Chemical Kinetics

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Learning Objectives

Learning Objectives for Chemical Kinetics

Define the average and instantaneous rate of a reaction.
Express the rate of a reaction in terms of change in concentration of either of the reactants or products with time.
Distinguish between elementary and complex reactions.
Differentiate between the molecularity and order of a reaction.
Define rate constant.
Discuss the dependence of rate of reactions on concentration, temperature, and catalyst.
Derive integrated rate equations for the zero and first order reactions.
Determine the rate constants for zeroth and first order reactions.
Describe collision theory.

Detailed Notes

Chemical Kinetics Notes

Overview

Chemical kinetics studies the rates of chemical reactions and the factors affecting them.
It helps understand how reactions occur and the speed at which they proceed.

Key Concepts

Rate of Reaction

Average Rate:
- Formula:
Instantaneous Rate:
- As B4t approaches 0,

Factors Affecting Reaction Rates

Concentration of reactants
Temperature
Presence of catalysts

Rate Laws

General Form:
- Rate = k[A]^m[B]^n
- Where k is the rate constant, and m and n are the orders of the reaction with respect to A and B.

Orders of Reaction

Zero Order: Rate is independent of the concentration of reactants.
First Order: Rate is directly proportional to the concentration of one reactant.
Second Order: Rate is proportional to the square of the concentration of one reactant or the product of the concentrations of two reactants.

Examples

Example 3.2: Decomposition of N₂O₅ in CCl₄ at 318K:
- Initial concentration: 2.33 mol L⁻¹
- After 184 minutes: 2.08 mol L⁻¹
- Average rate calculation provided.

Important Equations

First Order Reaction:
Half-life for First Order Reaction:
- t₁/₂ =

Experimental Data

Example Data Table: | Experiment | [A]/mol L⁻¹ | [B]/mol L⁻¹ | Initial Rate/mol L⁻¹ min⁻¹ | | --- | --- | --- | --- | | I | 0.1 | 0.1 | 6.0 X 10⁻³ | | II | 0.3 | 0.2 | 7.2 X 10⁻² | | III | 0.3 | 0.4 | 2.88 X 10⁻¹ | | IV | 0.4 | 0.1 | 2.40 X 10⁻² |

Conclusion

Understanding chemical kinetics is crucial for predicting how reactions occur and optimizing conditions for desired outcomes.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Reaction Orders: Students often confuse the order of a reaction with the stoichiometry of the reaction. Remember that the order is determined experimentally and can be different from the coefficients in the balanced equation.
Ignoring Temperature Effects: Many students overlook how temperature affects the rate constant (k). The Arrhenius equation shows that k increases with temperature, which can significantly impact reaction rates.
Incorrectly Applying Rate Laws: When deriving rate laws from experimental data, students may fail to account for all variables or misinterpret the data, leading to incorrect conclusions about the reaction order.
Confusing Average and Instantaneous Rates: Students sometimes mix up average rates with instantaneous rates. Ensure you understand the difference and how to calculate each.
Neglecting Units: Failing to include units when calculating rate constants or other quantities can lead to errors. Always check that your units are consistent.

Tips for Success

Practice with Data: Work through problems involving experimental data to derive rate laws and calculate rate constants. Familiarity with data interpretation is crucial.
Understand the Arrhenius Equation: Make sure you can apply the Arrhenius equation to calculate activation energy (Eₐ) and understand its implications on reaction rates.
Use Graphs Effectively: Be comfortable plotting graphs of concentration vs. time and log concentration vs. time to determine reaction orders and rate constants visually.
Review Kinetic Concepts Regularly: Regularly revisit key concepts in chemical kinetics, such as collision theory and the factors affecting reaction rates, to reinforce your understanding.
Work on Time Calculations: Practice calculating half-lives and time required for reactions to reach certain completion percentages, as these are common exam questions.

Practice & Assessment

Multiple Choice Questions

The rate will remain the same.

The rate will double.

The rate will quadruple.

The rate will halve.

Correct Answer: B

Solution:

For a first order reaction with respect to A, the rate is directly proportional to the concentration of A. Thus, if [A] is doubled, the rate will also double.

