Write the rate law expression for the reaction A + B + C → D + E , if the order of reaction is first, second and zero with respect to A , B and C , respectively.

The 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. It is given that a reaction is first order in A and second order in B and third in C. The rate law of the given reaction can be written as : Rate = k [ A ] 1 [ B ] 2 [ C ] 0

How many times the rate of reaction will increase if the concentration of A , B and C are doubled in the equation A + B + C → D + E ?(order of reaction is first, second and zero with respect to A,B and C)

The rate law of the given reaction can be written as : Rate = k [ A ] 1 [ B ] 2 [ C ] 0 If concentration of B is doubled then rate increases by four times and concentration of C is doubled then no change in the rate of reaction.

The rate of reaction becomes four times when the temperature changes from 293 K to 313 K . Calculate the energy of activation E a of the reaction assuming that it does not change with temperature. R = 8 . 314 J K - 1 mol - 1 .

Given: T 1 = 293 K and T 2 = 313 K k 2 = 4 k 1 . Solution: log 10 k 2 k 1 = E a 2 . 303 R 1 T 1 - 1 T 2 Value of gas constant R = 8 . 314 J mol - 1 K - 1 . Putting the given values in the formula: log 10 4 k 1 k 1 = E a 2 . 303 × 8 . 314 1 293 - 1 313 log 10 4 = E a 2 . 303 × 8 . 314 313 - 293 313 × 293 0 . 602 = E a 19 . 15 20 91709 E a = 1057248 . 86 20 = 52 . 86 × 10 3 J mol - 1 The value of activation energy comes out be 52 . 86 kJ mol - 1 .

For the reaction A + B → C + D, the initial rate for different reactions and initial concentration of reactants are given below: S.No Initial concentration Initial rate A mol L - 1 B mol L - 1 mol L - 1 s - 1 1 1 . 0 1 . 0 2 × 10 - 3 2 2 . 0 1 . 0 4 × 10 - 3 3 4 . 0 1 . 0 8 × 10 - 3 4 1 . 0 2 . 0 2 × 10 - 3 5 1 . 0 4 . 0 2 × 10 - 3 What is the overall order of reaction?

For the given reaction, rate = k [ A ] x [ B ] y . After substituting the values of initial rates and initial concentrations, the value of x and y can be calculated. Then ( x + y ) will give the order of the reaction. For experiment ( i ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 1 . 0 ] y For experiment ( ii ) 4 . 0 × 10 - 3 = [ 2 . 0 ] x [ 1 . 0 ] y For experiment ( iii ) 8 . 0 × 10 - 3 = [ 4 . 0 ] x [ 1 . 0 ] y For experiment ( iv ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 2 . 0 ] y For experiment ( v ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 4 . 0 ] y Determination of x . In order to get the value of x , divide equation (2) by (1) 4 . 0 × 10 - 3 2 . 0 × 10 − 3 = [ 2 . 0 ] x [ 1 . 0 ] y [ 1 . 0 ] x [ 1 . 0 ] y 2 1 = 2 x x = 1 Determination of y . In order to get the value of y , divide equation (4) by (1) 2 . 0 × 10 - 3 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 2 . 0 ] y [ 1 . 0 ] x [ 1 . 0 ] y y = 0 The overall order of reaction = 1 + 0 = 1

For the reaction A + B → C + D, the initial rate for different reactions and initial concentration of reactants are given below: S.No Initial concentration Initial rate A mol L - 1 B mol L - 1 mol L - 1 s - 1 1 1 . 0 1 . 0 2 × 10 - 3 2 2 . 0 1 . 0 4 × 10 - 3 3 4 . 0 1 . 0 8 × 10 - 3 4 1 . 0 2 . 0 2 × 10 - 3 5 1 . 0 4 . 0 2 × 10 - 3 Write the rate law equation.

For the given reaction, rate = k [ A ] x [ B ] y . After substituting the values of initial rates and initial concentrations, the value of x and y can be calculated. Then ( x + y ) will give the order of the reaction. For experiment ( i ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 1 . 0 ] y For experiment ( ii ) 4 . 0 × 10 - 3 = [ 2 . 0 ] x [ 1 . 0 ] y For experiment ( iii ) 8 . 0 × 10 - 3 = [ 4 . 0 ] x [ 1 . 0 ] y For experiment ( iv ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 2 . 0 ] y For experiment ( v ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 4 . 0 ] y Determination of x . In order to get the value of x , divide equation (2) by (1) 4 . 0 × 10 - 3 2 . 0 × 10 − 3 = [ 2 . 0 ] x [ 1 . 0 ] y [ 1 . 0 ] x [ 1 . 0 ] y 2 1 = 2 x x = 1 Determination of y . In order to get the value of y , divide equation (4) by (1) 2 . 0 × 10 - 3 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 2 . 0 ] y [ 1 . 0 ] x [ 1 . 0 ] y y = 0 The overall order of reaction = 1 + 0 = 1 The rate law (also known as the rate equation) 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. Rate law, r α A 1 B 0 r α A 1 r = k A 1

