1

### JEE Main 2016 (Online) 10th April Morning Slot

The rate law for the reaction below is given by the expression k [A] [B]

A + B $\to$ Product

If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be :
A
k
B
k/3
C
3k
D
9k

## Explanation

Rate constant only depends on temperature only and it is independed of concentration of reactants.
2

### JEE Main 2017 (Offline)

Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to :
(R = 8.314 J mol–1 K–1)
A
12
B
6
C
4
D
8

## Explanation

We know, from arrhenius equation,

k = A.${e^{{{ - {E_a}} \over {RT}}}}$

$\therefore$ k1 = A.${e^{{{ - {E_{{a_1}}}} \over {RT}}}}$ ......(1)

k2 = A.${e^{{{ - {E_{{a_2}}}} \over {RT}}}}$ ......(2)

On dividing equation (2) by (1), we get

${{{k_2}} \over {{k_1}}} = {e^{{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}}}$

$\Rightarrow$ $\ln \left( {{{{k_2}} \over {{k_1}}}} \right) = {{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}$ = ${{10,000} \over {8.314 \times 300}}$ = 4
3

### JEE Main 2017 (Online) 8th April Morning Slot

The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be Increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.
A
9.84 K
B
4.92 K
C
2.45 K
D
19.67 K

## Explanation

For reaction A, T1 = 300 K, T2 = 310 K, k2 = 2 k1

$\log {{{k_2}} \over {{k_1}}} = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$

$\therefore$ $\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$ ...(1)

For reaction B, T1 = 300 K, T2 = ?, k2 = 2k1, Ea2 = 2Ea1

$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right]$

From eq. (i) and (ii), we get

${{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$

$\Rightarrow$ $2\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = \left[ {{1 \over {300}} - {1 \over {310}}} \right]$

$\Rightarrow$ T2 = ${{{300 \times 310} \over {610}}}$ $\times$ 2 = 304.92 K

$\therefore$ Increased temperature = (304.92 – 300) = 4.92 K
4

### JEE Main 2017 (Online) 9th April Morning Slot

The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :

(Assume activation energy and preexponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol−1 K−1)
A
107.2 kJ mol$-$1
B
53.6 kJ mol$-$1
C
26.8 kJ mol$-$1
D
214.4 kJ mol$-$1