Chemical Kinetics Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A reaction has the rate law \( \text{Rate} = k[\text{A}]^2[\text{B}]^2 \). If the concentration of A is doubled and B is reduced to half, what happens to the rate?

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

New rate = \( k (2[\text{A}])^2 \left( \frac{[\text{B}]}{2} \right)^2 = k \times 4 [\text{A}]^2 \times \frac{[\text{B}]^2}{4} = k [\text{A}]^2 [\text{B}]^2 \).

Rate remains the same.

Increases by 4 times
Doubles
Remains same
Decreases to half
3

A reaction’s rate increases by 16 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: \( k (4[\text{A}])^n = 16 k [\text{A}]^n \), so \( 4^n = 16 \).

\( 4^n = 4^2 \), \( n = 2 \).

2
1
3
4
1

A reaction’s rate increases by 9 times when the temperature rises from 330 K to 350 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log 9 = 0.954 \), \( 0.954 = \frac{E_a}{2.303 \times 8.314} \left( \frac{20}{330 \times 350} \right) \).

\( E_a = \frac{0.954 \times 19.147 \times 115500}{20} \approx 105,500 \, \text{J mol}^{-1} \approx 105.5 \, \text{kJ mol}^{-1} \).

52.8 kJ mol\(^{-1}\)
79.1 kJ mol\(^{-1}\)
131.9 kJ mol\(^{-1}\)
105.5 kJ mol\(^{-1}\)
4

A reaction’s rate constant increases from \( 2.0 \times 10^{-4} \, \text{min}^{-1} \) at 290 K to \( 6.0 \times 10^{-4} \, \text{min}^{-1} \) at 300 K. What is the activation energy (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

\( \log \frac{6.0 \times 10^{-4}}{2.0 \times 10^{-4}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{290 \times 300} \right) \).

\( \log 3 = 0.477 \), \( E_a = \frac{0.477 \times 19.147 \times 87000}{10} \approx 79,400 \, \text{J mol}^{-1} \approx 79.4 \, \text{kJ mol}^{-1} \).

39.7 kJ mol\(^{-1}\)
59.6 kJ mol\(^{-1}\)
99.2 kJ mol\(^{-1}\)
79.4 kJ mol\(^{-1}\)
4

The molecularity of the elementary step \( \text{NO}_2 + \text{CO} \to \text{NO} + \text{CO}_2 \) is:

Molecularity is the number of molecules in an elementary step.

For \( \text{NO}_2 + \text{CO} \), two molecules collide, so molecularity = 2.

1
3
2
4
3

A reaction’s rate constant is \( 1.0 \times 10^{-2} \, \text{s}^{-1} \) at 27°C. If the activation energy is 50 kJ/mol, what is the rate constant at 37°C (\( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \))?

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

\( \log \frac{k_2}{1.0 \times 10^{-2}} = \frac{50000}{2.303 \times 8.314} \left( \frac{10}{300 \times 310} \right) \approx 0.281 \).

\( \frac{k_2}{1.0 \times 10^{-2}} = 10^{0.281} \approx 1.91 \), \( k_2 \approx 1.91 \times 10^{-2} \, \text{s}^{-1} \).

1.5 \(\times\) 10\(^{-2}\) s\(^{-1}\)
1.91 \(\times\) 10\(^{-2}\) s\(^{-1}\)
2.5 \(\times\) 10\(^{-2}\) s\(^{-1}\)
3.0 \(\times\) 10\(^{-2}\) s\(^{-1}\)
2

The rate of a reaction is given by \( \text{Rate} = k[\text{A}]^{1/2}[\text{B}] \). What is the unit of the rate constant?

Rate = \( k [\text{A}]^{1/2} [\text{B}] \), order = \( \frac{1}{2} + 1 = 1.5 \).

Unit of rate = \( \text{mol L}^{-1} \text{s}^{-1} \), unit of \( [\text{A}]^{1/2} [\text{B}] = (\text{mol L}^{-1})^{1/2} \cdot \text{mol L}^{-1} = \text{mol}^{1.5} \text{L}^{-1.5} \).

\( k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^{1.5} \text{L}^{-1.5}} = \text{mol}^{-0.5} \text{L}^{0.5} \text{s}^{-1} \).

s\(^{-1}\)
L mol\(^{-1}\) s\(^{-1}\)
mol\(^{-0.5}\) L\(^{0.5}\) s\(^{-1}\)
mol\(^{-1}\) L s\(^{-1}\)
3

A first-order gaseous reaction has an initial pressure of 0.6 atm. After 20 s, the total pressure is 0.9 atm. What is the rate constant?

For \( \text{A} \to 2\text{B} \), \( p_t = p_i + x \), \( x = 0.9 - 0.6 = 0.3 \, \text{atm} \).

\( p_A = p_i - x = 0.6 - 0.3 = 0.3 \, \text{atm} \).

\( k = \frac{2.303}{20} \log \frac{0.6}{0.3} = \frac{2.303 \times 0.301}{20} \approx 0.0346 \, \text{s}^{-1} \).

0.0231 s\(^{-1}\)
0.0288 s\(^{-1}\)
0.0462 s\(^{-1}\)
0.0346 s\(^{-1}\)
4

A first-order reaction is 40% complete in 20 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, \( \frac{[\text{R}]}{[\text{R}]_0} = 0.6 \), \( \frac{[\text{R}]_0}{[\text{R}]} = \frac{1}{0.6} \approx 1.667 \).

\( k = \frac{2.303}{20} \log 1.667 = \frac{2.303 \times 0.222}{20} \approx 0.0256 \, \text{min}^{-1} \).

0.0154 min\(^{-1}\)
0.0256 min\(^{-1}\)
0.0346 min\(^{-1}\)
0.0462 min\(^{-1}\)
2

A first-order reaction has an initial concentration of \( 0.8 \, \text{mol L}^{-1} \). After 30 minutes, the concentration reduces to \( 0.4 \, \text{mol L}^{-1} \). What is the rate constant?

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

Given: \( t = 30 \, \text{min} \), \( [\text{R}]_0 = 0.8 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.4 \, \text{mol L}^{-1} \).

\( k = \frac{2.303}{30} \log \frac{0.8}{0.4} = \frac{2.303}{30} \log 2 = \frac{2.303 \times 0.301}{30} \approx 0.0231 \, \text{min}^{-1} \).

0.0154 min\(^{-1}\)
0.0231 min\(^{-1}\)
0.0346 min\(^{-1}\)
0.0462 min\(^{-1}\)
2

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