For the reaction A + B → C + D, the initial rate for different reactions and initial concentration of reactants are given below: S.No Initial concentration Initial rate A mol L - 1 B mol L - 1 mol L - 1 s - 1 1 1 . 0 1 . 0 2 × 10 - 3 2 2 . 0 1 . 0 4 × 10 - 3 3 4 . 0 1 . 0 8 × 10 - 3 4 1 . 0 2 . 0 2 × 10 - 3 5 1 . 0 4 . 0 2 × 10 - 3 Write the rate law equation.

For the given reaction, rate = k [ A ] x [ B ] y . After substituting the values of initial rates and initial concentrations, the value of x and y can be calculated. Then ( x + y ) will give the order of the reaction. For experiment ( i ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 1 . 0 ] y For experiment ( ii ) 4 . 0 × 10 - 3 = [ 2 . 0 ] x [ 1 . 0 ] y For experiment ( iii ) 8 . 0 × 10 - 3 = [ 4 . 0 ] x [ 1 . 0 ] y For experiment ( iv ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 2 . 0 ] y For experiment ( v ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 4 . 0 ] y Determination of x . In order to get the value of x , divide equation (2) by (1) 4 . 0 × 10 - 3 2 . 0 × 10 − 3 = [ 2 . 0 ] x [ 1 . 0 ] y [ 1 . 0 ] x [ 1 . 0 ] y 2 1 = 2 x x = 1 Determination of y . In order to get the value of y , divide equation (4) by (1) 2 . 0 × 10 - 3 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 2 . 0 ] y [ 1 . 0 ] x [ 1 . 0 ] y y = 0 The overall order of reaction = 1 + 0 = 1 The rate law (also known as the rate equation) 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. Rate law, r α A 1 B 0 r α A 1 r = k A 1

For the reaction A + B → C + D, the initial rate for different reactions and initial concentration of reactants are given below: S.No Initial concentration Initial rate A mol L - 1 B mol L - 1 mol L - 1 s - 1 1 1 . 0 1 . 0 2 × 10 - 3 2 2 . 0 1 . 0 4 × 10 - 3 3 4 . 0 1 . 0 8 × 10 - 3 4 1 . 0 2 . 0 2 × 10 - 3 5 1 . 0 4 . 0 2 × 10 - 3 What is the overall order of reaction?

For the given reaction, rate = k [ A ] x [ B ] y . After substituting the values of initial rates and initial concentrations, the value of x and y can be calculated. Then ( x + y ) will give the order of the reaction. For experiment ( i ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 1 . 0 ] y For experiment ( ii ) 4 . 0 × 10 - 3 = [ 2 . 0 ] x [ 1 . 0 ] y For experiment ( iii ) 8 . 0 × 10 - 3 = [ 4 . 0 ] x [ 1 . 0 ] y For experiment ( iv ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 2 . 0 ] y For experiment ( v ) 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 4 . 0 ] y Determination of x . In order to get the value of x , divide equation (2) by (1) 4 . 0 × 10 - 3 2 . 0 × 10 − 3 = [ 2 . 0 ] x [ 1 . 0 ] y [ 1 . 0 ] x [ 1 . 0 ] y 2 1 = 2 x x = 1 Determination of y . In order to get the value of y , divide equation (4) by (1) 2 . 0 × 10 - 3 2 . 0 × 10 - 3 = [ 1 . 0 ] x [ 2 . 0 ] y [ 1 . 0 ] x [ 1 . 0 ] y y = 0 The overall order of reaction = 1 + 0 = 1

Show that for a first order reaction the time required to complete 75 % of reaction is about 2 times more than that required to complete 50 % of the reaction.

The relationship between the time taken and the rate constant of the reaction is given as, t = 2 . 303 k log a a - x . For the first order reaction, time required to complete 50 % of the reaction is, t = 2 . 303 k log a a - x ⇒ t 50 % = 2 . 303 k log 100 100 - 50 . t 50 % = 2 . 303 k log 100 50 ⇒ t 50 % = 2 . 303 k log 2 . For the first order reaction, time required to complete 75 % of the reaction is, t = 2 . 303 k log a a - x ⇒ t 75 % = 2 . 303 k log 100 100 - 75 . t 75 % = 2 . 303 k log 100 25 ⇒ t 75 % = 2 . 303 k log 4 . t 75 % = 2 . 303 k log 2 2 ⇒ t 75 % = 2 × 2 . 303 k log 2 . t 75 % = 2 × t 50 % . Thus, it will be twice the time taken for 50 % of the reaction.