0.0115 min⁻¹

0.0231 min⁻¹

0.0346 min⁻¹

0.0462 min⁻¹

Correct Answer: B

Solution:

For a first order reaction,

t_{1/2} = \frac{0.693}{k}

. Solving for

k

gives

k = 0.693/60 = 0.0115 \text{ min}^{-1}

1 minute

2 minutes

3 minutes

4 minutes

Correct Answer: B

Solution:

For a first order reaction,

t = \frac{1}{k} \ln \frac{[A]_0}{[A]}

. Here,

[A] = 0.25[A]_0

, so

t = \frac{1}{0.693} \ln 4 = 2 \text{ minutes}

125 kJ/mol

250 kJ/mol

52.897 kJ/mol

60 kJ/mol

Correct Answer: C

Solution:

The energy of activation

E_a

can be calculated from the equation given in the excerpt, which results in

E_a = 52.897

kJ/mol.

1.92 \times 10^{-4} \text{ s}^{-1}

2.30 \times 10^{-4} \text{ s}^{-1}

1.50 \times 10^{-4} \text{ s}^{-1}

2.10 \times 10^{-4} \text{ s}^{-1}

Correct Answer: A

Solution:

Using the Arrhenius equation and the given conditions, the activation energy

E_a

can be calculated. Then, substituting

T = 318 \text{ K}

into the Arrhenius equation

k = A e^{-E_a/RT}

gives

k = 1.92 \times 10^{-4} \text{ s}^{-1}

20°C

30°C

40°C

50°C

Correct Answer: B

Solution:

Using the Arrhenius equation, the temperature at which the rate constant becomes 1.5 X 10⁴ s⁻¹ is calculated to be approximately 30°C.

The rate remains unchanged

The rate doubles

The rate quadruples

The rate increases by a factor of 8

Correct Answer: C

Solution:

The rate law is rate = k[A]^2[B]. If [A] is doubled, the rate increases by a factor of 4 (since it is squared), and if [B] is halved, the rate decreases by a factor of 2. Overall, the rate increases by a factor of

4/2 = 2

233.3 kJ/mol

280.0 kJ/mol

179.9 kJ/mol

150.0 kJ/mol

Correct Answer: B

Solution:

The activation energy

E_a

is given directly in the exponential term as 28000 K, which corresponds to 280.0 kJ/mol.

125 kJ/mol

250 kJ/mol

100 kJ/mol

150 kJ/mol

Correct Answer: A

Solution:

The activation energy

E_a

can be calculated using the Arrhenius equation in logarithmic form. By comparing the given equation with

\log k = \log A - \frac{E_a}{2.303RT}

, we find

E_a = 1.25 \times 10^4 \times 2.303 \times R

, where

R = 8.314 \text{ J/mol K}

. Solving gives

E_a = 125 \text{ kJ/mol}

The rate will remain the same.

The rate will double.

The rate will halve.

The rate will quadruple.

Correct Answer: C

Solution:

The rate law is Rate = k[A][B]². Doubling [A] and halving [B] results in Rate = k(2[A])(0.5[B])² = 0.5k[A][B]², so the rate is halved.

4.0 \times 10^{-8}$$ mol L⁻¹ s⁻¹

8.0 \times 10^{-8}$$ mol L⁻¹ s⁻¹

2.0 \times 10^{-8}$$ mol L⁻¹ s⁻¹

1.0 \times 10^{-8}$$ mol L⁻¹ s⁻¹

Correct Answer: A

Solution:

Substituting the values into the rate law:

\text{rate} = 2.0 \times 10^{-6} \times (0.1) \times (0.2)^2 = 4.0 \times 10^{-8}

mol L⁻¹ s⁻¹.

\text{bar}^{-1/2} \text{ min}^{-1}

\text{bar}^{-3/2} \text{ min}^{-1}

\text{bar}^{1/2} \text{ min}^{-1}

\text{bar}^{-1} \text{ min}^{-1}

Correct Answer: A

Solution:

The rate constant k has units that make the rate equation dimensionally consistent. Here, Rate = k(P_{CH₃OCH₃})^{3/2}, so k must have units of \text{bar}^{-1/2} \text{ min}^{-1} to ensure the rate has units of bar/min.

