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  4. Chemical Kinetics Chapter-Wise Test 2

Chemical Kinetics Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A reaction has the rate law \( \text{Rate} = k[\text{A}]^2[\text{B}]^{1/2} \). If the concentration of A is doubled and B is quadrupled, what happens to the rate?

Initial rate = \( k [\text{A}]^2 [\text{B}]^{1/2} \).

New rate = \( k (2[\text{A}])^2 (4[\text{B}])^{1/2} = k \times 4 [\text{A}]^2 \times 2 [\text{B}]^{1/2} = 8 k [\text{A}]^2 [\text{B}]^{1/2} \).

Rate increases by 8 times.

Increases by 4 times
Increases by 8 times
Increases by 16 times
Doubles
2

A reaction has the rate law \( \text{Rate} = k[\text{A}][\text{B}]^{1/2} \). If the concentration of A is doubled and B is quadrupled, what happens to the rate?

Initial rate = \( k [\text{A}] [\text{B}]^{1/2} \).

New rate = \( k (2[\text{A}]) (4[\text{B}])^{1/2} = k \times 2 [\text{A}] \times 2 [\text{B}]^{1/2} = 4 k [\text{A}] [\text{B}]^{1/2} \).

Rate increases by 4 times.

Doubles
Triples
Increases by 4 times
Increases by 8 times
3

The molecularity of the elementary step \( \text{H}_2 + \text{Br}_2 \to 2\text{HBr} \) is:

Molecularity is the number of molecules in an elementary step.

For \( \text{H}_2 + \text{Br}_2 \), two molecules collide, so molecularity = 2.

1
3
2
4
3

A first-order reaction is 25% complete in 10 minutes. What is the rate constant?

For first-order, \( k = \frac{2.303}{t} \log \frac{[\text{R}]_0}{[\text{R}]} \).

25% complete means 75% remains, \( \frac{[\text{R}]}{[\text{R}]_0} = 0.75 \), \( \frac{[\text{R}]_0}{[\text{R}]} = \frac{1}{0.75} \approx 1.333 \).

\( k = \frac{2.303}{10} \log 1.333 = \frac{2.303 \times 0.125}{10} \approx 0.0288 \, \text{min}^{-1} \).

0.0231 min\(^{-1}\)
0.0346 min\(^{-1}\)
0.0288 min\(^{-1}\)
0.0192 min\(^{-1}\)
3

A second-order reaction has a rate constant of \( 0.04 \, \text{L mol}^{-1} \text{min}^{-1} \) and an initial concentration of \( 0.25 \, \text{mol L}^{-1} \). What is the half-life?

For second-order, \( t_{1/2} = \frac{1}{k [\text{A}]_0} \).

Given: \( k = 0.04 \, \text{L mol}^{-1} \text{min}^{-1} \), \( [\text{A}]_0 = 0.25 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.04 \times 0.25} = \frac{1}{0.01} = 100 \, \text{min} \).

50 min
75 min
100 min
125 min
3

The rate of a reaction triples when the concentration of the reactant is increased by 9 times. What is the order of the reaction?

Rate = \( k [\text{A}]^n \). Initial rate: \( r_1 = k [\text{A}]^n \).

New rate: \( r_2 = k (9[\text{A}])^n = 3 r_1 \), so \( 9^n = 3 \).

\( n \log 9 = \log 3 \), \( n \times 0.954 = 0.477 \), \( n \approx 0.5 \).

0.5
1
2
1.5
1

The rate constant of a reaction doubles when the temperature increases from 300 K to 310 K. What is the approximate activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

Using \( \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \).

Given: \( \frac{k_2}{k_1} = 2 \), \( T_1 = 300 \, \text{K} \), \( T_2 = 310 \, \text{K} \).

\( \log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{300 \times 310} \right) \).

\( 0.301 = \frac{E_a \times 10}{19.147 \times 93000} \), \( E_a \approx 52,900 \, \text{J mol}^{-1} \approx 52.9 \, \text{kJ mol}^{-1} \).

26.5 kJ mol\(^{-1}\)
52.9 kJ mol\(^{-1}\)
79.3 kJ mol\(^{-1}\)
105.8 kJ mol\(^{-1}\)
2

The rate law for a reaction is \( \text{Rate} = k[\text{A}]^2[\text{B}] \). What is the overall order of the reaction?

Overall order = sum of the powers of concentration terms in the rate law.

Given: \( \text{Rate} = k[\text{A}]^2[\text{B}]^1 \).

Order = \( 2 + 1 = 3 \).

3
2
1
4
1

A second-order reaction has a rate constant of \( 0.5 \, \text{L mol}^{-1} \text{min}^{-1} \). If the initial concentration is \( 0.2 \, \text{mol L}^{-1} \), what is the half-life?

For second-order, \( t_{1/2} = \frac{1}{k [\text{A}]_0} \).

Given: \( k = 0.5 \, \text{L mol}^{-1} \text{min}^{-1} \), \( [\text{A}]_0 = 0.2 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.5 \times 0.2} = \frac{1}{0.1} = 10 \, \text{min} \).

5 min
10 min
15 min
20 min
2

For a first-order reaction, the time for 99% completion is approximately how many times the half-life?

For 99% completion, \( \frac{[\text{R}]_0}{[\text{R}]} = 100 \), \( t = \frac{2.303}{k} \log 100 = \frac{4.606}{k} \).

Half-life \( t_{1/2} = \frac{0.693}{k} \).

\( \frac{t}{t_{1/2}} = \frac{4.606}{0.693} \approx 6.64 \approx 7 \) (rounded for NEET approximation).

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7
4

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