\/li>\n
                      19. Heating is required for the reaction between KMnO4<\/sub> and oxalate ion and is carried out at around ………..<\/li>\n
                      20. ……….. reactions are faster as compared to reactions involving solid or liquid reactants.<\/li>\n
                      21. The rate of the reaction ……….. with the increase in the concentration of the reactants.<\/li>\n
                      22. Higher the concentration of reactants greater is the possibility of and hence the ………..<\/li>\n
                      23. In the presence of catalyst the energy of activation is ……….. and hence greater number of molecules change over to products there by increasing the rate of the reaction.<\/li>\n
                      24. Bio availability of drugs within the body and this branch of study is called ………..<\/li>\n
                      25. ……….. has a half life of 2.5 hours within the body.<\/li>\n
                      26. The change in concentration of species per unit time gives the ……….. of the reaction.<\/li>\n
                      27. The rate constant is equal to the rate of the reaction when concentration of reactants is ………..<\/li>\n
                      28. Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of the reaction is ………..<\/li>\n
                      29. Acid hydroJysis of an ester is an example of ………..<\/li>\n
                      30. Molecularity of a chemical reaction will never be equal to ………..<\/li>\n<\/ol>\n

                        Answer:<\/p>\n

                          \n
                        1. moI L-1<\/sup> s-1<\/sup><\/li>\n
                        2. gas phase<\/li>\n
                        3. molecularity<\/li>\n
                        4. Molecularity<\/li>\n
                        5. order<\/li>\n
                        6. 3\/2 or 1.5<\/li>\n
                        7. Pseudo first order reaction<\/li>\n
                        8. Zero order reaction<\/li>\n
                        9. First order<\/li>\n
                        10. initial concentration<\/li>\n
                        11. directly<\/li>\n
                        12. zero order<\/li>\n
                        13. Collision theory<\/li>\n
                        14. Max Trautz, William lewis<\/li>\n
                        15. 10 collisions<\/li>\n
                        16. Activation energy<\/li>\n
                        17. 10\u00b0C<\/li>\n
                        18. Frequency factor (A)<\/li>\n
                        19. 60\u00b0C<\/li>\n
                        20. Gas phase<\/li>\n
                        21. increases<\/li>\n
                        22. collisions, rate<\/li>\n
                        23. lowered<\/li>\n
                        24. Pharmaco kinetics<\/li>\n
                        25. Paracetamol<\/li>\n
                        26. rate<\/li>\n
                        27. unity<\/li>\n
                        28. increased<\/li>\n
                        29. Pseudo first order reaction<\/li>\n
                        30. zero<\/li>\n<\/ol>\n

                          III. Match the following<\/p>\n

                          Match the list I and II using the code given below the list.<\/p>\n

                          Quetion 1.
                          \n\"Samacheer
                          \nAnswer:
                          \n(a) 3, 4, 1, 2<\/p>\n

                          Question 2.
                          \n\"Samacheer
                          \nAnswer:
                          \n(a) 3, 4, 1, 2<\/p>\n

                          Question 3.
                          \n\"Samacheer
                          \nAnswer:
                          \n(a) 3, 1, 4, 2<\/p>\n

                          Question 4.
                          \n\"Samacheer
                          \nAnswer:
                          \n(a) 3, 4, 2, 1<\/p>\n

                          IV. Assertion and Reason<\/p>\n

                          Question 1.
                          \nAssertion (A): Decomposition of hydrogen peroxide catalysed by I–<\/sup> is a bimolecular first order reaction.
                          \nReason (R): The above reaction take place in two steps, step 1 involves both H2<\/sub>O2<\/sub> and I and so it is bimolecular but order is determined experimentally as 1.
                          \n(a) Both A and R are correct and R is the correct explanation of A.
                          \n(b) Both A and R are correct but R is not the correct explanation of A
                          \n(c) A is correct but R is wrong
                          \n(d) A is wrong but R is correct
                          \nAnswer:
                          \n(a) Both A and R arc correct and R is the correct explanation of A<\/p>\n

                          Question 2.
                          \nAssertion (A): \"Samacheer\u00a0The overall order of the reaction is equal to 4.
                          \nReason (R): The experimental rate law is.
                          \nRate = K [Br–<\/sup>] [BrO3<\/sub>–<\/sup>] [H+<\/sup>]2<\/sup>.
                          \nSo 1 + 1 + 2 = 4 order value is 4.
                          \n(a) Both A and R are correct and R is the correct explanation of A.
                          \n(b) Both A and R are wrong
                          \n(c) A is correct but R is wrong
                          \n(d) A is wrong but R is correct
                          \nAnswer:
                          \n(a) Both A and R are correct and R is the correct explanation olA.<\/p>\n

                          Question 3.
                          \nAssertion (A): The rate of a reaction increases with the increase in the concentration of the reactants.
                          \nReason (R): The rate of the reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.
                          \n(a) Both A and R are correct and R is the correct explanation of A,
                          \n(b) Both A and R are correct but R is not the correct explanation of A
                          \n(c) A is correct but R is wrong
                          \n(d) A is wrong but R is correct
                          \nAnswer:
                          \n(a) Both A and R are correct and R is the correct explanation of A.<\/p>\n

