Chemical Kinetics Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A first-order reaction has a half-life of 30 minutes. What fraction of the reactant remains after 90 minutes?

Number of half-lives = \( \frac{t}{t_{1/2}} = \frac{90}{30} = 3 \).

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

1/2
1/4
1/8
1/16
3

A reaction’s rate increases by 4 times when the temperature rises from 310 K to 330 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log 4 = \frac{E_a}{2.303 \times 8.314} \left( \frac{20}{310 \times 330} \right) \).

\( 0.602 = \frac{E_a \times 20}{19.147 \times 102300} \), \( E_a \approx 58,900 \, \text{J mol}^{-1} \approx 58.9 \, \text{kJ mol}^{-1} \).

29.5 kJ mol\(^{-1}\)
44.2 kJ mol\(^{-1}\)
73.6 kJ mol\(^{-1}\)
58.9 kJ mol\(^{-1}\)
4

A zero-order reaction takes 120 s to reduce the concentration from \( 0.3 \, \text{mol L}^{-1} \) to \( 0.18 \, \text{mol L}^{-1} \). What is the rate constant?

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

Given: \( [\text{R}]_0 = 0.3 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.18 \, \text{mol L}^{-1} \), \( t = 120 \, \text{s} \).

\( k = \frac{0.3 - 0.18}{120} = \frac{0.12}{120} = 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

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

For the reaction \( 2\text{A} + 2\text{B} \to 3\text{C} \), the rate of formation of C is \( 0.15 \, \text{mol L}^{-1} \text{min}^{-1} \). What is the rate of disappearance of A?

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

Given: \( \frac{\Delta[\text{C}]}{\Delta t} = 0.15 \, \text{mol L}^{-1} \text{min}^{-1} \).

Rate = \( \frac{0.15}{3} = 0.05 \), \( -\frac{\Delta[\text{A}]}{\Delta t} = 2 \times 0.05 = 0.1 \, \text{mol L}^{-1} \text{min}^{-1} \).

0.05 mol L\(^{-1}\) min\(^{-1}\)
0.1 mol L\(^{-1}\) min\(^{-1}\)
0.15 mol L\(^{-1}\) min\(^{-1}\)
0.2 mol L\(^{-1}\) min\(^{-1}\)
2

A reaction’s rate increases by 64 times when the concentration of the reactant is increased by 4 times. What is the order of the reaction?

Rate = \( k [\text{A}]^n \). New rate: \( k (4[\text{A}])^n = 64 k [\text{A}]^n \), so \( 4^n = 64 \).

\( 4^n = 4^3 \), \( n = 3 \).

3
2
4
1
1

A first-order reaction has a half-life of 14 minutes. What percentage of the reactant decomposes in 42 minutes?

Number of half-lives = \( \frac{42}{14} = 3 \).

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 zero-order reaction has a rate constant of \( 2.0 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \). If the initial concentration is \( 0.1 \, \text{mol L}^{-1} \), how long will it take for 40% decomposition?

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

40% decomposition means \( [\text{R}] = 0.6 \times 0.1 = 0.06 \, \text{mol L}^{-1} \).

\( t = \frac{0.1 - 0.06}{2.0 \times 10^{-3}} = \frac{0.04}{2.0 \times 10^{-3}} = 20 \, \text{min} \).

20 min
30 min
40 min
50 min
1

A reaction has a rate constant of \( 0.08 \, \text{s}^{-1} \). What is its order?

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

Given: \( k = 0.08 \, \text{s}^{-1} \), which matches the unit for a first-order reaction.

Thus, the order is 1.

1
2
0
3
1

A reaction has the rate law \( \text{Rate} = k[\text{A}]^{2/3}[\text{B}]^{2/3} \). What is the overall order?

Overall order = \( \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \approx 1.33 \).

1
2
1.5
1.33
4

For the reaction \( 3\text{A} \to 2\text{B} \), the rate of formation of B is \( 0.02 \, \text{mol L}^{-1} \text{s}^{-1} \). What is the rate of disappearance of A?

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

Given: \( \frac{\Delta[\text{B}]}{\Delta t} = 0.02 \, \text{mol L}^{-1} \text{s}^{-1} \).

Rate = \( \frac{0.02}{2} = 0.01 \, \text{mol L}^{-1} \text{s}^{-1} \), so \( -\frac{\Delta[\text{A}]}{\Delta t} = 3 \times 0.01 = 0.03 \, \text{mol L}^{-1} \text{s}^{-1} \).

0.015 mol L\(^{-1}\) s\(^{-1}\)
0.03 mol L\(^{-1}\) s\(^{-1}\)
0.06 mol L\(^{-1}\) s\(^{-1}\)
0.01 mol L\(^{-1}\) s\(^{-1}\)
2

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