1

### AIPMT 2009

Half-life period of a first order reaction is 1386 seconds. The specific rate constant of the reaction is
A
0.5 $\times$ 10$-$2 s$-$1
B
0.5 $\times$ 10$-$3 s$-$1
C
5.0 $\times$ 10$-$2 s$-$1
D
5.0 $\times$ 10$-$3 s$-$1.

## Explanation

Specific rate constant

k = ${{0.693} \over {{t_{1/2}}}}$

= ${{0.693} \over {1386}}$

= 0.5 $\times$ 10-3 sec-1
2

### AIPMT 2009

For the reaction A + B $\to$ products, it is observed that

(i)  on doubling the initial concentration of A only, the rate of reaction is also doubled and
(ii)  on doubling the initial concentration of both A and B, there is a change by a factor of 8 in the rate of the reaction.

The rate of this reaction is given by
A
rate = k[A]2 [B]2
B
rate = k[A] [B]2
C
rate = k[A] [B]
D
rate = k[A]2 [B]

## Explanation

R = k[A]m[B]n ... (i)

2R = k[2A]m[B]n ... (ii)

8R = k[2A]m[2B]n ... (iii)

from (i), (ii) and (iii), m = 1, n = 2

So, rate = k[A][B]2
3

### AIPMT 2009

In the reaction,
BrO$_{3(aq)}^ -$ + 5Br$_{(aq)}^ -$ + 6H+ $\to$ 3Br2(l) + 3H2O(l).
The rate of appearance of bromine (Br2) is related to rate of disappearance of bromide ions as
A
${{d\left[ {B{r_2}} \right]} \over {dt}} = - {5 \over 3}{{d\left[ {B{r^ - }} \right]} \over {dt}}$
B
${{d\left[ {B{r_2}} \right]} \over {dt}} = {5 \over 3}{{d\left[ {B{r^ - }} \right]} \over {dt}}$
C
${{d\left[ {B{r_2}} \right]} \over {dt}} = {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}$
D
${{d\left[ {B{r_2}} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}$

## Explanation

Rate = ${1 \over 3}{{d\left[ {B{r_2}} \right]} \over {dt}} = - {1 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}$

$\therefore$ ${{d\left[ {B{r_2}} \right]} \over {dt}} = - {3 \over 5}{{d\left[ {B{r^ - }} \right]} \over {dt}}$
4

### AIPMT 2009

For the reaction, N2 + 3H2 $\to$ 2NH3, if
${{d\left[ {N{H_3}} \right]} \over {dt}}$ = 2 $\times$ 10$-$4 mol L$-$1 s$-$1,
the value of ${{ - d\left[ {{H_2}} \right]} \over {dt}}$ would be
A
4 $\times$ 10$-$4 mol L$-$1 s$-$1
B
6 $\times$ 10$-$4 mol L$-$1 s$-$1
C
1 $\times$ 10$-$4 mol L$-$1 s$-$1
D
3 $\times$ 10$-$4 mol L$-$1 s$-$1

## Explanation

N2 + 3H2 $\to$ 2NH3

$- {1 \over 3}{{d\left[ {{H_2}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}$

$\Rightarrow$ $- {{d\left[ {{H_2}} \right]} \over {dt}} = {3 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}$

= ${3 \over 2} \times 2 \times {10^{ - 4}}$

= $3 \times {10^{ - 4}}$ mol L$-$1 s$-$1