Chemical Kinetics Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A first-order reaction has a half-life of 35 minutes. What fraction of the reactant remains after 105 minutes?

Number of half-lives = \( \frac{105}{35} = 3 \).

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

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

For the reaction \( 2\text{A} \to \text{B} + \text{C} \), the rate of formation of C is \( 0.05 \, \text{mol L}^{-1} \text{s}^{-1} \). What is the rate of disappearance of A?

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

Given: \( \frac{\Delta[\text{C}]}{\Delta t} = 0.05 \, \text{mol L}^{-1} \text{s}^{-1} \).

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

0.025 mol L\(^{-1}\) s\(^{-1}\)
0.1 mol L\(^{-1}\) s\(^{-1}\)
0.05 mol L\(^{-1}\) s\(^{-1}\)
0.2 mol L\(^{-1}\) s\(^{-1}\)
2

A reaction’s rate constant doubles when the temperature increases from 20°C to 30°C. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( T_1 = 293 \, \text{K} \), \( T_2 = 303 \, \text{K} \).

\( \log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{293 \times 303} \right) \).

\( 0.301 = \frac{E_a \times 10}{19.147 \times 88879} \), \( E_a \approx 51,200 \, \text{J mol}^{-1} \approx 51.2 \, \text{kJ mol}^{-1} \).

25.6 kJ mol\(^{-1}\)
38.4 kJ mol\(^{-1}\)
64.0 kJ mol\(^{-1}\)
51.2 kJ mol\(^{-1}\)
4

A zero-order reaction takes 120 s to reduce the concentration from \( 0.06 \, \text{mol L}^{-1} \) to \( 0.03 \, \text{mol L}^{-1} \). What is the rate constant?

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

Given: \( [\text{R}]_0 = 0.06 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.03 \, \text{mol L}^{-1} \), \( t = 120 \, \text{s} \).

\( k = \frac{0.06 - 0.03}{120} = \frac{0.03}{120} = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \).

1.0 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\)
2.0 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\)
2.5 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\)
3.0 \(\times\) 10\(^{-4}\) mol L\(^{-1}\) s\(^{-1}\)
3

A first-order reaction has a rate constant of \( 0.0462 \, \text{min}^{-1} \). What is the time required for 75% completion?

75% completion means 25% remains, \( \frac{[\text{R}]}{[\text{R}]_0} = 0.25 \), \( \frac{[\text{R}]_0}{[\text{R}]} = 4 \).

\( t = \frac{2.303}{k} \log \frac{[\text{R}]_0}{[\text{R}]} = \frac{2.303}{0.0462} \log 4 = \frac{2.303 \times 0.602}{0.0462} \approx 30 \, \text{min} \).

15 min
20 min
30 min
40 min
3

A first-order reaction has a half-life of 18 minutes. What percentage of the reactant remains after 54 minutes?

Number of half-lives = \( \frac{54}{18} = 3 \).

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

Percentage remaining = \( \frac{1}{8} \times 100 = 12.5\% \).

25%
50%
12.5%
6.25%
3

The molecularity of the elementary step \( \text{H}_2 + 2\text{ICl} \to 2\text{HCl} + \text{I}_2 \) is:

Molecularity is the number of molecules in an elementary step.

For \( \text{H}_2 + 2\text{ICl} \), three molecules collide (1 H₂ + 2 ICl), so molecularity = 3.

1
3
2
4
2

A second-order reaction has a rate constant of \( 0.02 \, \text{L mol}^{-1} \text{min}^{-1} \) and an initial concentration of \( 0.4 \, \text{mol L}^{-1} \). What is the half-life?

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

Given: \( k = 0.02 \, \text{L mol}^{-1} \text{min}^{-1} \), \( [\text{A}]_0 = 0.4 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.02 \times 0.4} = \frac{1}{0.008} = 125 \, \text{min} \).

100 min
150 min
125 min
75 min
3

A first-order reaction has a rate constant of \( 0.0139 \, \text{min}^{-1} \). What is the half-life?

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

Given: \( k = 0.0139 \, \text{min}^{-1} \).

\( t_{1/2} = \frac{0.693}{0.0139} \approx 50 \, \text{min} \).

50 min
40 min
60 min
70 min
1

A reaction has the rate law \( \text{Rate} = k[\text{A}][\text{B}] \). If the concentration of A is tripled and B is reduced to one-third, what happens to the rate?

Initial rate = \( k [\text{A}] [\text{B}] \).

New rate = \( k (3[\text{A}]) \left( \frac{[\text{B}]}{3} \right) = k \times 3 \times \frac{1}{3} [\text{A}] [\text{B}] = k [\text{A}] [\text{B}] \).

Rate remains the same.

Triples
Decreases to one-third
Remains same
Doubles
3

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