Chemical Kinetics Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A second-order reaction has an initial concentration of \( 0.3 \, \text{mol L}^{-1} \) and a half-life of 20 s. What is the rate constant?

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

Given: \( t_{1/2} = 20 \, \text{s} \), \( [\text{A}]_0 = 0.3 \, \text{mol L}^{-1} \).

\( k = \frac{1}{20 \times 0.3} = \frac{1}{6} \approx 0.1667 \, \text{L mol}^{-1} \text{s}^{-1} \).

0.1 L mol\(^{-1}\) s\(^{-1}\)
0.2 L mol\(^{-1}\) s\(^{-1}\)
0.25 L mol\(^{-1}\) s\(^{-1}\)
0.1667 L mol\(^{-1}\) s\(^{-1}\)
4

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

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

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

\( \frac{\Delta[\text{C}]}{\Delta t} = 2 \times 0.02 = 0.04 \, \text{mol L}^{-1} \text{min}^{-1} \).

0.02 mol L\(^{-1}\) min\(^{-1}\)
0.04 mol L\(^{-1}\) min\(^{-1}\)
0.06 mol L\(^{-1}\) min\(^{-1}\)
0.01 mol L\(^{-1}\) min\(^{-1}\)
1

A zero-order reaction has a rate constant of \( 2.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \). How long will it take for the concentration to decrease from \( 0.12 \, \text{mol L}^{-1} \) to \( 0.03 \, \text{mol L}^{-1} \)?

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

Given: \( [\text{R}]_0 = 0.12 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.03 \, \text{mol L}^{-1} \), \( k = 2.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

\( t = \frac{0.12 - 0.03}{2.0 \times 10^{-3}} = \frac{0.09}{2.0 \times 10^{-3}} = 45 \, \text{s} \).

30 s
60 s
50 s
45 s
4

A reaction’s rate increases by 2.25 times when the temperature rises from 300 K to 310 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log 2.25 = 0.352 \), \( E_a = \frac{0.352 \times 19.147 \times 93000}{10} \approx 62,700 \, \text{J mol}^{-1} \approx 62.7 \, \text{kJ mol}^{-1} \).

31.4 kJ mol\(^{-1}\)
47.1 kJ mol\(^{-1}\)
78.4 kJ mol\(^{-1}\)
62.7 kJ mol\(^{-1}\)
3

A first-order reaction is 40% complete in 15 minutes. What is the rate constant?

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

40% complete means 60% remains, \( \frac{[\text{R}]}{[\text{R}]_0} = 0.6 \), \( \frac{[\text{R}]_0}{[\text{R}]} = \frac{1}{0.6} \approx 1.667 \).

\( k = \frac{2.303}{15} \log 1.667 = \frac{2.303 \times 0.222}{15} \approx 0.0341 \, \text{min}^{-1} \).

0.0231 min\(^{-1}\)
0.0341 min\(^{-1}\)
0.0462 min\(^{-1}\)
0.0154 min\(^{-1}\)
2

The rate constant of a reaction is \( 3.0 \times 10^{-3} \, \text{s}^{-1} \) at 27°C. If the activation energy is 55 kJ/mol, what is the rate constant at 37°C (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( T_1 = 300 \, \text{K} \), \( T_2 = 310 \, \text{K} \).

\( \log \frac{k_2}{3.0 \times 10^{-3}} = \frac{55000}{2.303 \times 8.314} \left( \frac{10}{300 \times 310} \right) \approx 0.309 \).

\( \frac{k_2}{3.0 \times 10^{-3}} = 10^{0.309} \approx 2.04 \), \( k_2 \approx 6.12 \times 10^{-3} \, \text{s}^{-1} \).

4.5 \(\times\) 10\(^{-3}\) s\(^{-1}\)
6.12 \(\times\) 10\(^{-3}\) s\(^{-1}\)
7.5 \(\times\) 10\(^{-3}\) s\(^{-1}\)
9.0 \(\times\) 10\(^{-3}\) s\(^{-1}\)
2

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

Molecularity is the number of molecules in an elementary step.

For \( \text{Cl}_2 + \text{H}_2 \), two molecules collide, so molecularity = 2.

1
2
3
4
2

A first-order reaction is 20% complete in 12 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, \( \frac{[\text{R}]}{[\text{R}]_0} = 0.8 \), \( \frac{[\text{R}]_0}{[\text{R}]} = \frac{1}{0.8} = 1.25 \).

\( k = \frac{2.303}{12} \log 1.25 = \frac{2.303 \times 0.097}{12} \approx 0.0186 \, \text{min}^{-1} \).

0.0154 min\(^{-1}\)
0.0186 min\(^{-1}\)
0.0231 min\(^{-1}\)
0.0288 min\(^{-1}\)
2

A first-order reaction is 25% complete in 15 minutes. What is the rate constant?

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

25% complete means 75% remains, so \( \frac{[\text{R}]}{[\text{R}]_0} = 0.75 \), \( \frac{[\text{R}]_0}{[\text{R}]} = \frac{1}{0.75} \).

\( k = \frac{2.303}{15} \log \frac{1}{0.75} = \frac{2.303}{15} \times 0.125 = 0.0192 \, \text{min}^{-1} \).

0.0154 min\(^{-1}\)
0.0192 min\(^{-1}\)
0.0231 min\(^{-1}\)
0.0306 min\(^{-1}\)
2

A first-order reaction has a half-life of 24 minutes. What percentage of the reactant remains after 72 minutes?

Number of half-lives = \( \frac{72}{24} = 3 \).

Fraction remaining = \( \left( \frac{1}{2} \right)^3 = \frac{1}{8} = 0.125 \).

Percentage remaining = \( 0.125 \times 100 = 12.5\% \).

25%
50%
6.25%
12.5%
4

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