The following rate data were obtained at 300 K for the reaction 2 A + B ⟶ C + D Experiment No. [ A ] molL - [ B ] molL - Rate of formation of D molL - min - 1. 0 . 1 0 . 1 5 . 0 × 10 - 3 2. 0 . 3 0 . 2 6 . 0 × 10 - 2 3. 0 . 3 0 . 4 2 . 4 × 10 - 1 4. 0 . 4 0 . 1 2 . 0 × 10 - 2 Calculate the rate of formation of D when: [ A ] = 0 . 5 mol L - 1 and [ B ] = 0 . 2 mol L - 1

Rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentrations of the reactants participating in it. The rate law can be expressed as, Rate = k [ A ] p [ B ] q On comparing the experiments 2 and 3 ( Rate ) 2 = k [ 0 . 3 ] p [ 0 . 2 ] q = 6 . 0 × 10 - 2 .......(i) ( Rate ) 3 = k [ 0 . 3 ] p [ 0 . 4 ] q = 2 . 4 × 10 - 1 ......(ii) Upon dividing equations (ii) and (i), we get [ 2 ] q = 4 or q = 2 On comparing experiments 4 and 1 ( Rate ) 4 = k [ 0 . 4 ] p [ 0 . 1 ] q = 2 . 0 × 10 - 2 .......(iii) ( Rate ) 1 = k [ 0 . 1 ] p [ 0 . 1 ] q = 5 . 0 × 10 - 3 ......(iv) Upon diving equations (iii) and (iv), we get [ 4 ] p = 4 or p = 1 Substitute the values of p and q in the equation (iv), the value of rate constant is calculated as, k [ 0 . 1 ] 1 [ 0 . 1 ] 2 = 5 . 0 × 10 - 3 k = 5 Rate of formation of D when [ A ] = 0 . 5 mol L - 1 and [ B ] = 0 . 2 mol L - 1 is calculated as, k [ A ] p [ B ] q = 5 × [ 0 . 5 ] 1 [ 0 . 2 ] 2 = 0 . 1 mol L - 1 min - 1 Rate = k [ A ] [ B ] 2 ; 0 . 1 mol L - min -

For the following reaction, 2 H 2 + 2 NO ⇌ 2 H 2 O + N 2 the following rate data was obtained Experiment NO mol 1 - . H 2 molL - rate mol L - s - 1 0 . 40 0 . 40 4 . 8 × 10 - 3 2 0 . 80 0 . 40 19 . 2 × 10 - 3 3 0 . 40 0 . 80 9 . 6 × 10 - 3 Determine the rate equation and calculate the value of rate constant, k.

Rate law is used for studying the relationship between the rate of the reaction and the concentration of the reactants. Let rate law for the given reaction be, r = k [ H 2 ] p [ NO ] q On substituting the given rate data in the rate law, we get 4 . 8 × 10 - 3 = k [ 0 . 40 ] p [ 0 . 40 ] q ........(i) 19 . 2 × 10 - 3 = k [ 0 . 40 ] p [ 0 . 80 ] q ........(ii) 9 . 6 × 10 - 3 = k [ 0 . 80 ] p [ 0 . 40 ] q ........(iii) Upon dividing equation (ii) by (i), we get [ 2 ] q = 4 or q = 2 Dividing equation (iii) by (i), we get [ 2 ] p = 2 or p = 1 Therefore, rate equation for the given reaction is r = k [ H 2 ] 1 [ NO ] 2 Upon substituting the values of p and q in equation (i), we get Rate constant, k = 4 . 8 × 10 - 3 [ 0 . 4 ] [ 0 . 4 ] 2 = 0 . 075 mol - 2 L 2 s - 1

The initial rate of reaction A + 5 B + 6 C ⟶ 3 D + 3 E has been determined by measuring the rate of disappearance of A under the following conditions: Experiment Initial Concentration molL - 1 Initial Rate A B C I 0 . 02 0 . 02 0 . 02 2 . 08 × 10 - 3 II 0 . 01 0 . 02 0 . 02 1 . 04 × 10 - 3 III 0 . 02 0 . 04 0 . 02 4 . 16 × 10 - 3 IV 0 . 02 0 . 02 0 . 04 8 . 32 × 10 - 3 Determine the order of reaction with respect to each reactant and overall order of the reaction.

