Chemical Kinetics Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A zero-order reaction has an initial concentration of \( 0.15 \, \text{mol L}^{-1} \) and a rate constant of \( 2.5 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \). What is the half-life?

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

Given: \( [\text{R}]_0 = 0.15 \, \text{mol L}^{-1} \), \( k = 2.5 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

\( t_{1/2} = \frac{0.15}{2 \times 2.5 \times 10^{-3}} = \frac{0.15}{5.0 \times 10^{-3}} = 30 \, \text{s} \).

15 s
20 s
40 s
30 s
4

The half-life of a first-order reaction is 20 minutes. What is the rate constant?

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

Given: \( t_{1/2} = 20 \, \text{min} \).

\( k = \frac{0.693}{20} = 0.03465 \, \text{min}^{-1} \approx 3.465 \times 10^{-2} \, \text{min}^{-1} \).

3.465 \(\times\) 10\(^{-2}\) min\(^{-1}\)
6.93 \(\times\) 10\(^{-2}\) min\(^{-1}\)
1.732 \(\times\) 10\(^{-2}\) min\(^{-1}\)
2.31 \(\times\) 10\(^{-2}\) min\(^{-1}\)
1

A reaction has a rate constant of \( 0.05 \, \text{L mol}^{-1} \text{s}^{-1} \). What is its order?

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

Given: \( k = 0.05 \, \text{L mol}^{-1} \text{s}^{-1} \), which matches the unit for a second-order reaction.

Thus, the order is 2.

2
1
0
3
1

A first-order reaction has a rate constant of \( 0.0115 \, \text{min}^{-1} \). What percentage of the reactant remains after 60 minutes?

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

\( \log \frac{[\text{R}]_0}{[\text{R}]} = \frac{0.0115 \times 60}{2.303} \approx 0.301 \), \( \frac{[\text{R}]_0}{[\text{R}]} = 10^{0.301} \approx 2 \).

Fraction remaining = \( \frac{1}{2} = 0.5 \), percentage = \( 50\% \).

25%
75%
33.3%
50%
4

A reaction’s rate increases by 2.25 times when the concentration of the reactant is increased by 1.5 times. What is the order of the reaction?

Rate = \( k [\text{A}]^n \). New rate: \( k (1.5[\text{A}])^n = 2.25 k [\text{A}]^n \), so \( 1.5^n = 2.25 \).

\( n \log 1.5 = \log 2.25 \), \( n \times 0.176 = 0.352 \), \( n \approx 2 \).

1
2
1.5
3
2

A reaction has the rate law \( \text{Rate} = k[\text{A}]^2[\text{B}] \). If the concentration of A is tripled and B is doubled, by how many times will the rate increase?

Initial rate = \( k [\text{A}]^2 [\text{B}] \).

New rate = \( k (3[\text{A}])^2 (2[\text{B}]) = k \times 9 [\text{A}]^2 \times 2 [\text{B}] = 18 k [\text{A}]^2 [\text{B}] \).

Rate increases by 18 times.

6
18
12
9
2

A second-order reaction has a rate constant of \( 0.02 \, \text{L mol}^{-1} \text{s}^{-1} \) and an initial concentration of \( 0.2 \, \text{mol L}^{-1} \). What is the half-life?

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

Given: \( k = 0.02 \, \text{L mol}^{-1} \text{s}^{-1} \), \( [\text{A}]_0 = 0.2 \, \text{mol L}^{-1} \).

\( t_{1/2} = \frac{1}{0.02 \times 0.2} = \frac{1}{0.004} = 250 \, \text{s} \).

250 s
200 s
300 s
150 s
1

A first-order reaction is 20% complete in 10 minutes. What is the rate constant?

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

20% complete means 80% remains, \( \frac{[\text{R}]}{[\text{R}]_0} = 0.8 \), \( \frac{[\text{R}]_0}{[\text{R}]} = 1.25 \).

\( k = \frac{2.303}{10} \log 1.25 = \frac{2.303 \times 0.0969}{10} \approx 0.0223 \, \text{min}^{-1} \).

0.0223 min\(^{-1}\)
0.0346 min\(^{-1}\)
0.0154 min\(^{-1}\)
0.0462 min\(^{-1}\)
1

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

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

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

Rate = \( \frac{0.06}{2} = 0.03 \, \text{mol L}^{-1} \text{s}^{-1} \).

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

A reaction has the rate law \( \text{Rate} = k[\text{A}]^2 \). If the concentration of A is increased by 5 times, by how many times will the rate increase?

Initial rate = \( k [\text{A}]^2 \).

New rate = \( k (5[\text{A}])^2 = k \times 25 [\text{A}]^2 = 25 \times \text{initial rate} \).

Rate increases by 25 times.

5
25
10
50
2

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