The rate of reaction becomes four times when the temperature changes from 293 K to 313 K . Calculate the energy of activation E a of the reaction assuming that it does not change with temperature. R = 8 . 314 J K - 1 mol - 1 .

Given: T 1 = 293 K and T 2 = 313 K k 2 = 4 k 1 . Solution: log 10 k 2 k 1 = E a 2 . 303 R 1 T 1 - 1 T 2 Value of gas constant R = 8 . 314 J mol - 1 K - 1 . Putting the given values in the formula: log 10 4 k 1 k 1 = E a 2 . 303 × 8 . 314 1 293 - 1 313 log 10 4 = E a 2 . 303 × 8 . 314 313 - 293 313 × 293 0 . 602 = E a 19 . 15 20 91709 E a = 1057248 . 86 20 = 52 . 86 × 10 3 J mol - 1 The value of activation energy comes out be 52 . 86 kJ mol - 1 .

How many times the rate of reaction will increase if the concentration of A , B and C are doubled in the equation A + B + C → D + E ?(order of reaction is first, second and zero with respect to A,B and C)

The rate law of the given reaction can be written as : Rate = k [ A ] 1 [ B ] 2 [ C ] 0 If concentration of B is doubled then rate increases by four times and concentration of C is doubled then no change in the rate of reaction.

MEDIUM

12th ICSE

IMPORTANT

Earn 100

A first order reaction is $50 %$ completed in $30$ minutes at $300 K$ and in $10$ minutes at $320 K$ . Calculate the activation energy of the reaction.

( $R = 8.314 J K^{- 1} {mol}^{- 1}$ ).

Important Questions on Chemical Kinetics

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2018>Q 1

Write the rate law expression for the reaction

A + B + C \to D + E

, if the order of reaction is first, second and zero with respect to

A, B

and

C

, respectively.

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2018>Q 2

How many times the rate of reaction will increase if the concentration of

A, B

and

C

are doubled in the equation

A + B + C \to D + E

?(order of reaction is first, second and zero with respect to A,B and C)

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2018>Q 3

The rate of reaction becomes four times when the temperature changes from

293 K

313 K

. Calculate the energy of activation

(E_{a})

of the reaction assuming that it does not change with temperature.

(R = 8.314 J K^{- 1} {mol}^{- 1})

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2018>Q 4

40 %

of a first order reaction is completed in

50

minutes. How much time will it take for the completion of

80 %

of this reaction?

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2019>Q 6

A catalyst is a substance which:

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2019>Q 7

For the reaction A + B → C + D, the initial rate for different reactions and initial concentration of reactants are given below:

S.No	Initial concentration		Initial rate
S.No	$[A] mol L^{- 1}$	$[B] mol L^{- 1}$	$mol L^{- 1} s^{- 1}$
$1$	$1.0$	$1.0$	$2 \times 10^{- 3}$
$2$	$2.0$	$1.0$	$4 \times 10^{- 3}$
$3$	$4.0$	$1.0$	$8 \times 10^{- 3}$
4	$1.0$	$2.0$	$2 \times 10^{- 3}$
5	$1.0$	$4.0$	$2 \times 10^{- 3}$

What is the overall order of reaction?

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2019>Q 8

For the reaction A + B → C + D, the initial rate for different reactions and initial concentration of reactants are given below:

S.No	Initial concentration		Initial rate
S.No	$[A] mol L^{- 1}$	$[B] mol L^{- 1}$	$mol L^{- 1} s^{- 1}$
$1$	$1.0$	$1.0$	$2 \times 10^{- 3}$
$2$	$2.0$	$1.0$	$4 \times 10^{- 3}$
$3$	$4.0$	$1.0$	$8 \times 10^{- 3}$
4	$1.0$	$2.0$	$2 \times 10^{- 3}$
5	$1.0$	$4.0$	$2 \times 10^{- 3}$

Write the rate law equation.

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2019>Q 9

25% of a first-order reaction is completed in 30 minutes. Calculate the time taken in minutes for the reaction to go to 90% completion.

A first order reaction is 50 % completed in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate the activation energy of the reaction. (R=8.314 J K-1mol-1).

Important Questions on Chemical Kinetics

Learn and Prepare for any exam you want !

A first order reaction is $50 %$ completed in $30$ minutes at $300 K$ and in $10$ minutes at $320 K$ . Calculate the activation energy of the reaction.

( $R = 8.314 J K^{- 1} {mol}^{- 1}$ ).