Chemical Kinetics Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

For the reaction \( 2\text{A} + \text{B} \to 3\text{C} \), the rate of disappearance of B is \( 0.03 \, \text{mol L}^{-1} \text{min}^{-1} \). 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: \( -\frac{\Delta[\text{B}]}{\Delta t} = 0.03 \, \text{mol L}^{-1} \text{min}^{-1} \).

\( \frac{\Delta[\text{C}]}{\Delta t} = 3 \times 0.03 = 0.09 \, \text{mol L}^{-1} \text{min}^{-1} \).

0.03 mol L\(^{-1}\) min\(^{-1}\)
0.09 mol L\(^{-1}\) min\(^{-1}\)
0.06 mol L\(^{-1}\) min\(^{-1}\)
0.12 mol L\(^{-1}\) min\(^{-1}\)
2

A zero-order reaction has a rate constant of \( 6.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \). If the initial concentration is \( 0.09 \, \text{mol L}^{-1} \), how long will it take for 66.67% decomposition?

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

66.67% decomposition means \( [\text{R}] = \frac{1}{3} \times 0.09 = 0.03 \, \text{mol L}^{-1} \).

\( t = \frac{0.09 - 0.03}{6.0 \times 10^{-4}} = \frac{0.06}{6.0 \times 10^{-4}} = 100 \, \text{s} \).

100 s
150 s
75 s
120 s
1

A reaction’s rate increases by 1.73 times when the temperature rises from 290 K to 300 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log 1.73 \approx 0.238 \), \( E_a = \frac{0.238 \times 19.147 \times 87000}{10} \approx 39,700 \, \text{J mol}^{-1} \approx 39.7 \, \text{kJ mol}^{-1} \).

19.9 kJ mol\(^{-1}\)
29.8 kJ mol\(^{-1}\)
39.7 kJ mol\(^{-1}\)
49.6 kJ mol\(^{-1}\)
3

A zero-order reaction has an initial concentration of \( 0.27 \, \text{mol L}^{-1} \) and a rate constant of \( 4.5 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \). What is the half-life?

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

Given: \( [\text{R}]_0 = 0.27 \, \text{mol L}^{-1} \), \( k = 4.5 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

\( t_{1/2} = \frac{0.27}{2 \times 4.5 \times 10^{-3}} = \frac{0.27}{9.0 \times 10^{-3}} = 30 \, \text{s} \).

20 s
30 s
40 s
15 s
2

A first-order gaseous reaction has an initial pressure of 1.2 atm. After 30 s, the total pressure is 1.8 atm. What is the rate constant?

For \( \text{A} \to 2\text{B} \), \( p_t = p_i + x \), \( x = 1.8 - 1.2 = 0.6 \, \text{atm} \).

\( p_A = p_i - x = 1.2 - 0.6 = 0.6 \, \text{atm} \).

\( k = \frac{2.303}{30} \log \frac{1.2}{0.6} = \frac{2.303 \times 0.301}{30} \approx 0.0231 \, \text{s}^{-1} \).

0.0154 s\(^{-1}\)
0.0192 s\(^{-1}\)
0.0306 s\(^{-1}\)
0.0231 s\(^{-1}\)
4

A first-order reaction has a half-life of 25 minutes. What percentage of the reactant remains after 75 minutes?

Number of half-lives = \( \frac{75}{25} = 3 \).

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

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

12.5%
25%
50%
6.25%
1

A first-order reaction has a rate constant of \( 0.01386 \, \text{min}^{-1} \). What is its half-life?

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

Given: \( k = 0.01386 \, \text{min}^{-1} \).

\( t_{1/2} = \frac{0.693}{0.01386} \approx 50 \, \text{min} \).

50 min
25 min
75 min
100 min
1

A second-order reaction has a rate constant of \( 0.05 \, \text{L mol}^{-1} \text{min}^{-1} \) and an initial concentration of \( 0.4 \, \text{mol L}^{-1} \). What is the time for 80% completion?

For second-order, \( \frac{1}{[\text{A}]} - \frac{1}{[\text{A}]_0} = kt \).

80% completion means \( [\text{A}] = 0.2 \times 0.4 = 0.08 \, \text{mol L}^{-1} \).

\( \frac{1}{0.08} - \frac{1}{0.4} = 0.05 \times t \), \( 12.5 - 2.5 = 0.05 t \), \( t = \frac{10}{0.05} = 200 \, \text{min} \).

150 min
250 min
100 min
200 min
4

The molecularity of the reaction \( 2\text{NO} + \text{O}_2 \to 2\text{NO}_2 \) is:

Molecularity is the number of molecules colliding in an elementary step.

For \( 2\text{NO} + \text{O}_2 \to 2\text{NO}_2 \), three molecules (2 NO + 1 O₂) collide, so molecularity = 3.

1
2
3
4
3

The rate of a reaction increases by 27 times when the concentration of the reactant is increased by 3 times. What is the order of the reaction?

Rate = \( k [\text{A}]^n \). New rate: \( k (3[\text{A}])^n = 27 k [\text{A}]^n \), so \( 3^n = 27 \).

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

3
2
1
4
1

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