Chemical Kinetics Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A reaction’s rate increases by 3.24 times when the temperature rises from 310 K to 320 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log 3.24 \approx 0.510 \), \( E_a = \frac{0.510 \times 19.147 \times 99200}{10} \approx 96,900 \, \text{J mol}^{-1} \approx 96.9 \, \text{kJ mol}^{-1} \).

48.5 kJ mol\(^{-1}\)
72.7 kJ mol\(^{-1}\)
96.9 kJ mol\(^{-1}\)
121.1 kJ mol\(^{-1}\)
3

A reaction’s rate constant increases from \( 4.0 \times 10^{-4} \, \text{min}^{-1} \) at 30°C to \( 1.2 \times 10^{-3} \, \text{min}^{-1} \) at 40°C. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( T_1 = 303 \, \text{K} \), \( T_2 = 313 \, \text{K} \).

\( \log \frac{1.2 \times 10^{-3}}{4.0 \times 10^{-4}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{303 \times 313} \right) \).

\( \log 3 = 0.477 \), \( E_a = \frac{0.477 \times 19.147 \times 94839}{10} \approx 86,600 \, \text{J mol}^{-1} \approx 86.6 \, \text{kJ mol}^{-1} \).

43.3 kJ mol\(^{-1}\)
64.9 kJ mol\(^{-1}\)
108.2 kJ mol\(^{-1}\)
86.6 kJ mol\(^{-1}\)
4

The molecularity of the elementary step \( \text{O}_3 \to \text{O}_2 + \text{O} \) is:

Molecularity is the number of molecules in an elementary step.

For \( \text{O}_3 \to \text{O}_2 + \text{O} \), one molecule decomposes, so molecularity = 1.

2
3
1
4
3

A zero-order reaction takes 90 s to reduce the concentration from \( 0.15 \, \text{mol L}^{-1} \) to \( 0.06 \, \text{mol L}^{-1} \). What is the rate constant?

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

Given: \( [\text{R}]_0 = 0.15 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.06 \, \text{mol L}^{-1} \), \( t = 90 \, \text{s} \).

\( k = \frac{0.15 - 0.06}{90} = \frac{0.09}{90} = 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

5.0 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\)
1.0 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
1.5 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
2.0 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
2

A reaction’s rate increases by 4 times when the temperature rises from 320 K to 340 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log 4 = 0.602 \), \( 0.602 = \frac{E_a}{2.303 \times 8.314} \left( \frac{20}{320 \times 340} \right) \).

\( E_a = \frac{0.602 \times 19.147 \times 108800}{20} \approx 62,700 \, \text{J mol}^{-1} \approx 62.7 \, \text{kJ mol}^{-1} \).

31.4 kJ mol\(^{-1}\)
47.1 kJ mol\(^{-1}\)
78.4 kJ mol\(^{-1}\)
62.7 kJ mol\(^{-1}\)
4

A zero-order reaction has a rate constant of \( 8.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \). How long will it take for the concentration to decrease from \( 0.16 \, \text{mol L}^{-1} \) to \( 0.04 \, \text{mol L}^{-1} \)?

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

Given: \( [\text{R}]_0 = 0.16 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.04 \, \text{mol L}^{-1} \), \( k = 8.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \).

\( t = \frac{0.16 - 0.04}{8.0 \times 10^{-4}} = \frac{0.12}{8.0 \times 10^{-4}} = 150 \, \text{s} \).

100 s
120 s
150 s
180 s
3

A reaction’s rate doubles when the temperature increases from 25°C to 35°C. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( T_1 = 298 \, \text{K} \), \( T_2 = 308 \, \text{K} \), \( \frac{k_2}{k_1} = 2 \).

\( \log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{298 \times 308} \right) \).

\( 0.301 = \frac{E_a \times 10}{19.147 \times 91684} \), \( E_a \approx 52,800 \, \text{J mol}^{-1} \approx 52.8 \, \text{kJ mol}^{-1} \).

26.4 kJ mol\(^{-1}\)
39.6 kJ mol\(^{-1}\)
52.8 kJ mol\(^{-1}\)
66.0 kJ mol\(^{-1}\)
3

A zero-order reaction has an initial concentration of \( 0.15 \, \text{mol L}^{-1} \) and takes 75 s to reach half concentration. What is the rate constant?

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

Given: \( [\text{R}]_0 = 0.15 \, \text{mol L}^{-1} \), \( t_{1/2} = 75 \, \text{s} \).

\( 75 = \frac{0.15}{2k} \), \( k = \frac{0.15}{2 \times 75} = 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

5.0 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\)
7.5 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\)
1.0 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
1.5 \(\times\) 10\(^{-3}\) mol L\(^{-1}\) s\(^{-1}\)
2

A first-order reaction has a rate constant of \( 0.0346 \, \text{min}^{-1} \). What is the half-life?

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

Given: \( k = 0.0346 \, \text{min}^{-1} \).

\( t_{1/2} = \frac{0.693}{0.0346} \approx 20 \, \text{min} \).

15 min
25 min
30 min
20 min
4

For the reaction \( \text{A} + 2\text{B} \to 3\text{C} \), the rate of disappearance of B is \( 0.06 \, \text{mol L}^{-1} \text{s}^{-1} \). What is the rate of formation of C?

Rate = \( -\frac{1}{2} \frac{\Delta[\text{B}]}{\Delta t} = \frac{1}{3} \frac{\Delta[\text{C}]}{\Delta t} \).

Given: \( -\frac{\Delta[\text{B}]}{\Delta t} = 0.06 \, \text{mol L}^{-1} \text{s}^{-1} \).

Rate = \( \frac{0.06}{2} = 0.03 \), \( \frac{\Delta[\text{C}]}{\Delta t} = 3 \times 0.03 = 0.09 \, \text{mol L}^{-1} \text{s}^{-1} \).

0.03 mol L\(^{-1}\) s\(^{-1}\)
0.09 mol L\(^{-1}\) s\(^{-1}\)
0.06 mol L\(^{-1}\) s\(^{-1}\)
0.12 mol L\(^{-1}\) s\(^{-1}\)
2

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