1

### 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%
2

### 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.
3

### 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
4

### AIPMT 2012 Prelims

In a zero-order reaction, for every 10oC rise of temperature, the rate is doubled. If the temperature is increased from 10oC to 100oC, the rate of the reaction will become
A
256 times
B
512 times
C
64 times
D
128 times

## Explanation

For energy 10° rise in temperature the rate of reaction doubles. So, rate = 2n
when, n = 1 rate = 21 = 2

when, temperature is increased from 10°C to 100°C change in temperature = 100 – 10 = 90°C

i.e., n = 9

So, rate = 29 = 512 times