Chemical Kinetics Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A first-order reaction has a rate constant of \( 0.01155 \, \text{min}^{-1} \). What is the half-life?

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

Given: \( k = 0.01155 \, \text{min}^{-1} \).

\( t_{1/2} = \frac{0.693}{0.01155} \approx 60 \, \text{min} \).

60 min
30 min
45 min
75 min
1

A reaction’s rate constant is \( 2.5 \times 10^{-3} \, \text{s}^{-1} \) at 25°C and \( 7.5 \times 10^{-3} \, \text{s}^{-1} \) at 35°C. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( T_1 = 298 \, \text{K} \), \( T_2 = 308 \, \text{K} \).

\( \log \frac{7.5 \times 10^{-3}}{2.5 \times 10^{-3}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{298 \times 308} \right) \).

\( \log 3 = 0.477 \), \( E_a = \frac{0.477 \times 19.147 \times 91684}{10} \approx 83,700 \, \text{J mol}^{-1} \approx 83.7 \, \text{kJ mol}^{-1} \).

41.9 kJ mol\(^{-1}\)
83.7 kJ mol\(^{-1}\)
62.8 kJ mol\(^{-1}\)
104.6 kJ mol\(^{-1}\)
2

A zero-order reaction has a rate constant of \( 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \). How long will it take for 80% of a \( 0.2 \, \text{mol L}^{-1} \) reactant to decompose?

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

80% decomposition means \( [\text{R}] = 0.2 \times 0.2 = 0.04 \, \text{mol L}^{-1} \).

\( t = \frac{0.2 - 0.04}{1.0 \times 10^{-3}} = \frac{0.16}{1.0 \times 10^{-3}} = 160 \, \text{min} \).

100 min
120 min
160 min
200 min
3

For the reaction \( 2\text{A} + \text{B} \to 3\text{C} \), the rate of disappearance of B is \( 0.01 \, \text{mol L}^{-1} \text{s}^{-1} \). What is the rate of formation of C?

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

Given: \( -\frac{\Delta[\text{B}]}{\Delta t} = 0.01 \, \text{mol L}^{-1} \text{s}^{-1} \).

\( \frac{\Delta[\text{C}]}{\Delta t} = 3 \times 0.01 = 0.03 \, \text{mol L}^{-1} \text{s}^{-1} \).

0.015 mol L\(^{-1}\) s\(^{-1}\)
0.03 mol L\(^{-1}\) s\(^{-1}\)
0.045 mol L\(^{-1}\) s\(^{-1}\)
0.06 mol L\(^{-1}\) s\(^{-1}\)
1

A zero-order reaction has a rate constant of \( 3.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \). If the initial concentration is \( 0.18 \, \text{mol L}^{-1} \), how long will it take for 50% decomposition?

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

Given: \( [\text{R}]_0 = 0.18 \, \text{mol L}^{-1} \), \( k = 3.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

\( t_{1/2} = \frac{0.18}{2 \times 3.0 \times 10^{-3}} = \frac{0.18}{6.0 \times 10^{-3}} = 30 \, \text{s} \).

30 s
60 s
45 s
90 s
1

A first-order reaction has an initial concentration of \( 1.0 \, \text{mol L}^{-1} \). After 40 minutes, the concentration reduces to \( 0.25 \, \text{mol L}^{-1} \). What is the rate constant?

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

Given: \( [\text{R}]_0 = 1.0 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.25 \, \text{mol L}^{-1} \), \( t = 40 \, \text{min} \).

\( k = \frac{2.303}{40} \log \frac{1.0}{0.25} = \frac{2.303 \times 0.602}{40} \approx 0.0346 \, \text{min}^{-1} \).

0.0231 min\(^{-1}\)
0.0346 min\(^{-1}\)
0.0462 min\(^{-1}\)
0.0577 min\(^{-1}\)
2

A reaction’s rate increases 8 times when the temperature rises from 300 K to 330 K. What is the approximate activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log 8 = \frac{E_a}{2.303 \times 8.314} \left( \frac{30}{300 \times 330} \right) \).

\( 0.903 = \frac{E_a \times 30}{19.147 \times 99000} \), \( E_a \approx 56,600 \, \text{J mol}^{-1} \approx 56.6 \, \text{kJ mol}^{-1} \).

28.3 kJ mol\(^{-1}\)
42.5 kJ mol\(^{-1}\)
70.8 kJ mol\(^{-1}\)
56.6 kJ mol\(^{-1}\)
4

A reaction has the rate law \( \text{Rate} = k[\text{A}]^{2.5} \). What is the overall order?

Overall order = \( 2.5 \).

2
3
1.5
2.5
4

A reaction’s rate increases by 1.414 times when the concentration of the reactant is doubled. What is the order of the reaction?

Rate = \( k [\text{A}]^n \). New rate: \( k (2[\text{A}])^n = 1.414 k [\text{A}]^n \), so \( 2^n = 1.414 \).

\( 1.414 \approx \sqrt{2} \), \( 2^n = 2^{1/2} \), \( n = 0.5 \).

1
0.5
2
1.5
2

A zero-order reaction reduces the concentration from \( 0.1 \, \text{mol L}^{-1} \) to \( 0.04 \, \text{mol L}^{-1} \) in 60 s. What is the rate constant?

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

Given: \( [\text{R}]_0 = 0.1 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.04 \, \text{mol L}^{-1} \), \( t = 60 \, \text{s} \).

\( k = \frac{0.1 - 0.04}{60} = \frac{0.06}{60} = 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

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\(^{-1}\)
3

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