1. Home
    2. >
  2. Neet Chemistry Class 12 Mock Test Papers
    3. >
  3. Chemistry Chapter Wise Test Papers
    4. >
  4. Chemical Kinetics Chapter-Wise Test 3

Chemical Kinetics Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

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

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

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

The decomposition of a gas follows first-order kinetics with an initial pressure of 2.0 atm. After 60 s, the total pressure is 2.8 atm. What is the rate constant?

For \( \text{A} \to \text{B} + \text{C} \), \( p_t = p_i + x \), \( x = 2.8 - 2.0 = 0.8 \, \text{atm} \).

\( p_A = p_i - x = 2.0 - 0.8 = 1.2 \, \text{atm} \).

\( k = \frac{2.303}{60} \log \frac{2.0}{1.2} = \frac{2.303 \times 0.176}{60} \approx 0.00675 \, \text{s}^{-1} \).

0.00462 s\(^{-1}\)
0.00675 s\(^{-1}\)
0.00892 s\(^{-1}\)
0.0115 s\(^{-1}\)
2

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

Molecularity is the number of molecules in an elementary step.

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

2
3
1
4
2

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

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

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

Rate increases by \( \frac{9}{2} = 4.5 \) times.

Increases by 9 times
Decreases to half
Increases by 4.5 times
Remains same
3

A reaction’s rate 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
4
1
1

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

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

New rate = \( k (2[\text{A}])^2 (3[\text{B}}) = k \times 4 [\text{A}]^2 \times 3 [\text{B}] = 12 k [\text{A}]^2 [\text{B}] \).

Rate increases by 12 times.

6 times
8 times
12 times
4 times
3

The decomposition of a reactant follows zero-order kinetics with a rate constant of \( 2.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \). If the initial concentration is \( 0.1 \, \text{mol L}^{-1} \), how long will it take for the concentration to decrease to \( 0.05 \, \text{mol L}^{-1} \)?

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

Given: \( k = 2.0 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \), \( [\text{R}]_0 = 0.1 \, \text{mol L}^{-1} \), \( [\text{R}] = 0.05 \, \text{mol L}^{-1} \).

\( t = \frac{0.1 - 0.05}{2.0 \times 10^{-3}} = \frac{0.05}{2.0 \times 10^{-3}} = 25 \, \text{s} \).

25 s
50 s
75 s
100 s
1

The decomposition of a gas follows first-order kinetics with an initial pressure of 0.8 atm. After 100 s, the total pressure is 1.2 atm. What is the rate constant?

For \( \text{A} \to \text{B} + \text{C} \), \( p_t = p_i + x \), \( x = p_t - p_i = 1.2 - 0.8 = 0.4 \, \text{atm} \).

\( p_A = p_i - x = 0.8 - 0.4 = 0.4 \, \text{atm} \).

\( k = \frac{2.303}{t} \log \frac{p_i}{p_A} = \frac{2.303}{100} \log \frac{0.8}{0.4} = \frac{2.303 \times 0.301}{100} \approx 6.93 \times 10^{-3} \, \text{s}^{-1} \).

4.62 \(\times\) 10\(^{-3}\) s\(^{-1}\)
5.78 \(\times\) 10\(^{-3}\) s\(^{-1}\)
8.66 \(\times\) 10\(^{-3}\) s\(^{-1}\)
6.93 \(\times\) 10\(^{-3}\) s\(^{-1}\)
4

A first-order gaseous reaction has an initial pressure of 1.2 atm. After 50 s, the total pressure is 1.6 atm. What is the rate constant?

For \( \text{A} \to \text{B} + \text{C} \), \( p_t = p_i + x \), \( x = 1.6 - 1.2 = 0.4 \, \text{atm} \).

\( p_A = p_i - x = 1.2 - 0.4 = 0.8 \, \text{atm} \).

\( k = \frac{2.303}{50} \log \frac{1.2}{0.8} = \frac{2.303 \times 0.176}{50} \approx 0.0081 \, \text{s}^{-1} \).

0.0058 s\(^{-1}\)
0.0115 s\(^{-1}\)
0.0069 s\(^{-1}\)
0.0081 s\(^{-1}\)
4

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

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

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

Rate increases by 9 times.

3
9
6
12
2

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0
error: Content is protected !!