NCERT Solutions Class 12 Chemistry Chapter 3 Chemical Kinetics PDF

NCERT Solution for Class 12 Chemistry Chapter 3 - Chemical Kinetics Intext Exercise 1. For the reaction R 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 un...

NCERT Solution for Class 12 Chemistry Chapter 3 - Chemical Kinetics Intext Exercise 1. For the reaction R 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. Ans: The average rate of the reaction can be calculated by dividing the change in the rate of decreasing the rate of reactant by the time taken. This is given below: d[R] Average rate = - dt This can be written as: [R]2 -[R]1 Average rate = - t 2 -t1 R 1 = 0.03 M R 2 = 0.02 M t 2 - t1 = 25 min Putting the values, we get: 0.02  0.03 Average rate = - 25 0.01 Average rate = -  4 × 104 M min 1 25 So, the average rate in minutes will be 4 × 104 M min 1 Now, to find the average rate in seconds we have to divide the above answer by 60. So, the answer will be: 4 × 104 Average rate = -  6. 66 × 106 M s 1 60 Therefore, the average rate in seconds will be 6.66 × 106 M s1 Class XII Chemistry www.vedantu.com 1 2. In a reaction, 2A Products, the concentration of A decreases from 0.5 mol L-1 to 0.4 mol L-1 in 10 minutes. Calculate the rate during this interval. Ans: The average rate of the reaction can be calculated by dividing the change in the rate of decreasing the rate of reactant by the time taken. This is given below: 1 Δ[A] Average rate = - 2 Δt This is due the fact that the reaction given is: 2A Products So, the average rate will be written as: 1 [A]2 -[A]1 Average rate = - 2 t 2 -t1 A1 = 0.5 M A 2 = 0.4 M t 2 - t1 = 10 min Putting the values, we get: 1 0.4  0.5 Average rate = - × 2 10 1 0.1 Average rate = - ×  5 × 103 M min 1 2 10 So, the average rate will be 5 × 103 M min1 3. For a reaction, A + B Product , the rate law is given by: r = k [A]1/2 [B]2. What is the order of the reaction? Ans: The order of the reaction can be calculated by adding the stoichiometry coefficients of the reactants in the given rate of the reaction. Given the rate is r = k [A]1/2 [B]2 So, the order will be: 1 Order = 2 + = 2.5 2 So, the order of the reaction is 2.5 Class XII Chemistry www.vedantu.com 2 4. The conversion of the molecules X to Y follows second order kinetics. If concentration of x is increased to three times how will it affect the rate of formation of Y? Ans: The reaction will be: X Y As the question says that this reaction follows the second order reaction, we can write the rate law equation as: Rate = k[X]2 = ka 2 If [X] = a mol/ L It is said that the concentration of X increases by three times, so we can write: [X] = 3a mol/ L Therefore, the rate of reaction will be: Rate = k (3a)2 = 9 ka 2 Thus, the rate of the reaction will increase by 9 times or the rate formation will increase by 9 times. 5. A first order reaction has a rate constant 1.15 × 10-3 s-1. How long will 5 g reactant take to reduce to 3 g? Ans: The initial amount of the reactant is given as 5 g. We can write: [R]0 = 5 g The final amount of the reactant is given as 3 g. We can write: [R] = 3 g We are also given the value of rate constant as: Rate constant = 1.15 × 10-3 s-1 We know that the reaction is a 1st order reaction, the time can be calculated by: 2.303 [R] t= log 0 k [R] Putting the values in the above formula, we get: 2.303 5 t= 3 log 1.15 x 10 3 Class XII Chemistry www.vedantu.com 3  2.303 t= × 0.2219 1.15 × 103 t = 444 seconds So, the time taken will be 444 seconds. 6. Time required to decompose SO 2Cl 2 to half its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. Ans: We are given that the decomposition of SO 2Cl2 is a first order reaction. So, we can write: 0.693 t1/2 = k It is mentioned that the time required for the initial amount to become half is 60 minutes. t1/2 = 60 min 0.693 Therefore, k = t1/2 Putting the value, we get: 0.693 k= = 1.925 × 10-4 s-1 60 × 60 Thus, the rate constant is 1.925 × 10-4 s-1. 