The rate constant of a reaction is 3 × 10 - 3 min - 1 at 27 ° C and at 47 ° C its value is 9 × 10 - 3 min - 1 , then calculate the energy of activation of the reaction. What will be its rate constant at 308 K ?

Given, For a first order reaction, k 1 = 3 × 10 - 3 min - 1 at T 1 = 27 ° C = 300 K , E a = ? , T 2 = 47 ° C = 320 K and k 2 = 9 × 10 - 3 min - 1 We know that, log k 2 k 1 = E a 2 . 303 R T 2 - T 1 T 1 T 2 log 9 × 10 - 3 3 × 10 - 3 = E a 2 . 303 × 8 . 314 320 - 300 320 × 300 Therefore, E a = 439 . 09 kJ . Again, logk 2 - logk 1 = E a 2 . 303 R T 2 - T 1 T 1 T 2 , therefore, logk 2 = E a 2 . 303 R T 2 - T 1 T 1 T 2 + logk 1 = 43909 2 . 303 × 8 . 314 308 - 300 308 × 300 + log 3 × 10 - 3 = - 2 . 324 Thus, k 2 = antilog - 2 . 324 = 4 . 742 × 10 - 3 min - 1 .

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Define activated complex.

Important Questions on Chemical Kinetics

EASY

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

An exothermic reaction

X \to Y

has an activation energy

30 kJ {mol}^{- 1}

. If energy change

ΔE

during the reaction is

- 20 kJ

, then the activation energy for the reverse reaction in

kJ

is (Integer answer)

HARD

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

If the first order rate constants for the decomposition of ethyl iodide at

600 K

and

700 K

are

2.0 \times 10^{- 5} s^{- 1}

and

2.0 \times 10^{- 4} s^{- 1}

respectively, the activation energy for this reaction in

{kJmol}^{- 1}

EASY

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

The unit of activation energy is

MEDIUM

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

Higher order

(> 3)

reactions are rare due to

EASY

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

Define activation energy.

HARD

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

Define activation energy.

MEDIUM

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

The specific rate of a reaction quadruples when temperature changes from

30^{o} C

50^{o} C

. Calculate energy of activation of the reaction in

{kJmol}^{- 1}

(to the closest integer).
(

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

)

MEDIUM

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

The rate constant of a reaction is

3 \times 10^{- 3} \min^{- 1}

27^{°} C

and at

47^{°} C

its value is

9 \times 10^{- 3} \min^{- 1}

, then calculate the energy of activation of the reaction. What will be its rate constant at

308 K

EASY

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

What is activation energy of a reaction?

MEDIUM

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

For an endothermic reaction, where

ΔH

represent the enthalpy of reaction in

kJ {mol}^{- 1}

, the minimum value for energy of activation (for forward reaction) will be

MEDIUM

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

For the reaction

A \to B,

the rate constant

k

(in

s^{- 1}

) is given by

\log_{10} k = 20.35 - \frac{(2.47 \times 10^{3})}{T}

The energy of activation in

kJ {mol}^{- 1}

is ________ . (Nearest integer)
[Given :

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

]

EASY

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

An increase in the concentration of the reactants of a reaction leads to change in

MEDIUM

Chemistry>Physical Chemistry>Chemical Kinetics>Collision Theory and Activation Energy

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.

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