In the experiments I and II, the initial concentrations of B and C are unchanged only the concentration of A is changed and hence from the rate expression, r = k [ A ] a [ B ] b [ C ] c = k [ A ] a , we get ( r ) 1 ( r ) 2 = 2 . 08 × 10 - 3 1 . 04 × 10 - 3 = [ A ] 1 a [ A ] 2 a = ( 0 . 02 ) a ( 0 . 01 ) a Or ( 2 ) 1 = ( 2 ) a or a = 1 Similarly, for experiments I and III ( r ) 3 ( r ) 1 = 4 . 16 × 10 - 3 1 . 04 × 10 - 3 = [ B ] 3 b [ B ] 1 b = ( 0 . 04 ) b ( 0 . 02 ) b Or ( 2 ) 1 = ( 2 ) b or b = 1 For experiments IV and I ( r ) 4 ( r ) 1 = 8 . 32 × 10 - 3 2 . 08 × 10 - 3 = [ C ] 4 c [ C ] 1 c = ( 0 . 04 ) c ( 0 . 02 ) c Or ( 2 ) 2 = ( 2 ) c or c = 2 Order of a reaction is defined as the sum of the exponents to which the molar concentrations are raised in the rate law or rate equation. Therefore, order of reaction with respect to A = 1 , B = 1 , C = 2 and overall order of the reaction = 1 + 1 + 2 = 4

The initial rate of reaction A + 5 B + 6 C ⟶ 3 D + 3 E has been determined by measuring the rate of disappearance of A under the following conditions: Experiment Initial Concentration molL - 1 Initial Rate A B C I 0 . 02 0 . 02 0 . 02 2 . 08 × 10 - 3 II 0 . 01 0 . 02 0 . 02 1 . 04 × 10 - 3 III 0 . 02 0 . 04 0 . 02 4 . 16 × 10 - 3 IV 0 . 02 0 . 02 0 . 04 8 . 32 × 10 - 3 What is the rate constant?

In the experiments I and II, the initial concentrations of B and C are unchanged only the concentration of A is changed and hence from the rate expression, r = k [ A ] a [ B ] b [ C ] c = k [ A ] a , we get ( r ) 1 ( r ) 2 = 2 . 08 × 10 - 3 1 . 04 × 10 - 3 = [ A ] 1 a [ A ] 2 a = ( 0 . 02 ) a ( 0 . 01 ) a Or ( 2 ) 1 = ( 2 ) a or a = 1 Similarly, for experiments I and III r = k [ A ] a [ B ] b [ C ] c = k [ B ] b , we get ( r ) 3 ( r ) 1 = 4 . 16 × 10 - 3 1 . 04 × 10 - 3 = [ B ] 3 b [ B ] 1 b = ( 0 . 04 ) b ( 0 . 02 ) b Or ( 2 ) 1 = ( 2 ) b or b = 1 For experiments IV and I r = k [ A ] a [ B ] b [ C ] c = k [ C ] c , we get ( r ) 4 ( r ) 1 = 8 . 32 × 10 - 3 2 . 08 × 10 - 3 = [ C ] 4 c [ C ] 1 c = ( 0 . 04 ) c ( 0 . 02 ) c Or ( 2 ) 2 = ( 2 ) c or c = 2 Rate constant is the proportionality constant in the rate law. It does not depend upon the initial concentration of the reactants and its units depends upon the order of the reaction. Rate constant, k = r [ A ] a [ B ] b [ C ] c = r [ A ] [ B ] [ C ] 2 Upon substituting the values of experiment I in the above equation, we get k = 2 . 08 × 10 - 3 ( 0 . 02 ) ( 0 . 02 ) ( 0 . 02 ) 2 = 2 . 08 × 10 - 3 16 × 10 - 8 = 1 . 3 × 10 4 M - 2 min - 1 Therefore, rate constant k = 1 . 3 × 10 4 M - 2 min - 1

The initial rate of reaction A + 5 B + 6 C ⟶ 3 D + 3 E has been determined by measuring the rate of disappearance of A under the following conditions: Experiment Initial Concentration molL - 1 Initial Rate A B C I 0 . 02 0 . 02 0 . 02 2 . 08 × 10 - 3 II 0 . 01 0 . 02 0 . 02 1 . 04 × 10 - 3 III 0 . 02 0 . 04 0 . 02 4 . 16 × 10 - 3 IV 0 . 02 0 . 02 0 . 04 8 . 32 × 10 - 3 Calculate the initial rate of the reaction when concentration of all the reactants is 0 . 01 M .

