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

A catalyst is a substance which:

shortens the time to reach equilibrium

changes the equilibrium constant of the reaction

increases the equilibrium constant of the reaction

supplies energy to the reaction

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?

Rate equation for first order reaction: Rate = 2 . 303 t log R 0 R At t = 50 min , R 50 = ( 1 − 0 . 4 ) R 0 = 0 . 6 R 0 At t = x min , R x = ( 1 − 0 . 8 ) R 0 = 0 . 2 R 0 Substituting, the values we get, 2 . 303 50 log R 0 0 . 6 R 0 = 2 . 303 x log R 0 0 . 2 R 0 1 50 log 1 . 66 = 1 x log 5 x = 0 . 69 × 50 0 . 22 = 156 . 81 min

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.

E M B I B E

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

EASY

12th ICSE

IMPORTANT

Earn 100

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.

Important Questions on Chemical Kinetics

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2019>Q 10

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.

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter 4 - Chemical Kinetics>ICSE-2020>Q 11

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.

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

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