                          Question 4.
                          \nAssertion (A): Powdered calcium carbonate reacts much faster with dilute
                          \nHCL than with the same mass of CaCO3<\/sub> as marble.
                          \nReason (R): For a given mass of a reactant, when the particle size decreases, surface area increases. Increase in surface area of the reactant leads to more collisions per litre per second and hence the rate of the reaction also increases.
                          \n(a) Both A and R are correct and R is the correct explanation of A.
                          \n(b) Both A and R arc correct but R is not correct explanation of A
                          \n(c) A is correct but R is wrong
                          \n(d) A is wrong but R is correct
                          \nAnswer:
                          \n(a) Both A and R are correct and R is the correct explanation of A.<\/p>\n

                          Question 5.
                          \nAssertion (A): Catalyst presence increases the rate of the reaction
                          \nReason (R): In the presence of a catalyst, energy of activation is lowered and hence greater number of molecules can across the energy harrier and change over to products thereby increasing the rate of the reaction.
                          \n(a) Both A and R are correct but R is not correct explanation of A
                          \n(b) Both A and R are correct and R is the correct explanation of A
                          \n(c) A is correct but R is wrong
                          \n(d) A is wrong but R is correct
                          \nAnswer:
                          \n(b) Both A and R are correct and R is the correct explanation of A<\/p>\n

                          Question 6.
                          \nAssertion (A): Doctors adviced to take paracetamol once in 6 hours during fever and body pain
                          \nReason (R): Paracetarnol has a half life of 2.5 hours within the body. After 10 hours (4 half lives) only 6.25% of drug remains. Based on this, doctors adviced to take it once in 6 hours.
                          \n(a) Both A and R are wrong
                          \n(b) A is correct but R is wrong
                          \n(c) A and R are correct and R is the correct explanation of A
                          \n(d) A and R are correct but R is not correct explanation of A
                          \nAnswer:
                          \n(a) Both A and R are correct and R is the correct explanation of A.<\/p>\n

                          Question 7.
                          \nAssertion (A): Order of the reaction can be zero or fractional
                          \nReason (R): We cannot determine order from balanced chemical equation
                          \n(a) Both A and R are correct but R is not correct explanation of A.
                          \n(b) Both A and R are correct and R is the correct explanation of A
                          \n(c) A is correct but R is wrong
                          \n(d) A is wrong but R is correct
                          \nAnswer:
                          \n(a) Both A and R are correct and R is not correct explanation of A<\/p>\n

                          Question 8.
                          \nAssertion (A): If the activation enery of a reaction is zero, temperature will have no effect on the rate constant
                          \nReason (R): Lower the activation energy, faster is the reaction.
                          \n(a) Both A and R are correct and R is the correct explanation of A.
                          \n(b) Both A and R are correct but R is not correct explanation of A
                          \n(c) A is correct but R is wrong
                          \n(d) A is wrong but R is correct
                          \nAnswer:
                          \n(b) Both A and R are correct but R is not correct explanation of A<\/p>\n

                          V. Find the odd one out<\/p>\n

                          Question 1.
                          \n\"Samacheer
                          \nAnswer:
                          \n\"Samacheer
                          \nHint: It is a fractional order reaction with order value 3\/2 where as others are first order reaction.<\/p>\n

                          Question 2.
                          \n\"Samacheer
                          \nAnswer:
                          \n\"Samacheer
                          \nHint: It is first order reaction whereas others are zero order reaction.<\/p>\n

                          Question 3.
                          \n(a) Decomposition of dinitrogen pentoxide
                          \n(b) Iodination of acetone in acid medium
                          \n(c) Decomposition of N20 on hot Pt surface
                          \n(d) photochemical reaction between H2 and CI2
                          \nAnswer:
                          \n(a) Decomposition of dinitrogen pentoxide
                          \nHint: It is a first order reaction whereas others are zero order reactions<\/p>\n

                          Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 2 Mark Questions and Answers<\/strong><\/p>\n

                          Question 1.
                          \nDefine rate of the reaction. Give the unit of rate.
                          \nAnswer:
                          \n1. In a chemical reaction, the change in the concentration of the species involved in a chemical reaction per unit time gives the rate of a reaction.
                          \nA \u2192 B
                          \nRate = –\\(\\frac { \\triangle [A] }{ dt }\\) (or) \\(\\frac { \\triangle [B] }{ dt }\\)<\/p>\n

                          2.
                          \n\"Samacheer
                          \n= mol L-1<\/sup> s-1<\/sup><\/p>\n

                          Question 2.
                          \nDefine molecularity of a reaction.
                          \nAnswer:
                          \nMolecularity of a reaction is the total number of reactant species that are involved in an elementary step.<\/p>\n

