The activation energy of a reaction is 94 . 14 kJ mol - 1 and the value of rate constant at 298 K is 18 x 10 - 5 s - 1 . Calculate the frequency factor or pre-exponential factor, A .

Arrhenius equation can be given as : logk = logA - E a 2 . 303 RT , where R is gas constant having value, 8 . 314 J mol - 1 K - 1 . logA = logk + E a 2 . 303 RT logA = log 18 × 10 - 5 + 94 . 14 × 1000 2 . 303 × 8 . 314 × 298 logA = - 3 . 744 + 16 . 5 = 12 . 756 A = antilog 12 . 756 A = 5 . 7 × 10 12 s - 1

The data given is for the reaction, N 2 O 5 g → NO 2 g + O 2 g : S.no. N 2 O 5 ( mol L - 1 ) Rate of disappearance of N 2 O 5 mol L - 1 min - 1 1 1 . 13 × 10 - 2 34 × 10 - 5 2 0 . 84 × 10 - 2 25 × 10 - 5 3 0 . 62 × 10 - 2 18 × 10 - 5 Determine the order of reaction.

Rate law expression for the reaction can be given as : Rate = k N 2 O 5 x . Where, x is the order of the reaction. 25 × 10 - 5 = k 0 . 84 × 10 - 2 x 18 × 10 - 5 = k 0 . 62 × 10 - 2 x 25 × 10 - 5 18 × 10 - 5 = k 0 . 84 × 10 - 2 x k 0 . 62 × 10 - 2 x 1 . 38 = 1 . 36 x x = 1 The order of reaction comes out to be one. Hence, the given reaction is of first order.

The data given is for the reaction, N 2 O 5 g → NO 2 g + O 2 g : S.no. N 2 O 5 ( mol L - 1 ) Rate of disappearance of N 2 O 5 mol L - 1 min - 1 1 1 . 13 × 10 - 2 34 × 10 - 5 2 0 . 84 × 10 - 2 25 × 10 - 5 3 0 . 62 × 10 - 2 18 × 10 - 5 Determine the rate law of reaction.

Rate law expression for the reaction can be given as : Rate = k N 2 O 5 x , where k is rate constant. 25 × 10 - 5 = k 0 . 84 × 10 - 2 x 18 × 10 - 5 = k 0 . 62 × 10 - 2 x 25 × 10 - 5 18 × 10 - 5 = k 0 . 84 × 10 - 2 x k 0 . 62 × 10 - 2 x 1 . 38 = 1 . 36 x x = 1 Rate law for the reaction can be given as : Rate = k N 2 O 5 .

The data given is for the reaction, N 2 O 5 g → NO 2 g + O 2 g : S.no. N 2 O 5 ( mol L - 1 ) Rate of disappearance of N 2 O 5 mol L - 1 min - 1 1 1 . 13 × 10 - 2 34 × 10 - 5 2 0 . 84 × 10 - 2 25 × 10 - 5 3 0 . 62 × 10 - 2 18 × 10 - 5 Determine the rate constant of reaction.

Rate law for the reaction can be given as : Rate = k N 2 O 5 , where k is rate constant. So, 25 × 10 - 5 = k 0 . 84 × 10 - 2 k = 25 × 10 - 5 0 . 84 × 10 - 2 = 2 . 97 × 10 - 2 min - 1 . Hence, the value of rate constant is calculated to be 2 . 97 × 10 - 2 min - 1

Under what condition a bimolecular reaction behaves kinetically first order reaction?

In a bimolecular reaction, when the amount of one of the reactants is present in excess, the reaction is kinetically first order because the concentration of reactant present in excess does not affect the order of reaction. Hence, it does not affect the rate of reaction and such reactions are called pseudo first order reactions. Example : Hydrolysis of sucrose yielding glucose and fructose.

MEDIUM

12th ICSE

IMPORTANT

Earn 100

Illustrate graphically the effect of catalyst on activation energy.

Important Questions on Chemical Kinetics

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 26

Catalysts have no effect on the equilibrium constant, why?

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 27

The rate of a chemical reaction doubles for every

10 ° C

rise in temperature. If the temperature is raised from

10 ° C

30 ° C

, what will be the increase in the rate of the reaction?

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 28

The activation energy of a reaction is

94.14 kJ {mol}^{- 1}

and the value of rate constant at

298 K

18 x 10^{- 5} s^{- 1}

. Calculate the frequency factor or pre-exponential factor,

A

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 29

For a reaction, the energy of activation is zero. What is the value of rate constant at

300 K

, if

k = 1.6 x 10^{6} s^{- 1}

280 K

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 30

The data given is for the reaction, $N_{2} O_{5} (g) \to {NO}_{2} (g) + O_{2} (g)$ :

S.no.	$N_{2} O_{5} (mol L^{- 1})$	Rate of disappearance of $N_{2} O_{5}$ $(mol L^{- 1} \min^{- 1})$
$1$	$1.13 \times 10^{- 2}$	$34 \times 10^{- 5}$
$2$	$0.84 \times 10^{- 2}$	$25 \times 10^{- 5}$
$3$	$0.62 \times 10^{- 2}$	$18 \times 10^{- 5}$

Determine the order of reaction.

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 30

The data given is for the reaction, $N_{2} O_{5} (g) \to {NO}_{2} (g) + O_{2} (g)$ :

S.no.	$N_{2} O_{5} (mol L^{- 1})$	Rate of disappearance of $N_{2} O_{5}$ $(mol L^{- 1} \min^{- 1})$
$1$	$1.13 \times 10^{- 2}$	$34 \times 10^{- 5}$
$2$	$0.84 \times 10^{- 2}$	$25 \times 10^{- 5}$
$3$	$0.62 \times 10^{- 2}$	$18 \times 10^{- 5}$

Determine the rate law of reaction.

EASY

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 30

The data given is for the reaction, $N_{2} O_{5} (g) \to {NO}_{2} (g) + O_{2} (g)$ :

S.no.	$N_{2} O_{5} (mol L^{- 1})$	Rate of disappearance of $N_{2} O_{5}$ $(mol L^{- 1} \min^{- 1})$
$1$	$1.13 \times 10^{- 2}$	$34 \times 10^{- 5}$
$2$	$0.84 \times 10^{- 2}$	$25 \times 10^{- 5}$
$3$	$0.62 \times 10^{- 2}$	$18 \times 10^{- 5}$

Determine the rate constant of reaction.

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 31

Derive the equation for the rate constant of a first order reaction and show that the time required for the completion of half of the first order reaction is independent of initial concentration.

Illustrate graphically the effect of catalyst on activation energy.

Important Questions on Chemical Kinetics

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