Page 1 :
CHEMICAL KINETICS- Previous HSE Questions And Answers, 1. An archeological substance contained wood had only 66.66% of the 14C found in a tree. Calculate the, age of the sample if the half life of 14C is 5730 years., (3), [ March 2008], Ans: We know that radioactive decay follows first order kinetics., For a first order reaction, k= 2.303 log [R]0, t, [R], Here t½ = 5730 years, [R]0 = 100 and [R] = 66.66, k = 0.693/t½ = 0.693/5730 = 1.21 x 10-4, Age of the sample, t = 2.303 log [R]0 = (2.303/1.21 x 10-4)log(100/66.66) = 3352.38 years, k, [R], 2. Unit of rate constant (k) of a reaction depends on the order of the reaction. If concentration is, expressed in mol L-1 and time in seconds (s), find the unit of k for zero, first and second order reaction., (3), [March 2009], Ans:, Reaction, Zero order reaction, First order reaction, Second order reaction, , Unit of rate, constant, Mol L-1s-1, s-1, Mol-1L s-1, , 3. The order of a reaction can be zero and even a fraction but Molecularity cannot be zero or a noninteger., i), What do you mean by the order of a reaction?, (1), ii), What is Molecularity of a reaction? (1), iii), The conversion of molecules A to B follows second order kinetics. If concentration of A is, increased to three times, how will it affect the rate of formation of B?, (2), [March 2010], Ans: i) Order of a reaction is the sum of the powers of the concentration terms of the reactants in the, rate law., ii) Molecularity of a reaction is the total number of reacting species collides simultaneously in a chemical, reaction., , iii) Let the initial concentration of A be x. Then the rate law for this reaction is r = k[x]2, When the concentration of A is increased to three times, the final concentration becomes 3x., Now the rate law is r1 = k[3x]2 = 9.k[x]2, So r1 = 9 x r, i.e. the rate formation of B is increased by 9 times., 4. The value of rate constant k of a reaction depends on temperature. From the values of k at two, different temperatures, the Arrhenius parameters Ea and A can be calculated., The rate constants of a reaction at 1000K and 1060K are 0.01M-1S-1 and 0.10M-1S-1 respectively. Find, the values of Ea and A., (3), [March 2010], Ans: We know that,, Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 1

Page 2 :
log k2/k1 = Ea [T2 - T1], 2.303 R T1.T2, -1 -1, Here T1 = 1000 K, k1 = 0.01M S , T2 = 1060 K, k2 = 0.1 M-1 s-1 and R = 8.314 J K-1 mol-1, log (0.1), =, Ea, [1060 – 1000], (0.01), 2.303 x 8.314 1000 x 1060, Ea = log (10) x 2.303 x 8.314 x 1000 x 1060 = 338266 J mol-1 = 338.266 kJ mol-1, 60, Also, from Arrhenius equation, k = A.e-Ea/RT, A = k/ e-Ea/RT, = 0.01/e –(338266/8.314 x 1000), OR,, From logarithmic form of Arrhenius equation, log k = log A – Ea/2.303RT,, log A = log k + Ea/2.303RT, = log (0.01) + 338266/(2.303x8.314x1000) = 15.67, A = Anti-log (15.67), = 4.67 x 1015, 5. The hydrolysis of an ester in acidic medium is a first order reaction., a) What do you mean by a first order reaction? (½ ), b) What is the relation between Rate constant and half life period of a first order reaction? (½ ), c) Half life period of a first order reaction is 20 seconds. How much time will it take to complete 90%, of the reaction? (3), [March 2011], Ans: a) Order of the reaction = 1. OR, it is a reaction in which rate of the reaction is directly, proportional to the concentration of the reactant. i.e. r = k[R]., b) Rate constant, k = 0.693/t½, c) Here t½ = 20 s., So k = 0.693/20 = 0.03465 s-1, For a first order reaction, k= 2.303 log [R]0, t, [R], Let [R]0 = 100. Then [R] = 100 – 90 = 10, So t = 2.303 log [R]0, k, [R], = (2.303/0.03465)x log(100/10), = 