                          Question 3.
                          \nDefine order of a chemical reaction.
                          \nAnswer:
                          \nOrder of a chemical reaction is the sum of powers of concentration terms involved in the experimentally determined rate law.<\/p>\n

                          Question 4.
                          \nDefine Half life period.
                          \nAnswer:
                          \nThe half life ola reaction is defined as the time required for the reactant concentration to reach one half its initial value.<\/p>\n

                          Question 5.
                          \nMention the factors affecting the reaction rate.
                          \nAnswer:
                          \nThe rate of the reaction is affected by the following factors.<\/p>\n

                            \n
                          1. Nature and state of the reactant<\/li>\n
                          2. Concentration of the reactant<\/li>\n
                          3. Surface area of the reactant<\/li>\n
                          4. Temperature of the reaction<\/li>\n
                          5. Presence of a catalyst<\/li>\n<\/ol>\n

                            Question 6.
                            \nHow is surface area of the reactant affect the rate of the reaction?
                            \nAnswer:<\/p>\n

                              \n
                            1. In heterogeneous reactions, the surface area of the solid reactants play an important role in deciding the rate.<\/li>\n
                            2. For a given mass of a reactant, when the particle size decreases surface area increases. Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased.<\/li>\n
                            3. For example, powdered calcium carbonate reacts much faster with dilute HCl than with the same mass of CaCO3<\/sub> as marble.<\/li>\n<\/ol>\n

                              Question 7.
                              \nParacetamol is prescribed to take once in 6 hours. Justify this statement.
                              \nAnswer:
                              \n1. Paracetamol is a well known antipyretic and analgesic that is prescribed in cases of fever and body pain.<\/p>\n

                              2. Paracetamol has a half life of 2.5 hours within the body. (Le) the plasma concentration of the drug is halved after 2.5 hours. So after 10 hours (4 half lives), only 6.25% of drug remains. Based on this, the dosage and frequency will be decided.<\/p>\n

                              3. In the case of paracetamol, it is usually prescribed to take once in 6 hours.<\/p>\n

                              Question 8.
                              \nFor a reaction, A + B \u2192 product; the rate law is given by r = k[A]1\/2<\/sup> [B]2<\/sup>. What is the order of the reaction?
                              \nAnswer:
                              \nOrder of the reaction = \\(\\frac { 1 }{ 2 }\\) + 2 = 2 \\(\\frac { 1 }{ 2 }\\) or 0.5<\/p>\n

                              Question 9.
                              \nThe conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will \u00a1t affect the rate of formation of Y?
                              \nAnswer:
                              \nFor the reaction, X \u2192 Y as it follows second order kinetics, therefore the rate law equation will be
                              \nRate = k[X]2<\/sup> = ka2<\/sup>
                              \nif [X] = a mol-1<\/sup>
                              \nif the concentration of X is increased three times, then
                              \n[X] = 3a mol L-1<\/sup>
                              \n\u2234 Rate = k (3a)2<\/sup> = 9 ka2<\/sup>
                              \nThus, the rate of the reaction will become 9 times. Hence the rate of formation of Y will increase by 9 times.<\/p>\n

                              Question 10.
                              \nTime required to decompose SO2<\/sub>Cl2<\/sub> to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
                              \nAnswer:
                              \nFor a first order reaction,
                              \n\"Samacheer<\/p>\n

                              Question 11.
                              \nWhat will be the effect of temperature on rate constant?
                              \nAnswer:
                              \nRate constant of a reaction is nearly doubled with rise in temperature by 10\u00b0C. The exact dependence of the rate constant on temperature is given by Arrhenius equation:
                              \nRate constant,
                              \n\"Samacheer<\/p>\n

                              Question 12.
                              \nA reaction is first order in A and second order in B.<\/p>\n

                                \n
                              1. Write the differential rate equation.<\/li>\n
                              2. How is the rate affected on increasing the concentration of B three times?<\/li>\n
                              3. How is the rate affected when the concentrations of both A and B arc doubled?<\/li>\n<\/ol>\n

                                Answer:<\/p>\n

                                  \n
                                1. \\(\\frac { dx }{ dt }\\) = k [A]1<\/sup> [B]2<\/sup><\/li>\n
                                2. If concentration of \u2018B\u2019 is tripled, then the rate will become 9 times.<\/li>\n
                                3. When concentration of both A and B are doubled, then the rate will become 8 times.<\/li>\n<\/ol>\n

                                  Question 13.
                                  \nDefine zero order reaction. Give the unit for its rate constant(k).
                                  \nAnswer:
                                  \nZero Order Reaction. The reaction in which the rate of reaction is independent of the concentration of the reactants is called zero order reaction.
                                  \nRate = k [A]0<\/sup>\u00a0\u21d2 k
                                  \nWhere k is the rate constant. Its unit is mol L-1<\/sup> s-1<\/sup><\/p>\n

                                  Question 14.
                                  \nWrite units of rate constant k for zero order, first order, second order and n order reaction.
                                  \nAnswer:
                                  \nOrder of Reaction<\/p>\n

