1

### 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.
2

### 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
3

### AIPMT 2003

If the rate of the reaction is equal to the rate constant, the order of the reaction is
A
0
B
1
C
2
D
3

## Explanation

As r = k[A]n

if n = 0

r = k[A]0

or r = k thus for zero order reactions rate is equal to the rate constant.
4

### AIPMT 2003

The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equation, $k = A \cdot {e^{ - E{}^ * /RT}}$. Activation energy (E$*$) of the reaction can be calculated by plotting
A
$k\,\,vs\,\,T$
B
$k\,\,vs\,\,{1 \over {\log T}}$
C
$\log \,k\,\,vs\,\,{1 \over T}$
D
$\log \,k\,\,vs\,{1 \over {\log T}}$

## Explanation

Arrhenius equation k = $A{e^{ - {{{E_a}} \over {RT}}}}$

$\Rightarrow$ log k = log A - ${{{{E_a}} \over {2.303RT}}}$

Comparing it with equation of straight line i.e.,

y = mx + C

On plotting log k vs ${1 \over T}$, we get a straight line, the slope indicates the value of activation energy.