66.46 s, 6. The value of rate constant k of a reaction depends on temperature. From the values of k at two, different temperatures, the Arrhenius parameters Ea and A can be calculated., a) The rate constants of a reaction at 600K and 900K are 0.02s -1 and 0.06s-1 respectively. Find the, values of Ea and A., (3), b) Write the unit of rate constant of a 2nd order reaction if concentration is in mol L-1 and time in s., (1), [SAY 2011], Ans: a) We know that, log k2/k1 = Ea [T2 - T1], 2.303 R T1.T2, Here T1 = 600 K, k1 = 0.02 s-1, T2 = 900 K, k2 = 0.06 s-1 and R = 8.314 J K-1 mol-1, Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 2

Page 3 :
log (0.06) =, Ea, [900 – 600], (0.02), 2.303 x 8.314 900 x 600, Ea = 0.4771 x 2.303 x 8.314 x 900 x 600 = 16443 J mol-1 = 16.443 kJ mol-1, 300, From logarithmic form of Arrhenius equation, log k = log A – Ea/2.303RT,, log A = log k + Ea/2.303RT, = log (0.02) + 16443/(2.303x8.314x600) = -0.267, A = Anti-log (-0.267), = 0.54, b) Mol-1L s-1, 7. Rate of a reaction is the change in concentration of any one of the reactants or any one of the, products in unit time., i) Express the rate of the following reaction in terms of reactants and products: 2HI, H2 + I2 (1½), 2, ii) If the rate expression for the above reaction is rate = k[HI] , what is the order of the reaction ? (½ ), iii) Define order of a reaction. (1), iv) Whether the Molecularity and order of the above reaction are the same? Give reason (1), [March 2012], Ans: i) rinst = – ½ d[HI] = d[H2] = d[I2], dt, dt, dt, OR rav. = – ½ ∆[HI] = ∆[H2] = ∆[I2], ∆t, ∆t, ∆t, ii), Order = 2, iii), It is the sum of the powers of the concentration terms of the reactants in the rate law., iv), Yes. Here the power of the concn. term in the rate law = 2, so order =2 . The no. of reactant, species = 2. So the molecularity = 2., 8. For a first order reaction half life period is independent of initial concentration of its reacting species., i), What is mean by half life period of a reaction?, (1), ii), By deriving the equation for t½ of first order reaction, prove that t½ is independent initial, concentration of reacting species. (3), [SAY 2012], Ans: i) It is the time taken for half of a reaction to complete., ii), For a first order reaction, k = 2.303 log[R]0, t, [R], When t = t1/2, [R] = [R]0/2, Substitute these values in the above equation, k = 2.303 log [R]0, t1/2, [R]0/2, Or, t1/2 = 2.303 log2 = 2.303 x 0.3010, k, k, Or,, t1/2 = 0.693, k, Thus for a first order reaction, half-life period is independent of initial concentration of the reacting species., 9. a) Zero order reaction means that the rate of a reaction is independent of the concentration of the, reactants., Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 3

Page 4 :
i), Write an example for a zero order reaction., (1), ii), Write the integral rate expression for the zero order reaction, R, P. (1), b) The temperature dependence of rate of a chemical reaction can be accurately explained by, Arrhenius equation. With the help of Arrhenius equation, calculate the rate constant for the first, order reaction C2H5I, C2H4 + HI at 700K. Energy of activation (Ea) for the reaction is 209, kJ/mol and rate constant at 600 K is 1.6x10-5 s-1 (R = 8.314 J/K/mol). (2), [March 2013], Ans: a) i) Decomposition of ammonia at the surface of pt metal at high pressure., ii) k= [R]0 - [R], t, b) We know that, log k2/k1 = Ea, [T2 - T1], 2.303 R T1.T2, Here T1 = 600 K, k1 = 1.6 x 10-5s-1, T2 = 700 K, Ea = 209kJ/mol = 209 x 103 J/mol and R = 8.314 J K-1 mol-1, log, k2, =, 209 x 103, [700 – 600], 1.6 x 10-5, 2.303 x 8.314 700 x 600, = 2.6, i.e log k2 – log(1.6 