                                    \n
                                  1. Zero order reaction<\/li>\n
                                  2. First order reaction<\/li>\n
                                  3. Second order reaction<\/li>\n
                                  4. nth order reaction<\/li>\n<\/ol>\n

                                    Unit of k:<\/p>\n

                                      \n
                                    1. mol L-1<\/sup> s-1<\/sup><\/li>\n
                                    2. s-1<\/sup><\/li>\n
                                    3. mol L s-1<\/sup><\/li>\n
                                    4. (mol\/ L)1-n<\/sup> s-1<\/sup><\/li>\n<\/ol>\n

                                      Question 15.
                                      \nWhat is the effect of catalyst on the activation energy? Why?
                                      \nAnswer:
                                      \nA Catalyst lower down the activation energy. It provides an alternate path to the reaction. It forms an unstable intermediate which readily changes into products.<\/p>\n

                                      Question 16.
                                      \nGive two differences between zero order and first order reaction.
                                      \nAnswer:
                                      \nZero Order:<\/p>\n

                                        \n
                                      1. Its \u2018k\u2019 has unit = mol L-1<\/sup> s-1<\/sup><\/li>\n
                                      2. Its t 1\/2 is directly proportional to initial conc. of reactant<\/li>\n<\/ol>\n

                                        First order:<\/p>\n

                                          \n
                                        1. Its \u2018k\u2019 has unit = time-1<\/sup> = sec-1<\/sup><\/li>\n
                                        2. Its half life is independent of the initial conc. of the reactant.<\/li>\n<\/ol>\n

                                          Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 3 Mark Questions and Answers<\/strong><\/p>\n

                                          Question 1.
                                          \nWrite the differences between the rate and rate constant of the reaction.
                                          \nAnswer:
                                          \nRate of a reaction:<\/p>\n

                                            \n
                                          1. It represents the speed at which the reactants are converted into products at any instant<\/li>\n
                                          2. It is measured as decrease in the concentration of the reactants (or) increase in the concentration of products<\/li>\n
                                          3. It depends on the initial concentration of reactants<\/li>\n<\/ol>\n

                                            Rate constant of a reaction:<\/p>\n

                                              \n
                                            1. It is a proportionality constant<\/li>\n
                                            2. It is equal to the rate of the reaction, when the concentration of each of the reactants is unity<\/li>\n
                                            3. It does not depend on the initial concentration of the reactants<\/li>\n<\/ol>\n

                                              Question 2.
                                              \nWhat are the examples of first order reaction?
                                              \nAnswer:<\/p>\n

                                                \n
                                              1. Decompostion of dinitrogen pentoxide
                                                \nN2<\/sub>O2(g)<\/sub> \u2192 2NO2(g)<\/sub> + \\(\\frac { 1 }{ 2 }\\) O2(g)<\/sub><\/li>\n
                                              2. Decomposition of thionylchloride
                                                \nSO2<\/sub>Cl2(g) <\/sub> \u2192 SO2(g)<\/sub> + CI2(g)<\/sub><\/li>\n
                                              3. Decomposition of H2<\/sub>O2<\/sub>\u00a0in aqueous solution
                                                \nH2<\/sub>O2(aq)<\/sub> \u2192 H2<\/sub>O(1)<\/sub> + \\(\\frac { 1 }{ 2 }\\) O2(g)<\/sub><\/li>\n
                                              4. Isomerisation of cyclopropane to propene<\/li>\n<\/ol>\n

                                                Question 3.
                                                \nFor the reaction R \u2192 P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
                                                \nAnswer:
                                                \n\"Samacheer<\/p>\n

                                                Question 4.
                                                \nIn a reaction, 2 A \u2192 products. the concentration of A decreases from 0.5 moI L-1<\/sup> to 0.4 mol L-1<\/sup> in 10 minutes. Calculate the rate during this interval.
                                                \nAnswer:
                                                \nAverage rate
                                                \n\"Samacheer<\/p>\n

                                                Question 5.
                                                \nA first order reaction has a rate constant, 1.15 x 10-3<\/sup> s-1<\/sup>. How long will 5 g of this reactant take to reduce to 3 g?
                                                \nAnswer:
                                                \nHere,
                                                \n[R]0<\/sub> = 5g
                                                \n[R] = 3 g
                                                \nk = 1.15 x 10-3<\/sup> s-1<\/sup> As the reaction is of first order,
                                                \n\"Samacheer<\/p>\n

                                                Question 6.
                                                \nTime required to decompose SO2<\/sub>CI2<\/sub> to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. For a first order reaction,
                                                \nAnswer:
                                                \n\"Samacheer<\/p>\n

                                                Question 7.
                                                \nA reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is-<\/p>\n

                                                  \n
                                                1. doubled<\/li>\n
                                                2. reduced to half.<\/li>\n<\/ol>\n

