Chemical Kinetics Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The half-life of a second-order reaction is 25 s when the initial concentration is \( 0.4 \, \text{mol L}^{-1} \). What is the rate constant?

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

Given: \( t_{1/2} = 25 \, \text{s} \), \( [\text{A}]_0 = 0.4 \, \text{mol L}^{-1} \).

\( k = \frac{1}{25 \times 0.4} = \frac{1}{10} = 0.1 \, \text{L mol}^{-1} \text{s}^{-1} \).

0.1 L mol\(^{-1}\) s\(^{-1}\)
0.2 L mol\(^{-1}\) s\(^{-1}\)
0.05 L mol\(^{-1}\) s\(^{-1}\)
0.4 L mol\(^{-1}\) s\(^{-1}\)
1

The rate constant of a reaction is \( 3.0 \times 10^{-2} \, \text{min}^{-1} \) at 300 K. If the activation energy is 60 kJ/mol, what is the rate constant at 310 K (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log \frac{k_2}{3.0 \times 10^{-2}} = \frac{60000}{2.303 \times 8.314} \left( \frac{10}{300 \times 310} \right) \approx 0.337 \).

\( \frac{k_2}{3.0 \times 10^{-2}} = 10^{0.337} \approx 2.17 \), \( k_2 \approx 6.51 \times 10^{-2} \, \text{min}^{-1} \).

4.0 \(\times\) 10\(^{-2}\) min\(^{-1}\)
5.0 \(\times\) 10\(^{-2}\) min\(^{-1}\)
6.5 \(\times\) 10\(^{-2}\) min\(^{-1}\)
8.0 \(\times\) 10\(^{-2}\) min\(^{-1}\)
3

A second-order reaction has a rate constant of \( 0.04 \, \text{L mol}^{-1} \text{s}^{-1} \) and an initial concentration of \( 0.2 \, \text{mol L}^{-1} \). What is the half-life?

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

Given: \( k = 0.04 \, \text{L mol}^{-1} \text{s}^{-1} \), \( [\text{A}]_0 = 0.2 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.04 \times 0.2} = \frac{1}{0.008} = 125 \, \text{s} \).

125 s
100 s
150 s
200 s
1

A second-order reaction has a rate constant of \( 0.03 \, \text{L mol}^{-1} \text{s}^{-1} \) 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: \( k = 0.03 \, \text{L mol}^{-1} \text{s}^{-1} \), \( [\text{A}]_0 = 0.25 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.03 \times 0.25} = \frac{1}{0.0075} \approx 133.33 \, \text{s} \).

133.33 s
100 s
150 s
200 s
1

A reaction’s rate increases by 3 times when the temperature rises from 290 K to 300 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log 3 = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{290 \times 300} \right) \).

\( 0.477 = \frac{E_a \times 10}{19.147 \times 87000} \), \( E_a \approx 79,400 \, \text{J mol}^{-1} \approx 79.4 \, \text{kJ mol}^{-1} \).

39.7 kJ mol\(^{-1}\)
59.6 kJ mol\(^{-1}\)
79.4 kJ mol\(^{-1}\)
99.2 kJ mol\(^{-1}\)
3

For a first-order reaction, how many half-lives are required for 75% completion?

For a first-order reaction, after \( n \) half-lives, \( [\text{R}] = [\text{R}]_0 \left( \frac{1}{2} \right)^n \).

75% completion means 25% remains, so \( \frac{[\text{R}]}{[\text{R}]_0} = 0.25 = \left( \frac{1}{2} \right)^2 \).

Thus, \( n = 2 \) half-lives.

1
2
3
4
2

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

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

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

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

50 s
100 s
75 s
125 s
2

A reaction has a rate constant with units \( \text{mol}^{-1} \text{L} \text{s}^{-1} \). What is the order of the reaction?

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

Given unit: \( \text{mol}^{-1} \text{L} \text{s}^{-1} = (\text{mol L}^{-1})^{-1} \text{s}^{-1} \).

\( 1 - n = -1 \), \( n = 2 \). Thus, the order is 2.

2
1
0
3
1

The rate of a reaction increases by 8 times when the concentration of the reactant is doubled. What is the order of the reaction?

Rate = \( k [\text{A}]^n \). New rate: \( k (2[\text{A}])^n = 8 k [\text{A}]^n \), so \( 2^n = 8 \).

\( 2^n = 2^3 \), \( n = 3 \).

3
2
1
4
1

A first-order reaction has a rate constant of \( 0.0173 \, \text{min}^{-1} \). What is the time for 80% completion?

80% completion means \( \frac{[\text{R}]}{[\text{R}]_0} = 0.2 \), \( \frac{[\text{R}]_0}{[\text{R}]} = 5 \).

\( t = \frac{2.303}{k} \log 5 = \frac{2.303 \times 0.699}{0.0173} \approx 93 \, \text{min} \).

60 min
80 min
100 min
93 min
4

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