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>\nTamilnadu 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
\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>\n12th 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
\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>\nChemical 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
\nAnswer:
\n
\nSolution:
\n
\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>\nSamacheer 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>\nSamacheer 12 Chemistry Solutions Question 5.<\/strong>
\nFor the reaction, 2NH3<\/sub> \u2192 N2<\/sub> + 3H2<\/sub>, if
\n
\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<\/p>\nSamacheer 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>\nChemical 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>\nSamacheer 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>\nChemical 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>\nCorrect 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
\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>\nSamacheer 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<\/p>\n12th 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<\/p>\nQuestion 12.
\n
\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<\/p>\nQuestion 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>\nQuestion 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>\nQuestion 15.
\nPredict the rate law of the following reaction based on the data given below:
\n2A + B \u2192 C + 3D
\n
\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
\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
\n\\(\\frac { 8 }{ 2 } \\) = 2m<\/sup>
\n\u2234m = 1
\n\u2234 rate = k [A]1<\/sup> [B]2<\/sup><\/p>\nQuestion 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>\nQuestion 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<\/p>\nQuestion 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 =
\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>\nQuestion 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<\/p>\nQuestion 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
\nAnswer:
\n
\nSolution:
\n
\n<\/p>\nQuestion 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<\/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<\/p>\nQuestion 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<\/p>\nII . 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>\n2. 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>\nQuestion 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<\/p>\nQuestion 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
\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- It is the sum of the powers of concentration terms involved in the experimentally determined rate law.<\/li>\n
- It can be zero (or) fractional (or) integer.<\/li>\n
- It is assigned for a overall reaction.<\/li>\n<\/ol>\n
Molecularity of a reaction:<\/p>\n
\n- It is the total number of reactant species that are involved in an elementary step.<\/li>\n
- It is always a whole number, cannot be zero or a fractional number.<\/li>\n
- 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
\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 + NO