                                                  Answer:
                                                  \n1. Reaction is second order with respect to the reactant
                                                  \n\u2234 Rate = k[A]2<\/sup> = ka2<\/sup>.
                                                  \nwhen [A] = 2a
                                                  \nRate k (2a)2<\/sup>
                                                  \n= 4ka2<\/sup>
                                                  \n= 4 times
                                                  \nTherefore, when concentration of the reactant is doubled the rate will become 4 times<\/p>\n

                                                  2. when [A] = \\(\\frac { 1 }{ 2 }\\) a
                                                  \nRate = k \\({ \\left( \\frac { 1 }{ 2 } a \\right) }^{ 2 }\\) = \\(\\frac { 1 }{ 4 }\\) ka2<\/sup> = \\(\\frac { 1 }{ 4 }\\) k
                                                  \nTherefore, rate will be reduced to one-fourth of the initial rate.<\/p>\n

                                                  Question 8.
                                                  \nThe rate constant for a first order reaction is 60 s1<\/sup>. How much time will It take to reduce the initial concentration of the reactant to its 1\/6th<\/sup> value?
                                                  \nAnswer:
                                                  \n\"Samacheer<\/p>\n

                                                  Question 9.
                                                  \nFor a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
                                                  \nAnswer:
                                                  \nFor a first order reaction, t = k = \\(\\frac { 2.303 }{ t }\\) log \\(\\frac { a }{ a-x }\\)
                                                  \n99% completion means that
                                                  \nx = 99% of a =0.99 a
                                                  \nt99%<\/sub> = \\(\\frac { 2303 }{ k }\\) log \\(\\frac { a }{ a-0.99a }\\) = \\(\\frac { 2.303 }{ k }\\) log 102<\/sup> = 2 x \\(\\frac { 2303 }{ k }\\)
                                                  \n90% completion means that
                                                  \nx = 90% of a = 0.90 a
                                                  \nt99%<\/sub> = \\(\\frac { 2303 }{ k }\\) log \\(\\frac { a }{ a-0.99a }\\) = \\(\\frac { 2.303 }{ k }\\) log 10 = \\(\\frac { 2.303 }{ k }\\)
                                                  \n\\(\\frac{t_{99 \\%}}{t_{90 \\%}}=\\left(\\frac{2 \\times 2303}{k}\\right) \/ \\frac{2.303}{k}=2\\)
                                                  \nor
                                                  \nt99%<\/sub> = 2 x t90%<\/sub><\/p>\n

                                                  Question 10.
                                                  \nCalculate the half life of a first order reaction whose rate constant is 200 s-1<\/sup>
                                                  \nAnswer:
                                                  \nHere rate constant
                                                  \nk= 200 s-1<\/sup>
                                                  \n\u2234 Half – life of a first order reaction is
                                                  \nt1\/2<\/sub> = \\(\\frac { 0.693 }{ k }\\) = \\(\\frac { 0.693 }{ 200 }\\) = 3.46 x 10-3<\/sup> sec<\/p>\n

                                                  Question 11.
                                                  \nThe decomposition of dinitrogen pentoxide (N2<\/sub>O) follows the first order rate law. Calculate the rate constant from the given data.
                                                  \nt = 800 sec, [N2<\/sub>O5<\/sub>] = 1.45 moI L-1<\/sup> = [A2<\/sub>]
                                                  \nt = 1600 sec
                                                  \n[N2<\/sub>O3<\/sub>] = 0.88 moI L-1<\/sup> = [A2<\/sub>]
                                                  \nAnswer:
                                                  \nApplying the formula,
                                                  \nk = \\(\\frac { 2.303 }{ \\left( { t }_{ 2 }-{ t }_{ 1 } \\right) }\\) log 10 \\(\\frac { \\left[ { A }_{ 1 } \\right] }{ \\left[ { A }_{ 2 } \\right] }\\)
                                                  \n= \\(\\frac { 2.303 }{ (1600 – 800) }\\) log 10 \\(\\frac { 1.45 }{ 0.88 }\\) = \\(\\frac { 2.303 }{ 800 }\\) x 0.2169
                                                  \n= 6.24 x 10-4<\/sup> sec-1<\/sup><\/p>\n

                                                  Question 12.
                                                  \nA second order reaction in which both the reactants have same concentration, is 20% completed in 500 seconds. How much time it will take for 60% completion?
                                                  \nAnswer:
                                                  \nThe second order equation when both the reactants have same concentration is
                                                  \nk = \\(\\frac { 1 }{ t }\\). \\(\\frac { x }{ a(a – x) }\\)
                                                  \nIf a = 100, x = 20, 1= 500 seconds
                                                  \nThen k = \\(\\frac { 1 }{ 500 }\\) x \\(\\frac { 20 }{ 100\u00a0 x\u00a0 (100-20) }\\)
                                                  \nWhen
                                                  \na = 100
                                                  \nx = 60
                                                  \nt = ?
                                                  \nt = \\(\\frac { 1 }{ k }\\) \\(\\frac { 60 }{ 100\u00a0 x\u00a0 40 }\\)
                                                  \nSubstituting the value of k,
                                                  \nt = \\(\\frac { 500\u00a0 x\u00a0 100\u00a0 x\u00a0 80 }{ 20 }\\) x \\(\\frac { 60 }{ 100\u00a0 x\u00a0 40 }\\)
                                                  \nor
                                                  \nt = 3000 seconds<\/p>\n