7. What will be the effect of temperature on rate constant? Ans: An increase of 10 degrees in temperature causes a reaction's rate constant to almost double in size. In any case, the Arrhenius equation gives the exact temperature dependency of a chemical reaction rate. The Arrhenius equation is given below: k = A e-E /Rt a Where, A is the Arrhenius factor or the frequency factor, T is the temperature, R is the gas constant, Class XII Chemistry www.vedantu.com 4 E a is the activation energy. 8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E a. Ans: The formula that can be used to solve the question is: k E a  T2 -T1  log 1 =   k2 2.303R  T1T2  T1 temperature given is absolute temperature and it is equal to 298 K. T2 = 298 + 10 = 308 K It is given that the rate of the chemical reaction double for an increase of 10 K, therefore, we can write the values of rate constant as: k1 = x k 2 = 2x Also, R = 8.314 J K-1 mol-1 Now, putting all the values in the formula, we get: 2x Ea  10  log = x 2.303 × 8.314  298 × 308  Ea  10  log2 =  2.303 × 8.314  298 × 308  2.303 × 8.314 × 298 × 308 × log 2 Ea = 10 Ea = 52897.78 J mol1 Ea = 52.89 kJ mol1 9. The activation energy for the reaction 2HI (g) H 2 +I 2 is 209.5 kJ mol-1 at (g) (g) 581 K. Calculate the fraction of molecules of molecules having energy equal to or greater than activation energy. Ans: We are given the activation energy as 209.5 kJ mol-1 T = 581 K Class XII Chemistry www.vedantu.com 5 R = 8.314 J K-1 mol-1 Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as: x = eE /RT a -E ln x = a RT -E a log x = 2.303 RT 209500 log x =  18.8323 2.303 × 8.314 × 581 Now, taking the antilog: x = Antilog (18.8323) x = 1.471 × 10-19 NCERT Exercise 1. From the rate expression for the following reactions, determine their order of reaction and the dimension of the rate constants. (i) 3 NO(g) N2O(g) Rate = k[NO]2 Ans: We are given: Rate = k[NO]2 From this we can see that the order of the reaction = 2 rate k= [No]2 Dimensions will be: mol L-1 s-1 k= (mol L-1 ) 2 mol L-1 s -1 = mol 2 L2 = L mol-1 s-1 Class XII Chemistry www.vedantu.com 6 (ii) H2O2 (aq)+3I- (aq)+2H+ 2H2O(l)+I -3 Rate=k[H2O2 ][I - ] Ans: Rate = k[H2O2 ][I ] From this we can see that the order of the reaction = 2 rate k= [H 2O 2 ][I  ] Dimensions will be: mol L-1 s-1 k= (mol L-1 ) 2 mol L-1 s -1 = mol 2 L2 = L mol-1 s-1 (iii) CH3CHO(g) CH4 (g)+ CO(g) Rate = k[CH3CHO]3/2 Ans: Rate = k[CH3 CHO]3/2 3 From this we can see that the order of the reaction = 2 rate k= 3 [CH 3 CHO]2 Dimensions will be: mol L-1 s-1 k= (mol L-1 )3/2 mol L-1 s -1 = 3 3 mol L 2 2 1 1 - = L2 mol 2 s -1 (iv) C2 H 5Cl(g) C2 H 4 (g)+ HCl(g) Rate = k[C2 H 5Cl] Ans: Rate = k[C2 H 5Cl] From this we can see that the order of the reaction = 1 Class XII Chemistry www.vedantu.com 7  rate k= [C2 H 5Cl] Dimensions will be: mol L-1 s-1 k= (mol L-1 ) = s-1 2. For a reaction: 2A+B A 2B The rate = k[A][B]2 with k = 2.0 x 10-6 mol -2 L2 s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1 , [B] = 0.2 mol L-1 Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1. Ans: We are given the rate of the reaction as: rate = k[A][B]2 Putting the values in this, we get the rate as: rate = 2.0 × 10-6 × 0.1 × (0.2)2 rate = 8.0 × 10-9 mol L1 s1 When [A] is reduced from 0.10 mol L1 to 0.06 mol L1. So, the amount of [A] reacted will be:  0.10  0.06  0.04 mol L1 Therefore, the concentration of B reacted will be: 1  × 0.04 = 0.02 mol L1 2 Hence, new [B] = 0.2 – 0.02 = 0.18 mol L1 Now, the new rate of the reaction will be: rate = 2.0 × 10-6 × 0.06 × (0.18)2 rate = 3.89 × 10-9 mol L1 s1 Therefore, the rate of the reaction is 3.89 × 10-9 mol L1 s1. Class XII Chemistry www.vedantu.com 8 3. The decomposition of NH 3 on the platinum surface there is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10-4 mol -1 L s-1 ? Ans: The following equation represents the breakdown of Ammonia on the platinum surface. 