In the experiments I and II, the initial concentrations of B and C are unchanged only the concentration of A is changed and hence from the rate expression, r = k [ A ] a [ B ] b [ C ] c = k [ A ] a , we get ( r ) 1 ( r ) 2 = 2 . 08 × 10 - 3 1 . 04 × 10 - 3 = [ A ] 1 a [ A ] 2 a = ( 0 . 02 ) a ( 0 . 01 ) a Or ( 2 ) 1 = ( 2 ) a or a = 1 Similarly, for experiments I and III r = k [ A ] a [ B ] b [ C ] c = k [ B ] b , we get ( r ) 3 ( r ) 1 = 4 . 16 × 10 - 3 1 . 04 × 10 - 3 = [ B ] 3 b [ B ] 1 b = ( 0 . 04 ) b ( 0 . 02 ) b Or ( 2 ) 1 = ( 2 ) b or b = 1 For experiments IV and I r = k [ A ] a [ B ] b [ C ] c = k [ C ] c , we get ( r ) 4 ( r ) 1 = 8 . 32 × 10 - 3 2 . 08 × 10 - 3 = [ C ] 4 c [ C ] 1 c = ( 0 . 04 ) c ( 0 . 02 ) c Or ( 2 ) 2 = ( 2 ) c or c = 2 Rate constant, k = r [ A ] a [ B ] b [ C ] c = r [ A ] [ B ] [ C ] 2 Upon substituting the values of experiment I in the above equation, we get k = 2 . 08 × 10 - 3 ( 0 . 02 ) ( 0 . 02 ) ( 0 . 02 ) 2 = 2 . 08 × 10 - 3 16 × 10 - 8 = 1 . 3 × 10 4 M - 2 min - 1 Therefore, rate constant k = 1 . 3 × 10 4 M - 2 min - 1 Given that the concentrations of all the reactants is 0 . 01 M i.e., [ A ] = [ B ] = [ C ] = 0 . 01 M Rate of the reaction is defined as the rate of change of concentration of any one of the reactants or the products at a given moment of time. Therefore, the initial rate of the reaction is, = 1 . 3 × 10 4 M - 3 min - 1 [ 0 . 01 ] [ 0 . 01 ] [ 0 . 01 ] 2 = 1 . 3 × 10 - 4 M min - 1

The initial rate of reaction A + 5 B + 6 C ⟶ 3 D + 3 E has been determined by measuring the rate of disappearance of A under the following conditions: Experiment Initial Concentration molL - 1 Initial Rate A B C I 0 . 02 0 . 02 0 . 02 2 . 08 × 10 - 3 II 0 . 01 0 . 02 0 . 02 1 . 04 × 10 - 3 III 0 . 02 0 . 04 0 . 02 4 . 16 × 10 - 3 IV 0 . 02 0 . 02 0 . 04 8 . 32 × 10 - 3 Calculate the initial rate of change in concentration of B and D.

