Chemical Kinetics Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A zero-order reaction has a rate constant of \( 6.0 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \). If the initial concentration is \( 0.18 \, \text{mol L}^{-1} \), how long will it take for 33.33% decomposition?

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

33.33% decomposition means \( [\text{R}] = \frac{2}{3} \times 0.18 = 0.12 \, \text{mol L}^{-1} \).

\( t = \frac{0.18 - 0.12}{6.0 \times 10^{-3}} = \frac{0.06}{6.0 \times 10^{-3}} = 10 \, \text{min} \).

10 min
15 min
20 min
25 min
1

A reaction’s rate increases by 25 times when the concentration of the reactant is increased by 5 times. What is the order of the reaction?

Rate = \( k [\text{A}]^n \). New rate: \( k (5[\text{A}])^n = 25 k [\text{A}]^n \), so \( 5^n = 25 \).

\( 5^n = 5^2 \), \( n = 2 \).

2
1
3
1.5
1

A second-order reaction has a rate constant of \( 0.2 \, \text{L mol}^{-1} \text{min}^{-1} \) and an initial concentration of \( 0.1 \, \text{mol L}^{-1} \). What is the time for 75% completion?

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

75% completion means \( [\text{A}] = 0.25 \times 0.1 = 0.025 \, \text{mol L}^{-1} \).

\( \frac{1}{0.025} - \frac{1}{0.1} = 0.2 \times t \), \( 40 - 10 = 0.2 t \), \( t = \frac{30}{0.2} = 150 \, \text{min} \).

100 min
120 min
150 min
180 min
3

A first-order gaseous reaction has an initial pressure of 1.0 atm. After 50 s, the total pressure is 1.5 atm. What is the rate constant?

For \( \text{A} \to 2\text{B} \), \( p_t = p_i + x \), \( x = 1.5 - 1.0 = 0.5 \, \text{atm} \).

\( p_A = p_i - x = 1.0 - 0.5 = 0.5 \, \text{atm} \).

\( k = \frac{2.303}{50} \log \frac{1.0}{0.5} = \frac{2.303 \times 0.301}{50} \approx 0.0139 \, \text{s}^{-1} \).

0.0092 s\(^{-1}\)
0.0115 s\(^{-1}\)
0.0162 s\(^{-1}\)
0.0139 s\(^{-1}\)
4

A reaction has the rate law \( \text{Rate} = k[\text{A}]^{1.5}[\text{B}] \). What is the overall order?

Overall order = \( 1.5 + 1 = 2.5 \).

1.5
2
3
2.5
4

A second-order reaction has a half-life of 40 s when the initial concentration is \( 0.1 \, \text{mol L}^{-1} \). What is the rate constant?

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

Given: \( t_{1/2} = 40 \, \text{s} \), \( [\text{A}]_0 = 0.1 \, \text{mol L}^{-1} \).

\( k = \frac{1}{40 \times 0.1} = \frac{1}{4} = 0.25 \, \text{L mol}^{-1} \text{s}^{-1} \).

0.125 L mol\(^{-1}\) s\(^{-1}\)
0.25 L mol\(^{-1}\) s\(^{-1}\)
0.5 L mol\(^{-1}\) s\(^{-1}\)
1.0 L mol\(^{-1}\) s\(^{-1}\)
2

A second-order reaction has a half-life of 60 s when the initial concentration is \( 0.1 \, \text{mol L}^{-1} \). What is the rate constant?

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

Given: \( t_{1/2} = 60 \, \text{s} \), \( [\text{A}]_0 = 0.1 \, \text{mol L}^{-1} \).

\( k = \frac{1}{60 \times 0.1} = \frac{1}{6} \approx 0.1667 \, \text{L mol}^{-1} \text{s}^{-1} \).

0.1667 L mol\(^{-1}\) s\(^{-1}\)
0.0833 L mol\(^{-1}\) s\(^{-1}\)
0.25 L mol\(^{-1}\) s\(^{-1}\)
0.5 L mol\(^{-1}\) s\(^{-1}\)
1

A second-order reaction has a rate constant of \( 0.07 \, \text{L mol}^{-1} \text{s}^{-1} \) 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 \( [\text{A}] = 0.2 \times 0.25 = 0.05 \, \text{mol L}^{-1} \).

\( \frac{1}{0.05} - \frac{1}{0.25} = 0.07 \times t \), \( 20 - 4 = 0.07 t \), \( t = \frac{16}{0.07} \approx 228.57 \, \text{s} \).

200 s
250 s
214.29 s
228.57 s
4

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

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

New rate = \( k (2[\text{A}]) \left( \frac{[\text{B}]}{2} \right)^2 = k (2[\text{A}]) \left( \frac{[\text{B}]^2}{4} \right) = \frac{2}{4} k [\text{A}][\text{B}]^2 = 0.5 \times \text{initial rate} \).

Rate becomes half.

Doubles
Remains same
Becomes half
Quadruples
3

A second-order reaction has a rate constant of \( 0.05 \, \text{L mol}^{-1} \text{s}^{-1} \) and an initial concentration of \( 0.4 \, \text{mol L}^{-1} \). What is the time for 50% completion?

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

Given: \( k = 0.05 \, \text{L mol}^{-1} \text{s}^{-1} \), \( [\text{A}]_0 = 0.4 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.05 \times 0.4} = \frac{1}{0.02} = 50 \, \text{s} \).

25 s
40 s
60 s
50 s
4

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