2NH3  (g) Pt N2 +3H2 (g) (g) Therefore, we can write the rate of the reaction as: 1 d[NH3 ] d[N 2 ] 1 d[H 2 ] Rate = - = = 2 dt dt 3 dt But we are given that the reaction is a zero order reaction. 1 d[NH3 ] d[N 2 ] 1 d[H 2 ] So, Rate = - = = = 2.5 × 104 mol L1 s 1 2 dt dt 3 dt Thus, the rate of production of N 2 will be: d[N 2 ] = 2.5 × 104 mol L1 s 1 dt And the rate of production of H 2 will be: d[H 2 ] = 3 × 2.5 × 104 mol L1 s 1  7.5 × 104 mol L1 s 1 dt 4. The decomposition of dimethyl ether leads to the formation of CH 4 , H 2 and CO and the reaction rate is given by Rate = k [CH3OCH3 ]3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate = k (PCH OCH )3/2 3 3 It the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants? Ans: If the pressure is measured in bar and time in minutes, then unit of rate will be: = bar min-1 Class XII Chemistry www.vedantu.com 9 We are given the rate of the reaction as: Rate = k [CH3OCH3 ]3/2 Rate Therefore, k = [CH3OCH3 ]3/2 So, we can write the units of rate constant as: bar min -1 k= 3/2 = bar -1/2 min -1 bar So, the units are bar -1/2 min-1. 5. Mention the factors that affect the rate of a chemical reaction. Ans: Factors that influence a reaction's speed include. (i) Reactant nature: The rate of reaction is affected by the kind of reactant. For example, ionic compound reactions are quicker than covalent compound reactions. (ii) The state of the reactants: Solid reactions are sluggish, liquid reactions are rapid, and gas reactions are very quick. (iii)Temperature: In addition, temperature has a significant impact on response rate. Every 10 degrees Celsius increase in temperature doubles the pace of reaction. (iv) Presence of catalyst: The rate of reaction is also affected by the presence of a catalyst in the reaction. It enhances the pace of reaction by increasing reaction surface area, by generating unstable intermediates with the substrate, and by offering a lower activation energy alternative path. 6. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is: (i) Doubled Ans: Let us assume that the concentration of the reactant be [A] = a Rate of the reaction will be: R = k[A]2 = ka 2 It is said that the concentration of A increases by two times, so we can write: [A] = 2a mol/ L Therefore, the rate of reaction will be: Rate = k (2a)2 = 4 ka 2 Class XII Chemistry www.vedantu.com 10 Thus, the rate of the reaction will increase by 4 times. (ii) Reduced to half It is said that the concentration of A reduced to half, so we can write: 1 [X] = a mol/ L 2 Therefore, the rate of reaction will be: 2 1  1 Rate = k  a  = ka 2 2  4 Thus, the rate of the reaction will reduce by ¼ times. 7. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively? Ans: An increase of 10 degrees in temperature causes a reaction's rate constant to almost double in size. In any case, the Arrhenius equation gives the exact temperature dependency of a chemical reaction rate. The Arrhenius equation is given below: k = A e-E /Rt a Where, A is the Arrhenius factor or the frequency factor, T is the temperature, R is the gas constant, E a is the activation energy. The formula can also be written as: k E a  T2 -T1  log 2 =   k1 2.303R  T1T2  Where k 2 is the rate constant at temperature T2 k 1 is the rate constant at temperature T1 8. Ina pseudo first order hydrolysis of ester in water, the following results were obtained Class XII