x 10-5) = 2.6, log k2 = 2.6 + log(1.6 x 10-5), = -2.195, So k2 = Anti log(-2.195) = 6.38 x 10-3 s-1, 10. The conversion of a molecule A to B follows second order kinetics., a) Write the rate equation for the second order reaction., (1), b) If the concentration of A is increased to four times, how will it affect the formation of B., (2), c) Indicate the order and Molecularity of the reaction given below:, C12H22O11 H+ C6H12O6 + C6H12O6, (1), (SAY 2013), 2, Ans: a) r = k[A], b) When the concentration is increased by 4 times, the new concentration of A = 4A, So, r = k[4A]2 = 16k[A]2, So the rate of formation of B is increased by 16 times., c) Order = 1 and molecularity = 2 [It is a pseudo first order reaction]., 11. a) Consider a general reaction aA + bB → cC + dD. The rate expression for the reaction is r = k[A] x[b]y, i), Establish the significance of (a+b) and (x+y) in terms of order and molecularity. (1), ii), Write any two differences between order and molecularity., (2), b) “Reactions with zero order are possible, but zero molecularity is not”. Justify the statement. (1), [March 14], Ans: a) i) (a+b) indicates molecularity and (x+y) indicates order., ii), Order, 1. It is the sum of the powers of the concentration terms in, the rate law expression, 2. It is an experimental quantity, 3. It can be zero or fractional, , Molecularity, It is the total number of reactant species, collide simultaneously in a chemical reaction, It is a theoretical quantity, It cannot be zero or fractional, , Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 4

Page 5 :
b) Zero order reaction means the rate of the reaction is independent of the concentration of the, reactants. So it is possible. But zero molecularity means there is no reactants. This is not possible., 12. a) Unit of rate constant (k) of a reaction depends on the order of the reactions. Values of ‘k’ of two, reactions are given below. Find the order of each reaction., i), k = 3 x 10-2 mol L-1 S-1, ii), k = 5 x 10-3 mol-1 L S-1, (1), b) i) Write integrated rate equation for a first order reaction., (1), ii) Write the relation between half life (t½) and rate constant (k) of a first order reaction., (1), -2 -1, iii) Rate constant of a reaction is 5 x 10 S . Find the half life (t½) of the reaction. (1) [SAY 2014], Ans: a) i) Zero order ii) Second order, b) i) For first order reaction, the integrated rate equation is:, k= 2.303 log [R]0, t, [R], ii), k = 0.693/t½, iii), Here k = 5 x 10-2 s-1., So t½ = 0.693/(5x10-2) = 13.86 s, 13. The terms order and molecularity are common in chemical kinetics., a) What do you mean by order and molecularity? (2), b) i) Write two factors influencing rate of a reaction., (1), ii) Write Arrhenius equation. (1), [March 2015], Ans: a) Refer the answer of the Question no. 3, b) i) Concentration of the reactants and temperature., ii) Arrhenius equation is k = A.e-Ea/RT, Or, log k = log A - Ea/2.303RT, 14. Integrated rate expression for rate constant of a first order reaction R → P is given by k = 2.303 log[R]0, t, [R], i) Derive an expression for half life period of first order reaction. (2), ii) A first order reaction has a rate constant 1.15 x 10-3 s-1. How long will 5 g of the reactant take to, reduce 3g? [SAY 2015], Ans: i) Refer the answer of the Question no. 8 (ii), ii) For a first order reaction, k= 2.303 log [R]0, t, [R], -3 -1, Here k = 1.15 x 10 s , [R]0 = 5g and [R] = 3g, So t = 2.303 log [R]0, k, [R], = (2.303/1.15 x 10-3) x log(5/3), = 440.5 s, 15. (i) The molecularity of the reaction 2NO + O2 → 2NO, is: a) 5 b) 2 c) 3, d) 0, (1), (ii) What do you mean by rate of a reaction ?, (1), (iii) What will be the effect of temperature on rate of a reaction ?, (1), Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 5

