1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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.

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