Chemistry www.vedantu.com 11  t/s 0 30 60 90 [Ester]mol L-1 0.55 0.31 0.17 0.085 (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. Ans: Between the time interval of 30 to 60 sec the average rate id reaction will be calculated as: d[Ester] = dt Putting the values from the data given in the question, we can write: 0.31  0.17 0.14 =  60  30 30 -3 = 4.67 × 10 mol L-1 s-1 (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. Ans: The rate law formula for pseudo first order reaction will be: 2.303 [R] k= log 0 t [R] Now, when t = 30 s, then we rate constant will be: 2.303 0.55 k1 = log  1.91 × 102 s 1 30 0.31 Now, when t = 60 s, then we rate constant will be: 2.303 0.55 k2 = log  1.96 × 102 s 1 60 0.17 Now, when t = 90 s, then we rate constant will be: 2.303 0.55 k3 = log  2.075 × 102 s 1 90 0.085 So, we can calculate the average rate constant as: Class XII Chemistry www.vedantu.com 12  k1 + k 2 + k 3 k= 3 =1.98 × 10-2 s-1 9. A reaction is first order in A and second order in B. (i) Write the differential rate equation Ans: The differential rate equation can be written as: d[R] - = k[A][B]2 dt (ii) How is the rate affected on increasing the concentration of B three times? Ans: The concentration of B is increased by 3 times, then B =3B Therefore, the rate will be: d[R] - = k[A][3B]2  9  [A][B]2 dt Therefore, the rate will increase by 9 times. (iii) How is the rate affected when the concentration of both A and B are doubled? Ans: The concentration of A is doubled, then A = 2A The concentration of B is doubled, then B = 2B Therefore, the rate will be: d[R] - = k[2A][2B]2  8  [A][B]2 dt Therefore, the rate will increase by 8 times. 10. In a reaction between A and B the initial rate of reaction (r0 ) was measured for different concentrations of A and B as given below: A/mol L-1 0.20 0.20 0.04 B/mol L-1 0.30 0.10 0.05 r0 /mol L-1 s-1 5.07 × 10-5 5.07 × 10-5 1.43 × 10-4 Class XII Chemistry www.vedantu.com 13  What is the order of the reaction with respect to A and B? Ans: Let us assume that the order of the reaction with respect to A be x and with respect to B be y. Therefore, we can write: r0 = k [A]x [B]y 5.07 × 10-5 = k [0.20]x [0.30]y.........(i) 5.07 × 10-5 = k [0.20]x [0.10]y.........(ii) 1.43 × 10-5 = k [0.40]x [0.05]y.........(iii) Let us divide (i) by (ii), we get: 5.07 × 10-5 k [0.20]x [0.30]y = 5.07 × 10-5 k [0.20]x [0.10]y [0.30]y =1 = [0.10]y x y  0.30   0.30  =  =   0.10   0.10  Therefore, y = 0 Now, dividing (iii) by (i), we get: 1.43 × 10-4 k [0.40]x [0.05]y = 5.07 × 10-5 k [0.20]x [0.30]y 1.43 × 10-4 k [0.40]x = 5.07 × 10-5 k [0.20]x Because y = 0 = 2.821 = 2x Now, taking log on both the sides. log 2.821 = x log 2 log 2.821 x= log 2 x = 1.496 x = 1.5 Thus, the order of the reaction according to A is 1.5 and according to B is 0. Class XII Chemistry www.vedantu.com 14 11. The following results have been obtained during the kinetic studies of the reaction. 2A + B C + D Experiment A/mol L-1 B/mol L-1 Initial rate of formation of D/mol L-1min-1 I 0.1 0.1 6.0 × 10-3 II 0.3 0.2 7.3 × 10-2 III 0.3 0.4 2.88 × 10-1 IV 0.4 0.1 2.40 × 10-2 Determine the rate law and the rate constant for the reaction. Ans: Let us assume that the order of the reaction with respect to A be x and with respect to B be y. Therefore, we can write: Rate = k [A]x [B]y 6.0 × 10-3 = k [0.1]x [0.1]y.........(i) 7.2 × 10-2 = k [0.3]x [0.2]y.........(ii) 2.88 × 10-1 = k [0.3]x [0.1]y.........(iii) 2.40 × 10-2 = k [0.4]x [0.1]y.........