                                                  Question 13.
                                                  \nA first order reaction is 20% completed in 10 minutes. Calculate the time taken for the reaction to go to 80% completion.
                                                  \nAnswer:
                                                  \nApplying the first order equation,
                                                  \nk = \\(\\frac { 2303 }{ t }\\) l0g \\(\\frac { { \\left[ R \\right] }_{ 0 } }{ \\left[ R \\right] }\\)
                                                  \nAt t = 10 min
                                                  \nR = 100 – 20
                                                  \nk = \\(\\frac { 2303 }{ t }\\) log 10 \\(\\frac { 100 }{ (100 – 20) }\\)
                                                  \nt = \\(\\frac { 2303 }{ 10 }\\) log 10 \\(\\frac { 100 }{ 80 }\\)
                                                  \n= 0.0223 min-1<\/sup><\/p>\n

                                                  Question 14.
                                                  \nFor a reaction: 2NH3(g)\u00a0<\/sub> \\(\\underrightarrow { Pt }\\)\u00a0 N2(g)<\/sub>+ 3H2(g)<\/sub> Rate = K<\/p>\n

                                                    \n
                                                  1. Write the order and molecularity of this reaction.<\/li>\n
                                                  2. Write the unit of K.<\/li>\n<\/ol>\n

                                                    Answer:<\/p>\n

                                                      \n
                                                    1. Order of reaction Zero order. Molecularity = 2<\/li>\n
                                                    2. Unit of K = mol L-1<\/sup> sec-1<\/sup><\/li>\n<\/ol>\n

                                                      Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 5 Marks Questions and Answers<\/strong><\/p>\n

                                                      Question 1.
                                                      \nHow would you calculate the order of the reaction 2NO + O2(g)<\/sub> \u2192 2NO2(g)<\/sub> by an experiment?
                                                      \n(or)
                                                      \nprove that 2NO + O2<\/sub> \u2192 2 NO2<\/sub> is a third order reaction.
                                                      \nAnswer:
                                                      \n2NO(g)<\/sub> + O2(g)<\/sub> \u2192 2NO2(g)<\/sub>
                                                      \nSeries of experiments are conducted by keeping the concentration of one of the reactants as constant and changing the concentration of the others.
                                                      \n\"Samacheer
                                                      \nRate = k [NO]m<\/sup>[O2<\/sub>]n<\/sup>
                                                      \nFor experiment 1, the rate law is
                                                      \nRate1<\/sub> = k [NO]m<\/sup> [O2<\/sub>]n<\/sup>
                                                      \n19.26 x 10-2<\/sup> = k [1.3]m<\/sup> [1.1]n<\/sup> ………………….(1)
                                                      \nFor experiment 2
                                                      \nRate2<\/sub> = k [NO]m<\/sup> [O2<\/sub>]n<\/sup>
                                                      \n38.40 x 10-2<\/sup> = k [1.3]m<\/sup> [2.2]n<\/sup> ……………….(2)
                                                      \nFor experiment 3
                                                      \nRate3<\/sub> = k [NO]m<\/sup> [O2<\/sub>]n<\/sup>
                                                      \n76.8 x 10-2<\/sup> = k [2.6]m<\/sup> [1.1]n<\/sup> ………………(3)
                                                      \n\"Samacheer
                                                      \n2 = \\({ \\left( \\frac { 2.2 }{ 1.1 } \\right) }^{ n }\\)
                                                      \n4 = 2m<\/sup>
                                                      \n\u21d2 n = 1
                                                      \nTherefore the reaction is first order with respect to O2<\/sub>
                                                      \n\"Samacheer
                                                      \n2 = \\({ \\left( \\frac { 2.6 }{ 1.3 } \\right) }^{ m }\\)
                                                      \n4 = 2m<\/sup>
                                                      \n\u21d2 m = 1
                                                      \nTherefore the reaction is second order with respect to NO
                                                      \nThe rate law is Rate–<\/sub> = k [NO]2<\/sup> [O2<\/sub>]1<\/sup>
                                                      \nThe overall order of the reaction = 2 + 1 = 3<\/p>\n

                                                      Question 2.
                                                      \nDerive the integrated rate law for a first order reaction?
                                                      \nAnswer:
                                                      \nA reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. First order reaction is A \u2192 product. Rate law can be expressed as, Rate = k [A]1<\/sup>. Where, k is the first order rate constant
                                                      \n\\(\\frac { -d[A] }{ dt }\\) = k[A]1<\/sup>
                                                      \n\\(\\frac { -d[A] }{ [A] }\\) = k.dt ……………………(1)
                                                      \nIntegrate the above equation (I) between the limits of time t = 0 and time equal to t, while the concentration varies from initial concentration [A0<\/sub>] to [A] at the later time.
                                                      \n\"Samacheer
                                                      \nThis equation (2) is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303
                                                      \n\"Samacheer<\/p>\n

