1
MCQ (Single Correct Answer)

### AIPMT 2005

The rate of reaction between two reactions A and B decreases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is
A
2
B
$-$2
C
1
D
$-$1

## Explanation

A + B $\to$ Product

Rate $\propto$ [A]x [B]y .......(1)

The rate of the reaction decreases by a factor of 4 if the concentration of reactant B is doubled.

${r \over 4}$ $\propto$ [A]x [2B]y ......(2)

From equation (1) and (2), we get

${\left( {{1 \over 2}} \right)^y} = 4$

$\Rightarrow$ y = -2

$\therefore$ Order of this reaction with respect to reactant B is -2.
2
MCQ (Single Correct Answer)

### AIPMT 2004

The rate of a first order reaction is 1.5 $\times$ 10$-$2 mol L$-$1 min$-$1 at 0.5 M concentration of the reactant. The half-life of the reaction is
A
0.383 min
B
23.1 min
C
8.73 min
D
7.53 min

## Explanation

For a first order reaction, A $\to$ products

Rate(r) = k[A]

$\Rightarrow$ k = ${r \over {\left[ A \right]}}$

$\Rightarrow$ k = ${{1.5 \times {{10}^{ - 2}}} \over {0.5}} = 3 \times {10^{ - 2}}$

So, ${t_{1/2}} = {{0.693} \over {3 \times {{10}^{ - 2}}}}$ = 23.1 min
3
MCQ (Single Correct Answer)

### AIPMT 2003

The activation energy for a simple chemical reaction A $\rightleftharpoons$ B is E$a$ in forward direction.
The activation energy for reverse reaction
A
is negative of E$a$
B
is always less than E$a$
C
can be less than or more than E$a$
D
is always double of E$a$

## Explanation

The energy of activation of reverse reaction is less than or more than energy of activation E$a$ of forward reaction

$\Delta H = {\left( {{E_a}} \right)_F} - {\left( {{E_a}} \right)_R}$

Because it depends upon the nature of reaction.

If ${\left( {{E_a}} \right)_F} > {\left( {{E_a}} \right)_R}$, reaction is endothermic.

or ${\left( {{E_a}} \right)_F} < {\left( {{E_a}} \right)_R}$, reaction is exothermic.
4
MCQ (Single Correct Answer)

### AIPMT 2003

The reaction A $\to$ B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B?
A
1 hour
B
0.5 hour
C
0.25 hour
D
2 hours

## Explanation

For first order reaction

k = ${{2.303} \over t}\log {{{{\left[ A \right]}_0}} \over {{{\left[ A \right]}_t}}}$

$\Rightarrow$ k = ${{2.303} \over 1}\log {{0.8} \over {0.2}}$ = 2.303 log 4 ....(1)

Let t1 hour is required for changing the concentration of A from 0.9 mole to 0.675 mole of B.

Remaining mole of A = 0.9 – 0.675 = 0.225

$\therefore$ k = ${{2.303} \over {{t_1}}}\log {{0.9} \over {0.225}}$ .....(2)

From equation (1) and (2)

${{2.303} \over {{t_1}}}\log {{0.9} \over {0.225}}$ = 2.303 log 4

$\Rightarrow$ ${{t_1}}$ = 1 hr

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