(iv) Let us divide (iv) by (i), we get: 2.40 × 10-2 k [0.4]x [0.1]y = 6.0 × 10-3 k [0.1]x [0.1]y [0.4]x =4 = [0.1]x x  0.4  =4=    0.1  Therefore, x = 1 Now, dividing (iii) by (ii), we get: 2.88 × 10-1 k [0.3]x [0.4]y = 7.2 × 10-2 k [0.3]x [0.2]y Class XII Chemistry www.vedantu.com 15  y  0.4  =4=    0.2  = 4 = 2y = 22 = 2 y y=2 Thus, the order of the reaction according to A is 1 and according to B is 2. So, the rate law is: Rate = k[A][B]2 Rate k= [A][B]2 Now, putting the values for each experiment, we get: From experiment I: 6.0 × 103 k= [0.1][0.1]2 = 6.0 L2 mol-2 min-1 From experiment II: 7.3 × 102 k= [0.3][0.2]2 = 6.0 L2 mol-2 min-1 From experiment III: 2.88 × 101 k= [0.3][0.4]2 = 6.0 L2 mol-2 min-1 From experiment IV: 2.40 × 102 k= [0.4][0.1]2 = 6.0 L2 mol-2 min-1 Therefore, the rate constant will be k = 6.0 L2 mol-2 min-1 12. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Class XII Chemistry www.vedantu.com 16  Experiment A/mol L-1 B/mol L-1 Initial rate D/mol L-1min-1 I 0.1 0.1 2.0 × 10-2 II --- 0.2 4.0 × 10-2 III 0.4 0.4 ---- IV ---- 0.2 2.0 × 10-2 Ans: We are given that, the reaction between A and B is first order with respect to A and zero order with respect to B. Therefore, the rate of the reaction is given by: Rate = k[A]1[B]0 So, we can write: Rate = k[A] According to the experiment I we can write: 2.0 × 10-2 = k(0.1) k = 0.2 min -1 According to the experiment II we can write: 4.0 × 102  (0.2) [A] [A] = 0.2 mol L-1 According to the experiment III we can write: Rate = 0.2 × 0.4 = 0.08 mol L-1 min-1 According to the experiment III we can write: 2.0 × 102  (0.2) [A] [A] = 0.1 mol L-1 13. Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s-1 Ans: Half-life of the reaction can be related with the rate constant of the reaction as: 0.693 t1/2 = k Putting the value of time, we get: Class XII Chemistry www.vedantu.com 17  0.693 k= = 3.46 × 10-3 s 200 So, the rate of the reaction is 3.46 × 10-3 s (ii) 2 min -1 Ans: Half-life of the reaction can be related with the rate constant of the reaction as: 0.693 t1/2 = k Putting the value of time, we get: 0.693 k= = 0.346 min 2 So, the rate of the reaction is 0.346 min. (iii) 4 years-1 Ans: Half-life of the reaction can be related with the rate constant of the reaction as: 0.693 t1/2 = k Putting the value of time, we get: 0.693 k= = 0.173 years 4 So, the rate of the reaction is 0.173 years. 14. The half-life for radioactive decay of 14 C is 5730 years. An archeological artifact containing wood had only 80% of the 14 C found in a living tree. Estimate the age of the sample. Ans: The given reaction in the above question is radioactive decay and the radioactive decay follows the first order kinetics. Therefore, the decay constant: 0.693 λ= t1/2 We are given the half-life time as 5730 years. 0.693 λ= years 1 5730 Class XII Chemistry www.vedantu.com 18 To find the age of the sample, we can write: 2.303 [R] t= log 0 λ [R] 80% of the wood is found so, the initial amount can be taken as 100 and the final amount as 80. Putting the values, we get: 2.303 100 t= log 0.693 80 5730 t = 1845 years Therefore, the age of the sample is 1845 years. 15. The experiment data for decomposition of N 2O 5 [2N 2O 5 4NO 2 +O 2 ] In gas phase at 318 K are given below: t/s 0 400 800 1200 1600 2000 2400 1800 3200 10-2 × [N2O5 ]/mol L-1 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 (i) Plot [ N 2O 5 ] against t Ans: The graph is given below: Class XII Chemistry www.vedantu.com 19 (ii) Find the half-life period for the reaction Ans: Time corresponding to the concentration, 1.630 × 10-2 mol L-1 =0.815 × 10-2 mol L-1 is the half-life. 2 From the graph, the half-life obtained is 1440 s. (iii) Draw the graph between log[N 2O5 ] and t Ans: T(s) 10-2 × [N2O5 ]/mol L-1 log[N 2O5 ] 0 1.63 -1.79 400 1.36 -1.87 800 1.14 -1.94 1200 0.93 -2.03 1600 0.78 -2.11 2000 0.64 -2.19 2400 0.53 -2.28 2800 0.43 -2.37 3200 0.35 -2.46 Class XII Chemistry www.vedantu.com 20 (iv) What is the rate law? Ans: The rate law of the reaction will be: Rate = k[N 2O5 ] (v) Calculate the rate constant. Ans: From the plot [N 2O 5 ] v/s t, is given by: k =- 2.303 Therefore, we obtain: k 0.67 =-  2.303 3200 = 4.82 × 10-4 mol L-1 s-1 (vi) Calculate the half-life period from k and compare it will (ii) Ans: Half-life is given by: 0.693 0.693 t1/2   s k 4.82 × 104 = 1438 seconds. The value of half-life calculated from the k is very close to that obtained from the graph. 