Chemical Kinetics Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A second-order reaction has a rate constant of \( 0.02 \, \text{L mol}^{-1} \text{min}^{-1} \) and an initial concentration of \( 0.5 \, \text{mol L}^{-1} \). What is the half-life?

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

Given: \( k = 0.02 \, \text{L mol}^{-1} \text{min}^{-1} \), \( [\text{A}]_0 = 0.5 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.02 \times 0.5} = \frac{1}{0.01} = 100 \, \text{min} \).

50 min
75 min
100 min
125 min
3

A reaction has a rate constant of \( 1.5 \times 10^{-2} \, \text{s}^{-1} \). What is its order if the unit of the rate constant is \( \text{s}^{-1} \)?

For a first-order reaction, the unit of \( k \) is \( \text{s}^{-1} \).

Given: \( k = 1.5 \times 10^{-2} \, \text{s}^{-1} \), which matches the unit for a first-order reaction.

Thus, the order is 1.

1
0
2
3
1

The rate law of a reaction is \( \text{Rate} = k[\text{A}]^{3/2} \). What is the overall order?

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

Given: \( \text{Rate} = k[\text{A}]^{3/2} \), order = \( \frac{3}{2} = 1.5 \).

1
2
3
1.5
4

A second-order reaction has a rate constant of \( 0.02 \, \text{L mol}^{-1} \text{s}^{-1} \) and an initial concentration of \( 0.25 \, \text{mol L}^{-1} \). What is the time for 80% completion?

For second-order, \( \frac{1}{[\text{A}]} - \frac{1}{[\text{A}]_0} = kt \).

80% completion means \( [\text{A}] = 0.2 \times 0.25 = 0.05 \, \text{mol L}^{-1} \).

\( \frac{1}{0.05} - \frac{1}{0.25} = 0.02 \times t \), \( 20 - 4 = 0.02 t \), \( t = \frac{16}{0.02} = 800 \, \text{s} \).

400 s
600 s
800 s
1000 s
3

A reaction has the rate law \( \text{Rate} = k[\text{A}][\text{B}] \). If the concentration of A is tripled and B is halved, what happens to the rate?

Initial rate = \( k [\text{A}][\text{B}] \).

New rate = \( k (3[\text{A}]) \left( \frac{[\text{B}]}{2} \right) = k \times 3 \times \frac{1}{2} [\text{A}][\text{B}] = 1.5 k [\text{A}][\text{B}] \).

Rate increases by 1.5 times.

Triples
Halves
Increases by 1.5 times
Remains same
3

For a reaction, the initial rate doubles when the concentration of the reactant is quadrupled. What is the order of the reaction?

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

New rate: \( r_2 = k (4[\text{A}])^n = 2 r_1 \), so \( 4^n = 2 \).

\( n \log 4 = \log 2 \), \( n \times 0.602 = 0.301 \), \( n = 0.5 \).

1
2
0
0.5
4

A reaction follows second-order kinetics with a rate constant of \( 0.02 \, \text{L mol}^{-1} \text{s}^{-1} \). If the initial concentration is \( 0.5 \, \text{mol L}^{-1} \), what is the half-life?

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

Given: \( k = 0.02 \, \text{L mol}^{-1} \text{s}^{-1} \), \( [\text{A}]_0 = 0.5 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.02 \times 0.5} = \frac{1}{0.01} = 100 \, \text{s} \).

50 s
75 s
100 s
125 s
2

A first-order reaction has a half-life of 20 minutes. How long will it take for 87.5% of the reactant to decompose?

87.5% decomposition means 12.5% remains, \( \frac{[\text{R}]}{[\text{R}]_0} = 0.125 = \left( \frac{1}{2} \right)^3 \).

Number of half-lives = 3, \( t = 3 \times t_{1/2} = 3 \times 20 = 60 \, \text{min} \).

40 min
50 min
70 min
60 min
4

A first-order reaction has a rate constant of \( 0.0289 \, \text{min}^{-1} \). What percentage of the reactant decomposes in 48 minutes?

\( \log \frac{[\text{R}]_0}{[\text{R}]} = \frac{k t}{2.303} \).

\( \log \frac{[\text{R}]_0}{[\text{R}]} = \frac{0.0289 \times 48}{2.303} \approx 0.602 \), \( \frac{[\text{R}]_0}{[\text{R}]} = 10^{0.602} \approx 4 \).

Fraction remaining = \( \frac{1}{4} = 0.25 \), percentage decomposed = \( (1 - 0.25) \times 100 = 75\% \).

50%
60%
80%
75%
4

A first-order reaction has a half-life of 15 minutes. What percentage of the reactant decomposes in 45 minutes?

Number of half-lives = \( \frac{45}{15} = 3 \).

Fraction remaining = \( \left( \frac{1}{2} \right)^3 = \frac{1}{8} = 0.125 \).

Percentage decomposed = \( (1 - 0.125) \times 100 = 87.5\% \).

50%
75%
90%
87.5%
4

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