                                                      Question 3.
                                                      \nExplain the effect of temperature on reaction rate based on Arrhenius theory.
                                                      \nAnswer:
                                                      \n1. Generally, the rate of a reaction increases with increasing temperature. However, there are very few exceptions.<\/p>\n

                                                      2. As a rough rule, for many reactions near room temperature, reaction rate tends to double when the temperature is increased by 10\u00b0C.<\/p>\n

                                                      3. A large number of reactions are known which do not take place at room temperature but occur readily at higher temperature. Example – Reaction between H2<\/sub> and O2<\/sub> to form H2<\/sub>O takes place only when an electric spark is passed.<\/p>\n

                                                      4. Arrhenius suggested that the rates of most reactions vary with temperature in such a way that the rate constant is directly proportional to
                                                      \n\"Samacheer and he proposed a relation between the rate constant and temperature.
                                                      \n\"Samacheer
                                                      \nwhere
                                                      \nk = frequency factor
                                                      \nR = gas constant ,
                                                      \nEa<\/sub> = Activation energy
                                                      \nT = Absolute temperature (in kelvin)
                                                      \nThe factor A does not vary significantly with temperature and hence it may be taken as a constant.<\/p>\n

                                                      5. Taking logarithm on both side of the equation (1)
                                                      \n\"Samacheer<\/p>\n

                                                      6. The plot of Ink vs\u00a0\\((\\frac { 1 }{ T })\\) is a straight line with negative slope\\(\\frac { { E }_{ a } }{ RT }\\). If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy.
                                                      \n\"Samacheer
                                                      \nThis equation can be used to calculate E from rate Ea<\/sub> constants k1<\/sub> & k2<\/sub> at temperature T1<\/sub> and T2<\/sub><\/p>\n

                                                      Question 4.
                                                      \nExplain about the factors that affecting the reaction rate.
                                                      \nAnswer:
                                                      \nThe rate of a reaction is affected by the following factors.
                                                      \n1. Nature and state of the reactant
                                                      \n(a) A chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.<\/p>\n

                                                      (b) Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine.<\/p>\n

                                                      2. Concentration of the reactant
                                                      \nThe rate of the reaction increases with the increase in the concentration of the reactants. According to collision theory, the rate of the reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.<\/p>\n

                                                      3. Effect of surface area of the reactant:
                                                      \nIn heterogeneous reactions, the surface areas of the solid reactants plays an important role in deciding the rate. For a given mass of a reactant, when the particle size decreases surface area increases.<\/p>\n

                                                      Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased. For example, powdered calcium carbonate reacts much faster with dilute HCI than with the same mass of CaCOl as marble<\/p>\n

                                                      4. Temperature:
                                                      \nFor many reactions near room temperature, the reaction rate tends to double when the temperature is increased by 10\u00b0C . For eg, Reaction between H2<\/sub> and O2<\/sub> to form H2<\/sub>O take place only when an electric spark is passed. So when the temperature increases, the rate of the reaction also increases.<\/p>\n

                                                      5. Effect of presence of catalyst
                                                      \n(a) A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change. They may participate in the reaction, but again regenerated and the end of the reaction.<\/p>\n

                                                      (b) In the presence of a catalyst, the energy of activation is lowered and hence greater number of molecules can cross the energy barrier and change over to products,thereby increasing the rate of the reaction.<\/p>\n

                                                      Question 5.
                                                      \nThe decomposition of A into product has value of k as 4.5 x 103<\/sup> s-1<\/sup> at 10\u00b0C and energy of activation 60 kJ mol-1<\/sup>. At what temperature would k be 1.5 x 104<\/sup> s-1<\/sup> ?
                                                      \nAnswer:
                                                      \nk1<\/sub> = 4.5 x 103<\/sup> s-1<\/sup>
                                                      \nT1<\/sub> = 10 + 273 K = 283 K
                                                      \nk2<\/sub> = 1.5 x 104<\/sup> s-1<\/sup>
                                                      \nT2<\/sub> = ?
                                                      \nEa<\/sub> = 6o kJ mol-1<\/sup>
                                                      \nAccording to Arrhenius equation
                                                      \n\"Samacheer<\/p>\n