16. The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? Ans: As we know that: 2.303 [R] t= log 0 k [R] The initial value of the reactant has become 1/16th. Now, putting the values, we get: 2.303 16 t= log 60 1 2.303 t= log16 = 4.62 × 10-2 s 60 Class XII Chemistry www.vedantu.com 21 Therefore, the time required will be 4.62 × 102 s. 17. During nuclear explosion, one of the products is 90 Sr with a half-life of 28.1 years. If 1 μg of 90 Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically. Ans: As radioactive disintegration follows first order kinetics. 0.693 0.693 Decay constant of 90 Sr (k) =   2.466 × 102 y 1 t1/2 28.1 To calculate the amount left after 10 years. a = 1 μg t = 10 years k=2.466 × 10-2 y-1 (a-x) = ? 2.303 a k= log t a-x 2.303 1 2.466 × 102 = log 10 a-x log(a-x) = -0.1071 (a-x) = Antilog -0.1071 = 0.7814 μg To calculate the amount left after 60 years. 2.303 a k= log t a-x 2.303 1 2.466 × 102 = log 60 a-x log(a-x) = -0.6425 (a-x) = Antilog -0.6425 = 0.2278 μg 18. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. Class XII Chemistry www.vedantu.com 22 Ans: For first order reaction, we can write: 2.303 a t= log k a-x 99% completion means that x = 99% of a = 0.99 a So, we can write: 2.303 a t 99% = log k a-0.99a 2.303 t 99% = log 102 k 2.303 t 99% = 2 × k 90% completion means that x = 90% of a = 0.90 a 2.303 a t 99% = log k a-0.90a 2.303 t 99% = log 10 k 2.303 t 99% = k Now, we can take the ratio as given below:  2 x 2.303  t 99%  k    t 90%  2.303     k  t 99%  2 × t 90% Hence, the time required for 99% completion of a first order reaction is twice the time required for the completion of 90% of the reaction. 19. A first order reaction takes 40 min for 30% decomposition. Calculate t 1/2. Ans: 30% decomposition means that x = 30% of a = 0.30 a Since, the reaction is of 1st order, we can write: 2.303 a k= log t a-x Class XII Chemistry www.vedantu.com 23 Time is given as 40 min. So, putting the values, we get: 2.303 a k= log 40 a-0.30a 2.303 10 k= log min 1 40 7 2.303 k= × 0.1549 min 1  8.918 × 10 3 min 1 40 Now, we can calculate the half-life period as we have the rate constant value. We can write: 0.693 0.693 t1/2    7.77 min k 8.918 × 103 So, the half-life is 7.77 min. 20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained: T (sec) P (mm of Hg) 0 35.0 360 54.0 720 63.0 Calculate the rate constant. Ans: The decomposition of azoisopropane to hexane and nitrogen at 54.3 k is represent by the following equation. Total pressure after time t, we will be: Pt = (Po -p) + p + p Pt =Po +p p=Pt  Po Now, we can substitute the value of p for the pressure of reactant at time t Class XII Chemistry www.vedantu.com 24 =Po -p =Po -(Pt -Po ) = 2Po - Pt Now, we can apply the rate constant formula of 1st order reaction. 2.303 P k= log t 2Po -Pt When t = 360 s, Putting the values, we get: 2.303 35.0 k= log  2.175 × 103 s 1 360 2 × 35 - 54 When t = 720 s, Putting the values, we get: 2.303 35.0 k= log  2.235 × 103 s 1 720 2 × 35 - 63 Now, we can find the average value: (2.175 × 10-3 )+(2.235 × 10-3 ) -1 k= s 2 k = 2.20 × 10-3 s-1 21. The following data were obtained during the first order thermal decomposition of SO 2Cl 2 at a constant volume. SO 2Cl 2 SO 2 +Cl 2 (g) (g) (g) Experiment Time/ s Total pressure/ atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm. Ans: The given reaction shows the thermal decomposition of SO 2Cl2 at constant volume. Class XII Chemistry www.vedantu.com 25 Total pressure after time t, we will be: Pt = (Po -p) + p + p Pt =Po +p p=Pt  Po Now, we can substitute the value of p for the pressure of reactant at time t =Po -p =Po -(Pt -Po ) = 2Po - Pt Now, we can apply the rate constant formula of 1st order reaction. 