1
MCQ (Single Correct Answer)

AIPMT 2005

The rate of reaction between two reactions A and B decreases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is
A
2
B
$$-$$2
C
1
D
$$-$$1

Explanation

A + B $$ \to $$ Product

Rate $$ \propto $$ [A]x [B]y .......(1)

The rate of the reaction decreases by a factor of 4 if the concentration of reactant B is doubled.

$${r \over 4}$$ $$ \propto $$ [A]x [2B]y ......(2)

From equation (1) and (2), we get

$${\left( {{1 \over 2}} \right)^y} = 4$$

$$ \Rightarrow $$ y = -2

$$ \therefore $$ Order of this reaction with respect to reactant B is -2.
2
MCQ (Single Correct Answer)

AIPMT 2004

The rate of a first order reaction is 1.5 $$ \times $$ 10$$-$$2 mol L$$-$$1 min$$-$$1 at 0.5 M concentration of the reactant. The half-life of the reaction is
A
0.383 min
B
23.1 min
C
8.73 min
D
7.53 min

Explanation

For a first order reaction, A $$ \to $$ products

Rate(r) = k[A]

$$ \Rightarrow $$ k = $${r \over {\left[ A \right]}}$$

$$ \Rightarrow $$ k = $${{1.5 \times {{10}^{ - 2}}} \over {0.5}} = 3 \times {10^{ - 2}}$$

So, $${t_{1/2}} = {{0.693} \over {3 \times {{10}^{ - 2}}}}$$ = 23.1 min
3
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.
4
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

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