1

### AIPMT 2008

The rate constants k1 and k2 for two different reactions are 1016 $\cdot$ e$-$2000/T and 1015 $\cdot$ e$-$1000/T, respectively.
The temperature at which k1 = k2 is
A
2000 K
B
${{1000} \over {2.303}}K$
C
1000 K
D
${{2000} \over {2.303}}K$

## Explanation

k1 = 1016 ${e^{ - {{2000} \over T}}}$

k2 = 1015 ${e^{ - {{1000} \over T}}}$

On taking log of both the equations, we get

$\log {k_1} = 16 - {{2000} \over {2.303T}}$

$\log {k_2} = 15 - {{1000} \over {2.303T}}$

As k1 = k2

$16 - {{2000} \over {2.303T}}$ = $15 - {{1000} \over {2.303T}}$

$\Rightarrow$ T = ${{1000} \over {2.303}}K$
2

### AIPMT 2007

The reaction of hydrogen and iodine monochloride is given as :
H2(g) + 2ICl(g) $\to$ 2HCl(g) + I2(g)
This reaction is of first order with respect to H2(g) and ICl(g),
following mechanisms were proposed.

Mechanism A :
H2(g) + 2ICl(g) $\to$ 2HCl(g) + I2(g)
Mechanism B :
H2(g) + ICl(g) $\to$ HCl(g) + HI(g) ; slow
HI(g) + ICl(g) $\to$ HCl(g) + I2(g) ; fast

Which of the above mechanism(s) can be consistent with the given information about the reaction?
A
A and B both
B
Neither A nor B
C
A only
D
B only

## Explanation

The slow step is the rate determining step and it involves 1 molecule of H2(g) and 1 molecule of ICl(g) . Hence the rate will be,

r = k[H2(g)] [ICl(g)]

$\therefore$ The reaction is 1st order with respect to H2(g) and ICl(g).
3

### 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 = ${a \over 2}$

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

= ${{\ln 2} \over k}$
4

### 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.