1
MCQ (Single Correct Answer)

AIPMT 2005

For a first order reaction A $$ \to $$ B the reaction rate a reactant concentration of 0.01 M is found to be 2.0 $$ \times $$ 10$$-$$5 mol L$$-$$1 s$$-$$1. The half-life period of the reaction is
A
30 s
B
220 s
C
300 s
D
347 s

Explanation

Given [A] = 0.01 M

Rate = 2.0 × 10–5 mol L–1 S–1

For a first order reaction

Rate = k[A]

k = $${{2 \times {{10}^{ - 5}}} \over {0.01}} = 2 \times {10^{ - 3}}$$

$$ \Rightarrow $$ $${t_{1/2}} = {{0.693} \over {2 \times {{10}^{ - 3}}}} = 347\,\sec $$
2
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.
3
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
4
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.

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