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1

### AIPMT 2009

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

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