Page 6 :
(iv) A first order reaction is found to have a rate constant, k = 5.5 x 10 -14 s-1. Find out the half-life of, the reaction. (1), [March 2016], Ans: i) 3, ii), It is the change in concentration of any one of the reactant or product in unit time., iii), When temperature increases, rate of the reaction also increases., iv), Here k = 5.5 x 10-14 s-1., For a first order reaction, t½ = 0.693/k = 0.693/(5.5 x 10-14) = 1.26 x 1013 s, 16. Rate of a reaction is the change in concentration of any one of the reactants or products in unit time., a) Express the rate of the following reaction in terms of reactants and products, 2NO(g) + O2(g), → 2NO2(g), (1), b) (i) N2O5(g) → 2NO2(g) + ½ O2(g) is a first order reaction. Find the unit of k., (1), (ii) Calculate the time required for the completion of 90% of a first order reaction. (k = 0.2303 s -1), (2) [SAY 2016], Ans: a) r = – ½ d[NO] = – d[O2] = ½ d[NO2], dt, dt, dt, OR, r = – ½ ∆[NO] = – ∆[O2]= ½ ∆[NO2], ∆t, ∆t, ∆t, -1, b) (i) For a first order reaction, unit of k = s, (ii) For a first order reaction, k= 2.303 log [R]0, t, [R], For 90% completion, [R]0 = 100 and [R] = 100 – 90 = 10, So t = 2.303 log [R]0, k, [R], = (2.303/0.2303)x log(100/10), = 10 s, 17. a) Plot a graph showing variation in the concentration of reactants against time for a zero order, reaction. (1), b) What do you mean by zero order reaction?, (1), c) The initial concentration of the first order reaction, N 2O5(g), 2 NO2(g) + ½ O2(g), was 1.24 x 10-2, mol L-1 at 300 K. The concentration of N2O5 after 1 hour was 0.20 x 10-2 mol L-1. Calculate the rate, constant of the reaction at 300 K. (2), [March 2017], Ans: (a), , Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 6

Page 7 :
(b) If the order of a reaction is zero, it is called zero order reaction. Or, these are reactions in which the, rate of reaction is independent of concentration of the reactants. Or, r =k[R] 0, (c) For a first order reaction, k= 2.303 log [R]0, t, [R], Here [R]0 = 1.24 x 10-2 mol/L, [R] = 0.2 x 10-2 mol/L, t = 1 hr = 60 min., K = 2.303 log 1.24 x 10-2, = 0.0304 min-1, -2, 60, 0.2 x 10, 18. The effect of temperature on rate of reaction is given by Arrhenius equation., i) Write Arrhenius equation. (1), ii) Define activation energy (Ea), (1), iii) Rate constant k2 of a reaction at 310K is two times of its rate constant k1 at 300 K. Calculate, activation energy of the reaction. (log 2 = 0.3010 and log 1 = 0) [SAY 2017], Ans: i) Arrhenius equation is k = A.e-Ea/RT, ii) It is the minimum amount of kinetic energy required for effective collision during a reaction., iii) We know that, log k2/k1 = Ea [T2 - T1], 2.303 R T1.T2, Here T1 = 300 K, k1 = x, T2 = 310 K, k2 = 2x and R = 8.314 J K-1 mol-1, log 2x, =, Ea, [310 – 300], x, 2.303 x 8.314 300 x 310, So Ea = (2.303x 8.314 x 300 x 310 x log 2)/10 = 53598 J/mol, 19. Identify the order of reaction if the unit of rate constant is mol L-l s-l., (1), Ans: Zero order, 20. For hydrolysis of methyl acetate in aqueous solution, the following results were observed., t/s, 0, 30, 60, CH3COOH, 0.60, 0.30, 0.15, C/mol L-1, Show that it follows pseudo first order reaction as the concentration of water remains constant. (3), [March 2018], Ans: Here the concentration of water remains constant. So for being pseudo first order, the reaction, should be first order w.r.t the concn. of ester (i.e. methyl acetate). The rate constant for pseudo first, order reaction is:, k= 2.303 log [R]0, t, [R], where k = k’[H2O], Here [R]0 = 0.6 mol/L., When t = 30 s, [R] = 0.3 mol/L, So, k1 = 2.303 log (0.6/0.3) = 0.0231 s-1, 30, When t = 60 s, [R] = 0.15 mol/L, So, k2 = 2.303 log(0.6/0.15) = 0.0231 s-1, 60, Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 7

