1

### AIPMT 2015

The rate constant of the reaction A $\to$ B is 0.6 $\times$ 10$-$3 mol L$-$1 s$-$1. If the concentration of A is 5 M, then concentration of B after 20 minutes is
A
3.60 M
B
0.36 M
C
0.72 M
D
1.08 M

## Explanation

For zero order reaction unit of rate constant is mole per second.

$\therefore$ For zero order

x = kt

$\Rightarrow$ x = 0.6 × 10–3 × 20 × 60 = 0.72 M
2

### NEET 2013 (Karnataka)

A reaction is 50% complete in 2 hours and 75% complete in 4 hours. The order of reaction is
A
1
B
2
C
3
D
0

## Explanation

For a first order reaction,

t75% = 2 $\times$ t50%
3

### NEET 2013 (Karnataka)

For a reaction between A and B the order with respect to A is 2 and the other with respect to B is 3. The concentrations of both A and B are doubled, the rate will increase by a factor of
A
12
B
16
C
32
D
10

## Explanation

Rate1 = k[A]2[B]3

when concentrations of both A and B are doubled then

Rate2 = k[2A]2[2B]3 = 32 k[A]2[B]3

$\therefore$ Rate will increase by a factor of 32.
4

### NEET 2013

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20oC to 35oC?
(R = 8.314 J mol$-$1 K$-$1)
A
34.7 kJ mol$-$1
B
15.1 kJ mol$-$1
C
342 kJ mol$-$1
D
269 kJ mol$-$1

## Explanation

$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$

Initial temperature, T1 = 20 + 273 = 293 K

Final temperature, T2 = 35 + 273 = 308 K

R = 8.314 JK–1 mol–1

As rate becomes double on raising temperature

$\therefore$ r2 = 2r1

As rate constant, k $\infty$ r

k2 = 2k1

$\therefore$ $\log 2 = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {293}} - {1 \over {308}}} \right)$

$\Rightarrow$ $0.301 = {{{E_a}} \over {19.147}} \times {{15} \over {293 \times 308}}$

$\Rightarrow$ E$a$ = 34673 J mol–1 = 34.7 kJ mol–1