1

### AIPMT 2007

In a first-order reaction A $\to$ B, if k is rate constant and initial concentration of the reactant A is 0.5 M, then the half-life is
A
${{\log 2} \over k}$
B
${{\log 2} \over {k\sqrt {0.5} }}$
C
${{\ln 2} \over k}$
D
${{0.693} \over {0.5k}}$

## Explanation

For first order reaction

k = ${{2.303} \over t}\log {a \over {a - x}}$

at ${t_{1/2}}$, x = ${a \over 2}$

${t_{1/2}}$ = ${{2.303} \over k}\log {a \over {a - {a \over 2}}}$

= ${{\ln 2} \over k}$
2

### AIPMT 2007

If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately
(log 4 = 0.60, log 5 = 0.69)
A
45 minutes
B
60 minutes
C
40 minutes
D
50 minutes

## Explanation

For a first order reaction,

k = ${{2.303} \over t}\log {a \over {a - x}}$

k = ${{2.303} \over {60}}\log {{100} \over {40}}$

= ${{2.303} \over {60}}\log 2.5$

= 0.0153

Also, ${t_{1/2}}$ = ${{2.303} \over k}\log {{100} \over {50}}$

= ${{2.303} \over {0.0153}}\log 2$

= 45 min.
3

### AIPMT 2006

For the reaction, 2A + B $\to$ 3C + D, which of the following does not express the reaction rate?
A
$- {{d\left[ A \right]} \over {2dt}}$
B
$- {{d\left[ C \right]} \over {3dt}}$
C
$- {{d\left[ B \right]} \over {dt}}$
D
${{d\left[ D \right]} \over {dt}}$

## Explanation

Given,

2A + B $\to$ 3C + D

Rate of reaction =

$- {1 \over 2}{{d\left[ A \right]} \over {dt}} = - {{d\left[ B \right]} \over {dt}}$ = ${1 \over 3}{{d\left[ C \right]} \over {dt}} = {{d\left[ C \right]} \over {dt}}$

4

### AIPMT 2006

Consider the reaction :  N2(g) + 3H2(g) $\to$ 2NH3(g)

The equality relationship between ${{d\left[ {N{H_3}} \right]} \over {dt}}$ and $- {{d\left[ {{H_2}} \right]} \over {dt}}$ is
A
${{d\left[ {N{H_3}} \right]} \over {dt}} = - {{d\left[ {{H_2}} \right]} \over {dt}}$
B
${{d\left[ {N{H_3}} \right]} \over {dt}} = - {1 \over 3}{{d\left[ {{H_2}} \right]} \over {dt}}$
C
$+ {{d\left[ {N{H_3}} \right]} \over {dt}} = - {2 \over 3}{{d\left[ {{H_2}} \right]} \over {dt}}$
D
$+ {{d\left[ {N{H_3}} \right]} \over {dt}} = - {3 \over 2}{{d\left[ {{H_2}} \right]} \over {dt}}$

## Explanation

N2(g) + 3H2(g) $\to$ 2NH3(g)

Rate = ${{ - d\left[ {{N_2}} \right]} \over {dt}} = - {1 \over 3}{{d\left[ {{H_2}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}$

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