2.303 P k= log t 2Po -Pt When the t = 100 s 2.303 0.5 k= log 100 2 × 0.5 - 0.6 -3 -1 k = 2.231 × 10 s When Pt = 0.65 atm Therefore, pressure of SO 2Cl2 at time t total pressure is 0.65 atm, is PSO Cl =2Po -Pt 2 2 = 2 × 0.50 – 0.65 = 0.35 atm Therefore, the rate of equation, when total pressure is 0.65 atm, is given by: Rate = k(PSO Cl ) 2 2 Rate = (2.33 x 10-3 )(0.354) = 7.8 × 10-4atm s-1 22. The rate constant for the decomposition of N 2O 5 at various temperature is Class XII Chemistry www.vedantu.com 26  given below: T/ oc 0 20 40 60 80 5 10 × k/s 0.0787 1.70 25.7 178 2140 Draw a graph between ln K and 1/T and calculate the values of A and E a. Predict the rate constant at 30o and 50o C. Ans: As the data is given we can write: T/ oc 0 20 40 60 80 T/K 273 293 313 333 353 1 -1 3.66 × 10-3 3.41 × 10-3 3.19 × 10-3 3.0 × 10-3 2.83 × 10-3 /k T 105 × k/s 0.0787 4.075 25.7 178 2140 ln K -7.147 -4.075 -1.359 -0.577 3.063 The graph is draw below: Slope of the line, will be given as: y2 -y1 = 12.301 K x 2 -x1 Class XII Chemistry www.vedantu.com 27 According to the Arrhenius equation, E Slope = - a R = E a = -slope × R = (-12.301) (8.314) =102.27 KJ mol-1 Again, E ln k = ln A - a RT E ln A = ln k + a RT As T = 273 K and ln k = -7.147 Applying this in the formula, we get: 102.27 × 103 ln A = -7.147 -  37.911 8.314 × 273 So, A=2.91 × 106 When T = 30 + 273 K = 303 K 1 = 0.0033K = 3.3 × 10-3K T 1 Now, at = 0.0033K = 3.3 × 10-3K T ln k = - 2.8 Therefore, k = 6.08 × 10-2 s-1 When T = 50 + 273 K = 323 K 1 = 0.0031K = 3.1 × 10-3K T 1 Now, at = 0.0031K = 3.1 × 10-3K T ln k = - 0.5 Therefore, k = 0.607 s-1 23. The rate constant for the decomposition of hydrocarbons is 2.418 × 10-5 s-1 Class XII Chemistry www.vedantu.com 28  at 546 K. If the energy of activation is 179.9 kJ /mol, what will be the value of the pre-exponential factor? Ans: We are given some values as: K = 2.418 × 10-5 s-1 T = 546 K Ea = 179.9 kJ mol-1  179.9 × 103 J mol1 We the Arrhenius equation is: k = A e-E /RT a In the log form, this can be written as: E ln k = ln A - a RT Ea log k = log A - 2.303 RT 179.9 × 103 log k = log (2.418 × 10 5 ) - 2.303 × 8.314 × 546 = (0.3835 – 5) + 17.2082 = 12.5917 Therefore, A = antilog (12.5917) A = 3.912 × 1012 s-1 24. Consider a certain reaction A Products with k = 2.0 × 10-2 s-1. Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0 mol L-1. Ans: We are given some values, that are given below: k = 2.0 × 10-2 s-1 t = 100 s [A]o =1.0 mol L-1 As we can see that the units of k is given in s-1 , this means that the reaction is a first order reaction. Therefore, we can write: 2.303 [A] k= log 0 t [A] Putting the values, we get: Class XII Chemistry www.vedantu.com 29  2.303 1.0 2.0 × 102 = log 100 [A] 2.303 2.0 × 102 = (-log[A]) 100 2.0 × 102 × 100 (-log[A]) = 2.303  2.0 × 102 × 100  [A] =antilog    2.303   0.135 mol L1 Therefore, the remaining amount of A is 0.135 mol L1. 25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t 1/2 = 3.00 hours. What fraction of a sample of sucrose remains after 8 hours? Ans: The given reaction is a first order reaction. So, we can write: 2.303 [R] k= log 0 t [R] We are given a half-life of 3 hours. Therefore, we can write: 0.693 k= t1/2 So, putting the values in this, we get: 0.693 k= = 0.231 h -1 3 Now, we can put this value of rate constant in the first order reaction formula. 