Page 8 :
Since k1 = k2, it is a pseudo first order reaction., 21. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the, energy of activation of the reaction assuming that it does not change with temperature., (3), Ans: We know that, log k2/k1 = Ea [T2 - T1], 2.303 R T1.T2, Here T1 = 293 K, k1 = x, T2 = 313 K, k2 = 4x and R = 8.314 J K-1 mol-1, log 4x, =, Ea, [313 – 293], x, 2.303 x 8.314 293 x 313, So, Ea = (2.303 x 8.314 x 293 x 313 x log 4)/20 = 52854 J/mol, 22. What is the order of a reaction, if its half life is independent of initial concentration? (1) [SAY 2018], Ans: First order., 23. For the reaction, 2NO(g) + O2(g), 2NO2(g), the rate law is given as, Rate = k[NO]2 [O2]. The order of, the reaction with respect to O2 is …………. (1), Ans: With respect to O2, the order of the reaction = 1, 24. Examine the graph given below. Identify the integrated rate equation and the order of the reaction, corresponding to it., , (2), Ans: It’s a zero order reaction., Its integrated rate equation is k = [R]0 – [R]/t., 25. The rate constant of a reaction at 293K is 1.7 x 105 s-1. When the temperature is increased by 20 K, the, rate constant is increased to 2.57 x 106 s-1. Calculate Ea and A of the reaction., (3), [March 2019], Ans: We know that,, , log k2/k1 = Ea [T2 - T1], 2.303 R T1.T2, 5 -1, Here T1 = 293K, k1 = 1.7 x 10 s , T2 = 293 + 20 =313 K, k2 = 2.57 x 106 s-1 and R = 8.314 J K-1 mol-1, log (2.57 x 106), =, Ea, [313 – 293], (1.7 x 105), 2.303 x 8.314 313 x 293, Ea = 1.179 x 2.303 x 8.314 x 313 x 293 = 103514 J mol-1 = 103.514 kJ mol-1, 20, Also, from Arrhenius equation, k = A.e-Ea/RT, A = k/ e-Ea/RT, = 1.7 x 105/e –(103514/8.314x293), OR,, From logarithmic form of Arrhenius equation, log k = log A – Ea/2.303RT,, log A = log k + Ea/2.303RT, = log (1.7 x 105) + 103514/(2.303x8.314x293) = 23.68, Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 8

Page 9 :
A = Anti-log (23.68), = 4.8 x 1023, 26. Differentiate molecularity and order of a reaction. (2), Ans: Refer the answer of the Question no. 11 (a), 27. Write the Arrhenius equation and identify the terms in it. (2), [SAY 2019], -Ea/RT, Ans: Arrhenius equation is k = A.e, Where k – Rate constant of the reaction, A – Arrhenius factor, Ea – Energy of activation, R – Universal, gas constant and T – absolute temperature., 28. For a reaction A + B, C + D, the rate equation is r = K [A]3/2[B]1/2. Give the overall order and, molecularity of the reaction. (2), Ans: Overall Order = 2, Molecularity = 2, 29. The temperature dependence of the rate of a chemical reaction can be explained by Arrhenius, equation., a) Give Arrhenius equation., b) The rate of a chemical reaction doubles for an increase of 10K in absolute temperature from 300K., Calculate the activation energy (Ea)? [R = 8.314 J/K/mol and log 2 = 0.3010] (3), [March 2020], Ans: a) The Arrhenius equation is k = A.e-Ea/RT, Or, logk = logA – Ea/2.303RT, b) We know that,, , log k2/k1 =, Ea, [T2 - T1], 2.303 R T1.T2, Here T1 = 300K, T2 = 300 + 10 =310 K, and R = 8.314 J K-1 mol-1, Suppose k1 = x, then k2 = 2x, Then,, log 2x, =, Ea, [310 – 300], x, 2.303 x 8.314 300 x 310, Ea = 0.3010 x 2.303 x 8.314 x 300 x 310 = 53598.6 J mol-1 = 53.6 kJ mol-1, 10, ----------------------------------------------------------------------------------------------------------------------, , Chemical Kinetics – Prepared by ANIL KUMAR K L, HSST CHEMISTRY GHSS ASHTAMUDI, , Page 9