Rate is halved

Rate is reduced to 36%

Rate remains the same

Rate doubles

Correct Answer: B

Solution:

The rate of reaction is proportional to the concentration of

A

. Reducing

[A]

from 0.1 to 0.06 mol L⁻¹ reduces the rate to 36% of its original value.

1.925 \times 10^{-4}$$ s⁻¹

2.925 \times 10^{-4}$$ s⁻¹

3.925 \times 10^{-4}$$ s⁻¹

4.925 \times 10^{-4}$$ s⁻¹

Correct Answer: A

Solution:

Using the Arrhenius equation and given data, the rate constant

k

at 318K is calculated to be

1.925 \times 10^{-4}

s⁻¹.

1.36 x 10⁻³

1.36 x 10⁻²

1.36 x 10⁻⁴

1.36 x 10⁻¹

Correct Answer: A

Solution:

The average rate of reaction is calculated as

\frac{\Delta [N_2O_5]}{\Delta t} = \frac{2.33 - 2.08}{184} = 1.36 \times 10^{-3}

mol L⁻¹ min⁻¹.

0.0462 seconds

0.1155 seconds

0.2309 seconds

0.462 seconds

Correct Answer: B

Solution:

For a first order reaction,

t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]} ight)

. Here,

[A]_0/[A] = 16

, so

t = \frac{1}{60} \ln(16) \approx 0.1155

seconds.

Activation energy

Collision frequency

Temperature dependence

Reaction order

Correct Answer: B

Solution:

The pre-exponential factor in the Arrhenius equation represents the collision frequency, which is related to the number of collisions between reactant molecules.

1.5

2.25

3.375

4.5

Correct Answer: C

Solution:

The initial rate is

k[A]^3

. If

[A]

is increased by 50%, the new concentration is

1.5[A]

. The new rate is

k(1.5[A])^3 = 3.375k[A]^3

, so the rate increases by a factor of 3.375.

The rate doubles

The rate quadruples

The rate remains the same

The rate halves

Correct Answer: A

Solution:

The rate of reaction is directly proportional to the concentration of A, so doubling the concentration of A will double the rate.

1146 years

2284 years

5730 years

11460 years

Correct Answer: A

Solution:

Using the half-life formula, the age of the sample is estimated to be 1146 years.

The rate remains the same.

The rate doubles.

The rate increases nine times.

The rate triples.

Correct Answer: C

Solution:

According to the rate law Rate =

k[A][B]^2

, if

[B]

is tripled, the rate increases by

3^2 = 9

times.

E_a = 1.25 \times 10^4 \times R

E_a = 1.25 \times 10^4 \times 2.303 \times R

E_a = 1.25 \times 10^4 \times 0.693 \times R

E_a = 1.25 \times 10^4 \times 0.5 \times R

Correct Answer: B

Solution:

The activation energy

E_a

is related to the slope of the Arrhenius plot

\log k = 14.34 - \frac{1.25 \times 10^4}{T}

by the equation

E_a = 1.25 \times 10^4 \times 2.303 \times R

, where

R

is the gas constant.

Correct Answer: B

Solution:

The rate law can be expressed as

\text{rate} = k[A]^m[B]^n

. If doubling

[A]

quadruples the rate,

m = 2

. If doubling

[B]

doubles the rate,

n = 1

. Therefore, the overall order is

m+n = 2+1 = 3

Correct Answer: C

Solution:

The overall order of the reaction is the sum of the powers of the concentration terms in the rate law, which is 1 (for [A]) + 2 (for [B]²) = 3.

0.02 M/min

0.03 M/min

0.04 M/min

0.05 M/min

Correct Answer: A

Solution:

For a zero-order reaction,

[A] = [A]_0 - kt

. Here,

0.4 = 0.8 - k \times 20

, so

k = 0.02 \text{ M/min}

First order in A and first order in B

Second order in A and zero order in B

First order in A and second order in B

Zero order in A and first order in B

Correct Answer: A

Solution:

Based on the given data, the reaction is first order with respect to both A and B.