In the experiments I and II, the initial concentrations of B and C are unchanged only the concentration of A is changed and hence from the rate expression, r = k [ A ] a [ B ] b [ C ] c = k [ A ] a , we get ( r ) 1 ( r ) 2 = 2 . 08 × 10 - 3 1 . 04 × 10 - 3 = [ A ] 1 a [ A ] 2 a = ( 0 . 02 ) a ( 0 . 01 ) a Or ( 2 ) 1 = ( 2 ) a or a = 1 Similarly, for experiments I and III r = k [ A ] a [ B ] b [ C ] c = k [ B ] b , we get ( r ) 3 ( r ) 1 = 4 . 16 × 10 - 3 1 . 04 × 10 - 3 = [ B ] 3 b [ B ] 1 b = ( 0 . 04 ) b ( 0 . 02 ) b Or ( 2 ) 1 = ( 2 ) b or b = 1 For experiments IV and I r = k [ A ] a [ B ] b [ C ] c = k [ C ] c , we get ( r ) 4 ( r ) 1 = 8 . 32 × 10 - 3 2 . 08 × 10 - 3 = [ C ] 4 c [ C ] 1 c = ( 0 . 04 ) c ( 0 . 02 ) c Or ( 2 ) 2 = ( 2 ) c or c = 2 Rate constant, k = r [ A ] a [ B ] b [ C ] c = r [ A ] [ B ] [ C ] 2 Upon substituting the values of experiment I in the above equation, we get k = 2 . 08 × 10 - 3 ( 0 . 02 ) ( 0 . 02 ) ( 0 . 02 ) 2 = 2 . 08 × 10 - 3 16 × 10 - 8 = 1 . 3 × 10 4 M - 2 min - 1 Therefore, rate constant k = 1 . 3 × 10 4 M - 2 min - 1 Initial rate of change in concentration of B is - d [ B ] dt = 5 × 1 . 3 × 10 - 4 = 6 . 5 × 10 - 4 M min - 1 Initial rate of change in concentration of D is, - d [ D ] dt = 3 × 1 . 3 × 10 - 4 = 3 . 9 × 10 - 4 M min - 1

The initial concentration of N 2 O 5 in the following first order reaction, N 2 O 5 g ⟶ 2 NO 2 g + 1 2 O 2 g was 1 . 24 × 10 - 2 mol L - 1 at 318 K . The concentration of N 2 O 5 after 60 minutes was 0 . 20 × 10 - 2 mol L - 1 . Calculate the rate constant of the reaction at 318 K .

The rate equation for the first order reaction can be written as : k = 2 . 303 t log [ a ] [ a - x ] Where a is the initial concentration of the reactant and x is the number of moles per litre decomposed after a time t . Given that, initial concentration= 1 . 24 × 10 - 2 mol L - 1 The concentration after 60 minutes is = 0 . 20 × 10 - 2 mol L - 1 Therefore, rate constant of the reaction, k = 2 . 303 t log [ a ] [ a - x ] k = 2 . 303 60 log 1 . 24 × 10 - 2 0 . 2 × 10 - 2 k = 0 . 0304 min - 1 Rate constant of the reaction at 318 K is 0 . 0304 min - 1

K L Chugh Solutions for Chapter: Chemical Kinetics, Exercise 5: Follow Up Problems

Author:K L Chugh

K L Chugh Chemistry Solutions for Exercise - K L Chugh Solutions for Chapter: Chemical Kinetics, Exercise 5: Follow Up Problems

Attempt the practice questions on Chapter 4: Chemical Kinetics, Exercise 5: Follow Up Problems with hints and solutions to strengthen your understanding. NEW SYSTEMATIC MODERN CHEMISTRY solutions are prepared by Experienced Embibe Experts.

Questions from K L Chugh Solutions for Chapter: Chemical Kinetics, Exercise 5: Follow Up Problems with Hints & Solutions

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 19

The following rate data were obtained at $300 K$ for the reaction $2 A + B ⟶ C + D$

Experiment No.	$[A]$ ${molL}^{-}$	$[B]$ ${molL}^{-}$	Rate of formation of $D {molL}^{-} \min^{-}$
1.	$0.1$	$0.1$	$5.0 \times 10^{- 3}$
2.	$0.3$	$0.2$	$6.0 \times 10^{- 2}$
3.	$0.3$	$0.4$	$2.4 \times 10^{- 1}$
4.	$0.4$	$0.1$	$2.0 \times 10^{- 2}$

Calculate the rate of formation of D when: $[A] = 0.5 mol L^{- 1}$ and $[B] = 0.2 mol L^{- 1}$

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 20

The rate of reaction, $2 NO + {Cl}_{2} ⟶ 2 NOCl$ is doubled when concentration of ${Cl}_{2}$ is doubled and it becomes $8$ times when concentrations of both $NO$ and ${Cl}_{2}$ are doubled. Deduce the order of this reaction.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 21

For the following reaction, $2 H_{2} + 2 NO ⇌ 2 H_{2} O + N_{2}$ the following rate data was obtained