2.303 [R] 0.231 = log 0 8 [R] [R] 0.231 × 8 log 0 = [R] 2.303 [R] log 0 = 0.8024 [R] Class XII Chemistry www.vedantu.com 30 [R]0 = antilog (0.8024) [R] [R]0 = 6.3445 [R] Or we can write: [R] = 0.158 [R]0 Therefore, the fraction of sample of sucrose that remains after 8 hours is 0.158. 26. The decomposition of hydrocarbon follows the equation k = (4.5 × 1011 s-1 )e-280000K/T. Calculate E a. Ans: According to the Arrhenius equation, k = Ae-E /RT a We are given the equation as: k = (4.5 × 1011 s-1 )e-280000K/T Therefore, the formula can be written as: E 28000 K - a =- RT T This can be written as: E a = 28000 × R Ea = 28000 × 8.314 = 232.79 kJ mol-1 Therefore, the value of E a is 232.79 kJ mol-1 27. The rate constant for the first order decomposition of H 2O 2 is given by the following equation: log k = 14.34 - 1.25 × 104 K/T Calculate E a for this reaction and at what temperature will its half-period be 256 minutes? Ans: According to the Arrhenius equation, k = Ae-E /RT a Class XII Chemistry www.vedantu.com 31 This can be written as: E ln k = ln A - a RT In the log form it can be written as: Ea log k = log A - 2.303 RT We are given: log k = 14.34 - 1.25 × 104 K/T Comparing these two, we get: Ea 1.25 × 104 K = 2.303 RT T Ea = 2.303 R × 1.25 × 104 K Ea = 2.303 × 8.314 × 1.25 × 104 K Ea = 239.34 kJ mol1 We are given half-life time as 256 minutes. 0.693 k= t1/2 0.693 k= = 4.51 × 10-5 s-1 256 × 60 Now, we have the value of rate constant, we can put in the equation: 1.25 × 104 log(4.51 × 10-5 ) = 14.34 - T T = 669 K 28. The decomposition of A into product has value of k as 4.5 × 103 s-1 at 10oC and energy of activation 60 kJ mol -1. At what temperature would k be 1.5 × 104 s-1 ? Ans: We are some information: k1 = 4.5 × 103 T1 = 10 + 273 = 283 K Class XII Chemistry www.vedantu.com 32 k 2 = 1.5 × 104 T2  ? Ea = 60 kJ mol-1 Applying Arrhenius equation: k E a  T2 -T1  log 2 =   k1 2.303R  T1T2  Putting the values, we can write: 1.5 × 104 60  T2 -283  log =   4.5 × 103 2.303 × 8.314  283T2   T -283  log 3.333 = 3133.63  2   283T2  0.5228  T2 -283    3133.63  283T2  0.0472T2 = T2 - 283 T2 = 297 K Or we can write: T2 = 24oC 29. The time required for 10% completion of a first order reaction at 298 k is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010 s-1 , Calculate k at 318 K and E a. Ans: There are two cases in this question. As the reaction given is first order reaction, we can use: 2.303 a k= log t a-x For case 1: 2.303 a k 298K = log t1 a - 0.10a Class XII Chemistry www.vedantu.com 33  2.303 10 k 298K = log t1 9 2.303 k 298K = × 0.0458 t1 0.1055 t1 = k 298K For case 2: 2.303 a k 308K = log t2 a - 0.25a 2.303 4 k 308K = log t2 3 2.303 k 308K = × 0.125 t2 0.2879 t2 = k 308K But t1 = t 2 Hence, 0.1055 0.2879 = k 298K k 308K k 308K = 2.7289 k 298K Now, applying the Arrhenius equation, k E a  T2 -T1  log 308K =   k 298K 2.303R  T1T2  Ea  308-298  log(2.7289) =   2.303 × 8.314  298 × 308  Ea = 76.623 kJ mol-1 Now, the calculation of k at 318 K Class XII Chemistry www.vedantu.com 34  Ea log k = log A - 2.303RT 7623 log k = log (4 × 1010 ) - 2.303 × 8.314 × 318 log k = 10.6021 - 12.5843 = -1.9822 k = Antilog (-1.9822) = antilog (2.0178) = 1.042 × 10-2 s-1 Therefore, k is 1.042 × 10-2 s-1. 30. 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. Ans: We are given that: k 2 = 4 k1 T1 = 293K T2 = 313K According the Arrhenius equation, we get: k E a  T2 -T1  log 2 =   k1 2.303R  T1T2  Putting the values, we get: 4k Ea  313-293  log 1 =   k1 2.303 × 8.314  293 × 313  Ea  313-293  0.6021 =   2.303 × 8.314  293 × 313  0.6021 × 2.303 × 8.314 × 293 × 313 Ea = 20 -1 Ea = 52863.00 J mol Ea = 52.863 kJ mol-1 Therefore, the required activation energy is 52.863 kJ mol-1. Class XII Chemistry www.vedantu.com 35