239.5 kJ/mol

104.2 kJ/mol

125.0 kJ/mol

208.3 kJ/mol

Correct Answer: D

Solution:

The activation energy

E_a

can be calculated using the Arrhenius equation in the form

\log k = \log A - \frac{E_a}{2.303RT}

. Comparing with the given equation,

E_a = 1.25 \times 10^4 \times 2.303 \times R

, where

R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}

. Thus,

E_a = 208.3 \text{ kJ/mol}

120 minutes

180 minutes

240 minutes

300 minutes

Correct Answer: C

Solution:

For a first order reaction, the time required to reduce the concentration to 1/16th is four half-lives. Since the half-life is 60 minutes, the time required is

4 \times 60 = 240

minutes.

0.023 s

0.046 s

0.092 s

0.184 s

Correct Answer: B

Solution:

For a first order reaction, the time to reduce to 1/16th is given by

t = \frac{4 \times 0.693}{k}

. Substituting

k = 60

s⁻¹, we get

t = \frac{4 \times 0.693}{60} = 0.046

The rate constant will increase by a factor of 1.5.

The rate constant will increase by a factor of 2.0.

The rate constant will increase by a factor of 2.5.

The rate constant will increase by a factor of 3.0.

Correct Answer: A

Solution:

Using the Arrhenius equation

k = Ae^{-E_a/RT}

, the ratio of rate constants at two temperatures is given by

\frac{k_2}{k_1} = e^{\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)}

. Substituting

E_a = 50 \text{ kJ/mol}

R = 8.314 \text{ J/mol K}

T_1 = 300 \text{ K}

, and

T_2 = 310 \text{ K}

, we find

\frac{k_2}{k_1} \approx 1.5

52.9 kJ/mol

35.0 kJ/mol

70.0 kJ/mol

60.0 kJ/mol

Correct Answer: A

Solution:

Using the Arrhenius equation and the given temperature change, the activation energy

E_a

can be calculated. The factor of 4 increase in rate corresponds to a specific

E_a

value, which is 52.9 kJ/mol.

The rate increases by a factor of 6

The rate increases by a factor of 18

The rate increases by a factor of 12

The rate increases by a factor of 36

Correct Answer: D

Solution:

The initial rate is

k[A]^2[B]

. When

[A]

is tripled and

[B]

is doubled, the new rate becomes

k(3[A])^2(2[B]) = 18k[A]^2[B]

. Thus, the rate increases by a factor of 36.

4.0 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}

8.0 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}

2.0 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}

1.6 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}

Correct Answer: A

Solution:

Substitute the given values into the rate expression:

\text{Rate} = (2.0 \times 10^{-6} \text{ mol}^{-2} \text{ L}^2 \text{ s}^{-1})(0.1 \text{ mol L}^{-1})(0.2 \text{ mol L}^{-1})^2 = 4.0 \times 10^{-9} \text{ mol L}^{-1} \text{ s}^{-1}

The rate constant decreases with increasing temperature.

The rate constant is independent of temperature.

The rate constant increases with increasing temperature.

The rate constant remains constant with temperature changes.

Correct Answer: C

Solution:

The rate constant generally increases with increasing temperature due to the increased kinetic energy of the molecules, leading to more frequent and effective collisions.

0.02 mol L^-1 min^-1

0.03 mol L^-1 min^-1

0.04 mol L^-1 min^-1

0.05 mol L^-1 min^-1

Correct Answer: A

Solution:

For a zero-order reaction, the rate is constant and given by

k = \frac{[R]_0 - [R]}{t}

. Substituting the given values:

k = \frac{0.5 - 0.3}{10} = 0.02 \text{ mol L}^{-1} \text{ min}^{-1}

50 kJ/mol

100 kJ/mol

125 kJ/mol

250 kJ/mol

Correct Answer: C

Solution:

The activation energy

E_a

can be calculated using the Arrhenius equation. Here, the slope of the plot of

\log k

versus

1/T

gives

-E_a/2.303R

. Solving for

E_a

gives 125 kJ/mol.

1.36 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}

6.79 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1}

3.40 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1}

2.72 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}

Correct Answer: B

Solution:

The average rate is calculated as \frac{[N₂O₅]{initial} - [N₂O₅]{final}}{time} = \frac{2.33 - 2.08}{184} = 6.79 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1}.

