Chemical Kinetics Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A zero-order reaction has an initial concentration of \( 0.12 \, \text{mol L}^{-1} \) and a rate constant of \( 2.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \). What is the half-life?

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

Given: \( [\text{R}]_0 = 0.12 \, \text{mol L}^{-1} \), \( k = 2.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

\( t_{1/2} = \frac{0.12}{2 \times 2.0 \times 10^{-3}} = \frac{0.12}{4.0 \times 10^{-3}} = 30 \, \text{s} \).

20 s
30 s
40 s
15 s
2

A second-order reaction has a rate constant of \( 0.08 \, \text{L mol}^{-1} \text{s}^{-1} \) and an initial concentration of \( 0.5 \, \text{mol L}^{-1} \). What is the time for 75% completion?

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

75% completion means \( [\text{A}] = 0.25 \times 0.5 = 0.125 \, \text{mol L}^{-1} \).

\( \frac{1}{0.125} - \frac{1}{0.5} = 0.08 \times t \), \( 8 - 2 = 0.08 t \), \( t = \frac{6}{0.08} = 75 \, \text{s} \).

50 s
100 s
62.5 s
75 s
4

A first-order gaseous reaction has an initial pressure of 1.5 atm. After 25 s, the total pressure is 2.0 atm. What is the rate constant?

For \( \text{A} \to \text{B} + \text{C} \), \( p_t = p_i + x \), \( x = 2.0 - 1.5 = 0.5 \, \text{atm} \).

\( p_A = p_i - x = 1.5 - 0.5 = 1.0 \, \text{atm} \).

\( k = \frac{2.303}{25} \log \frac{1.5}{1.0} = \frac{2.303 \times 0.176}{25} \approx 0.0162 \, \text{s}^{-1} \).

0.0115 s\(^{-1}\)
0.0203 s\(^{-1}\)
0.0162 s\(^{-1}\)
0.0092 s\(^{-1}\)
3

A first-order reaction has a rate constant of \( 0.0173 \, \text{min}^{-1} \). What percentage of the reactant decomposes in 80 minutes?

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

\( \log \frac{[\text{R}]_0}{[\text{R}]} = \frac{0.0173 \times 80}{2.303} \approx 0.602 \), \( \frac{[\text{R}]_0}{[\text{R}]} = 10^{0.602} \approx 4 \).

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

50%
60%
80%
75%
4

A first-order reaction is 10% complete in 5 minutes. What is the rate constant?

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

10% complete means 90% remains, \( \frac{[\text{R}]}{[\text{R}]_0} = 0.9 \), \( \frac{[\text{R}]_0}{[\text{R}]} = \frac{1}{0.9} \approx 1.111 \).

\( k = \frac{2.303}{5} \log 1.111 = \frac{2.303 \times 0.0458}{5} \approx 0.0211 \, \text{min}^{-1} \).

0.0211 min\(^{-1}\)
0.0346 min\(^{-1}\)
0.0154 min\(^{-1}\)
0.0462 min\(^{-1}\)
1

A first-order reaction has a half-life of 16 minutes. What percentage of the reactant decomposes in 48 minutes?

Number of half-lives = \( \frac{48}{16} = 3 \).

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

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

75%
50%
87.5%
93.75%
3

A reaction’s rate constant increases from \( 4.0 \times 10^{-4} \, \text{s}^{-1} \) at 280 K to \( 1.2 \times 10^{-3} \, \text{s}^{-1} \) at 290 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log \frac{1.2 \times 10^{-3}}{4.0 \times 10^{-4}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{280 \times 290} \right) \).

\( \log 3 = 0.477 \), \( E_a = \frac{0.477 \times 19.147 \times 81200}{10} \approx 74,200 \, \text{J mol}^{-1} \approx 74.2 \, \text{kJ mol}^{-1} \).

37.1 kJ mol\(^{-1}\)
55.6 kJ mol\(^{-1}\)
74.2 kJ mol\(^{-1}\)
92.8 kJ mol\(^{-1}\)
2

The molecularity of the elementary step \( \text{NO} + \text{O}_3 \to \text{NO}_2 + \text{O}_2 \) is:

Molecularity is the number of molecules in an elementary step.

For \( \text{NO} + \text{O}_3 \), two molecules collide, so molecularity = 2.

1
2
3
4
2

A first-order reaction has a rate constant of \( 0.0289 \, \text{min}^{-1} \). What is the time for 50% completion?

For first-order, \( t_{1/2} = \frac{0.693}{k} \).

Given: \( k = 0.0289 \, \text{min}^{-1} \).

\( t_{1/2} = \frac{0.693}{0.0289} \approx 24 \, \text{min} \).

20 min
30 min
24 min
28 min
3

A zero-order reaction has a rate constant of \( 5.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \). If the initial concentration is \( 0.2 \, \text{mol L}^{-1} \), how long will it take for 20% decomposition?

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

20% decomposition means \( [\text{R}] = 0.8 \times 0.2 = 0.16 \, \text{mol L}^{-1} \).

\( t = \frac{0.2 - 0.16}{5.0 \times 10^{-4}} = \frac{0.04}{5.0 \times 10^{-4}} = 80 \, \text{s} \).

80 s
100 s
60 s
120 s
1

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0