Experiment	$[NO] (mol 1^{-}) .$	$[H_{2}] ({molL}^{-})$	$rate (mol L^{-} s^{-})$
1	$0.40$	$0.40$	$4.8 \times 10^{- 3}$
2	$0.80$	$0.40$	$19.2 \times 10^{- 3}$
3	$0.40$	$0.80$	$9.6 \times 10^{- 3}$

Determine the rate equation and calculate the value of rate constant, k.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 22

The initial rate of reaction $A + 5 B + 6 C$ $⟶ 3 D + 3 E$ has been determined by measuring the rate of disappearance of A under the following conditions:

Experiment	Initial Concentration $({molL}^{- 1})$			Initial Rate
	A	B	C
I	$0.02$	$0.02$	$0.02$	$2.08 \times 10^{- 3}$
II	$0.01$	$0.02$	$0.02$	$1.04 \times 10^{- 3}$
III	$0.02$	$0.04$	$0.02$	$4.16 \times 10^{- 3}$
IV	$0.02$	$0.02$	$0.04$	$8.32 \times 10^{- 3}$

Determine the order of reaction with respect to each reactant and overall order of the reaction.

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 22

The initial rate of reaction $A + 5 B + 6 C$ $⟶ 3 D + 3 E$ has been determined by measuring the rate of disappearance of A under the following conditions:

Experiment	Initial Concentration $({molL}^{- 1})$			Initial Rate
	A	B	C
I	$0.02$	$0.02$	$0.02$	$2.08 \times 10^{- 3}$
II	$0.01$	$0.02$	$0.02$	$1.04 \times 10^{- 3}$
III	$0.02$	$0.04$	$0.02$	$4.16 \times 10^{- 3}$
IV	$0.02$	$0.02$	$0.04$	$8.32 \times 10^{- 3}$

What is the rate constant?

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 22

The initial rate of reaction $A + 5 B + 6 C$ $⟶ 3 D + 3 E$ has been determined by measuring the rate of disappearance of A under the following conditions:

Experiment	Initial Concentration $({molL}^{- 1})$			Initial Rate
	A	B	C
I	$0.02$	$0.02$	$0.02$	$2.08 \times 10^{- 3}$
II	$0.01$	$0.02$	$0.02$	$1.04 \times 10^{- 3}$
III	$0.02$	$0.04$	$0.02$	$4.16 \times 10^{- 3}$
IV	$0.02$	$0.02$	$0.04$	$8.32 \times 10^{- 3}$

Calculate the initial rate of the reaction when concentration of all the reactants is $0.01 M .$

HARD

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 22

The initial rate of reaction $A + 5 B + 6 C$ $⟶ 3 D + 3 E$ has been determined by measuring the rate of disappearance of A under the following conditions:

Experiment	Initial Concentration $({molL}^{- 1})$			Initial Rate
	A	B	C
I	$0.02$	$0.02$	$0.02$	$2.08 \times 10^{- 3}$
II	$0.01$	$0.02$	$0.02$	$1.04 \times 10^{- 3}$
III	$0.02$	$0.04$	$0.02$	$4.16 \times 10^{- 3}$
IV	$0.02$	$0.02$	$0.04$	$8.32 \times 10^{- 3}$

Calculate the initial rate of change in concentration of B and D.

MEDIUM

12th West Bengal Board

IMPORTANT

NEW SYSTEMATIC MODERN CHEMISTRY>Chapter 4 - Chemical Kinetics>Follow Up Problems>Q 23

The initial concentration of $N_{2} O_{5}$ in the following first order reaction, $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$
was $1.24 \times 10^{- 2} mol L^{- 1}$ at $318 K$ . The concentration of $N_{2} O_{5}$ after $60$ minutes was $0.20 \times 10^{- 2} mol L^{- 1} .$ Calculate the rate constant of the reaction at $318 K .$

K L Chugh Solutions for Chapter: Chemical Kinetics, Exercise 5: Follow Up Problems

K L Chugh Chemistry Solutions for Exercise - K L Chugh Solutions for Chapter: Chemical Kinetics, Exercise 5: Follow Up Problems

Questions from K L Chugh Solutions for Chapter: Chemical Kinetics, Exercise 5: Follow Up Problems with Hints & Solutions

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