                                                      Question 6.
                                                      \nFor a decomposition reaction the values of rate constant k at two different temperatures are given below:
                                                      \nk1<\/sub> = 2.15 x 10 L mol-1<\/sup> s-1<\/sup> at 650 K
                                                      \nk2<\/sub> = 2.39 x 10 L mol-1<\/sup> s-1<\/sup> at 700 K
                                                      \nCalculate the value of activation energy for this reaction. (R = 8.314 J K-1<\/sup> mol-1<\/sup> )
                                                      \nAnswer:
                                                      \nHere
                                                      \nk1<\/sub> = 2.15 x 10 L mol-1<\/sup> s-1<\/sup> at 650 K
                                                      \nT1<\/sub> = 650 K
                                                      \nT2<\/sub> = 700K and
                                                      \nk2<\/sub> = 2.39 x 10 L mol-1<\/sup> s-1<\/sup> at 700 K
                                                      \nR = 8.314 J K-1<\/sup> mol-1<\/sup>
                                                      \nUsing the formula
                                                      \n\"Samacheer<\/p>\n

                                                      Question 7.
                                                      \nFor a certain chemical reaction variation In concentration, In IRI Vs time (s) plot is given below.
                                                      \nFor this reaction write\/draw:<\/p>\n

                                                        \n
                                                      1. What is the order of the reaction?<\/li>\n
                                                      2. What is the units of rate constant (k)?<\/li>\n
                                                      3. Give the relationship between k and t1\/2<\/sub> (half-life period).<\/li>\n
                                                      4. What does the slope of above line indicate?<\/li>\n
                                                      5. Draw the plot of log [R0<\/sub>]\/[R] vs time (s)<\/li>\n<\/ol>\n

                                                        \"Samacheer<\/p>\n

                                                        Answer:
                                                        \n1. First order
                                                        \n2. time-1<\/sup> (s-1<\/sup>)
                                                        \n3. \"Samacheer
                                                        \n4.\u00a0 Rate constat (k) of reaction
                                                        \n5. \"Samacheer<\/p>\n

                                                        Question 8.
                                                        \nA substance reacts according to the first order rate law and the specific reaction rate for the reaction \u00a1s 1 x 10-2<\/sup> s-1<\/sup>. If the initial concentration is 1.0 M.<\/p>\n

                                                          \n
                                                        1. What is the initial rate?<\/li>\n
                                                        2. What \u00a1s the reaction rate after 1 minute?<\/li>\n<\/ol>\n

                                                          Answer:<\/p>\n

                                                          1. Initial rate of a first order reaction
                                                          \n= k C
                                                          \n= l x 10-2<\/sup> x 1.0
                                                          \n= l x 10-2<\/sup> mol L-1<\/sup> s-1<\/sup><\/p>\n

                                                          2. Concentration after 60 seconds is calculated by applying the first order kinetic equation,
                                                          \n\"Samacheer
                                                          \nRate of reaction after 1 minute
                                                          \n= k x C
                                                          \n= l x 10-2<\/sup> x 0.5489
                                                          \n5.489 x 10-3<\/sup> mol L-1<\/sup> s-1<\/sup><\/p>\n

                                                          Question 9.
                                                          \nA first order reaction is 50% completed in 30 minutes at 27\u00b0C and in 10 minutes at 47\u00b0C. Calculate the reaction rate constant at 27\u00b0C and the energy of activation of the reaction in kJ mol-1<\/sup>
                                                          \nAnswer:
                                                          \nFor a first order reaction
                                                          \n\"Samacheer<\/p>\n

                                                          Now applying the following equation:
                                                          \n\"Samacheer<\/p>\n

                                                          Question 10.
                                                          \nIn Arrhenius equation for a certain reaction, the values of A and E (activation energy) are 4 x 1013<\/sup> sec-1<\/sup> and 98.6 KJ mol-1<\/sup> respectively. If the reaction is of first order, at what temperature will its half life period be 10 minutes?
                                                          \nAnswer:
                                                          \nAccording to the Arrhenius equation.
                                                          \n\"Samacheer
                                                          \nFor a first order reaction
                                                          \nt1\/2<\/sub> = \\(\\frac { 0.693 }{ k }\\) \\(\\frac { 0.693 }{ 600 }\\)
                                                          \nSo,
                                                          \nk = \\(\\frac { 0.693 }{ 600 }\\) sec-1<\/sup> (t1\/2<\/sub> = 10 min = 600 sec)
                                                          \n= 1.1 x 10-3<\/sup> sec-1<\/sup>
                                                          \nHence, log (1.1 x 10-3<\/sup>)
                                                          \n\"Samacheer<\/p>\n

                                                          Common Errors<\/p>\n

                                                          Common Errors:<\/p>\n

                                                            \n
                                                          1. Order and molecularity may get confused<\/li>\n
                                                          2. Unit of first order Rate constant and zero order may get con fused.<\/li>\n
                                                          3. t1\/2<\/sub> – Half liefe period may be difficult to remember<\/li>\n<\/ol>\n

                                                            Rectifications:<\/p>\n

                                                              \n
                                                            1. Order and molecularity are same for the single step process. But for reactions of more than one step, they may be different.<\/li>\n
                                                            2. First order sec-1<\/sup>, Zero order – mol litre-1<\/sup> sec-1<\/sup><\/li>\n
                                                            3. t1\/2<\/sub> = 0.693 \/ k1<\/sub> for first order reaction.<\/li>\n<\/ol>\n

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