52.9 kJ/mol

44.0 kJ/mol

58.3 kJ/mol

60.5 kJ/mol

Correct Answer: A

Solution:

Using the equation

\ln \left(\frac{r_2}{r_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)

, where

r_2/r_1 = 4

T_1 = 293 \text{ K}

T_2 = 313 \text{ K}

, and

R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}

, solving gives

E_a = 52.9 \text{ kJ/mol}

10 seconds

20 seconds

30 seconds

40 seconds

Correct Answer: C

Solution:

For a first order reaction, the time taken to reduce the concentration to 1/8th is calculated using the formula

t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}

. Here,

[R]_0/[R] = 8

, and

k = 0.1 \text{ s}^{-1}

. Thus,

t = \frac{2.303}{0.1} \log 8 = 30 \text{ seconds}

4.0 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}

8.0 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}

2.0 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}

1.0 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}

Correct Answer: A

Solution:

The initial rate is calculated as Rate = k[A][B]² =

2.0 \times 10^{-6} \times 0.1 \times (0.2)^2 = 4.0 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}.

The rate constant remains the same.

The rate constant doubles.

The rate constant quadruples.

The rate constant decreases.

Correct Answer: C

Solution:

The rate constant increases with temperature, and in this case, it quadruples as the rate of the reaction quadruples.

0.5

Correct Answer: A

Solution:

The rate law is given by

\text{Rate} = k[A][B]^2

. If

[A]

is doubled, the rate becomes

2[A][B]^2

. If

[B]

is halved, the rate becomes

2[A](0.5[B])^2 = 2[A][B]^2/4 = 0.5[A][B]^2

. Therefore, the overall change in rate is a factor of 2.

52.897 kJ/mol

104.5 kJ/mol

209.5 kJ/mol

418.0 kJ/mol

Correct Answer: A

Solution:

Using the Arrhenius equation and the given temperature change, the energy of activation

E_a

is calculated to be 52.897 kJ/mol.

The rate will double.

The rate will remain unchanged.

The rate will increase by a factor of 4.

The rate will decrease by a factor of 2.

Correct Answer: A

Solution:

According to the rate law

\text{Rate} = k[X]^2[Y]

, if

[X]

is doubled, the rate increases by a factor of

2^2 = 4

. If

[Y]

is halved, the rate decreases by a factor of 2. Thus, the overall change in rate is

4/2 = 2

, so the rate will double.

s⁻¹

mol L⁻¹ s⁻¹

L mol⁻¹ s⁻¹

mol² L⁻² s⁻¹

Correct Answer: A

Solution:

For a first order reaction, the unit of the rate constant

k

is s⁻¹.

1 minute

0.5 minutes

2 minutes

0.693 minutes

Correct Answer: A

Solution:

For a first order reaction,

t_{1/2} = \frac{0.693}{k}

. Here,

k = 0.693 \text{ min}^{-1}

, so

t_{1/2} = \frac{0.693}{0.693} = 1

minute.

0.0115 s

0.0231 s

0.115 s

1.15 s

Correct Answer: A

Solution:

The half-life of a first order reaction is given by

t_{1/2} = \frac{0.693}{k}

. Substituting

k = 60

s⁻¹ gives

t_{1/2} = 0.0115

The rate constant increases by a factor of 2.5

The rate constant increases by a factor of 3.0

The rate constant increases by a factor of 4.0

The rate constant remains unchanged

Correct Answer: A

Solution:

The Arrhenius equation is given by

k = A e^{-E_a/RT}

. Given that the rate constant doubles when the temperature increases from 298 K to 308 K, we can use this information to calculate the rate constant at 318 K. The activation energy is 52.897 kJ/mol, and using the Arrhenius equation, the rate constant at 318 K increases by a factor of approximately 2.5.

1.5 \times 10^{-3} \text{ s}^{-1}

2.0 \times 10^{-3} \text{ s}^{-1}

1.0 \times 10^{-3} \text{ s}^{-1}

2.5 \times 10^{-3} \text{ s}^{-1}

Correct Answer: A

Solution:

Using the Arrhenius equation and the given conditions, the rate constant

k

at 318 K can be calculated. The equal time condition implies a relationship between the rate constants at different temperatures, allowing for the determination of

k

at 318 K.

0.135 mol L⁻¹

0.818 mol L⁻¹

0.670 mol L⁻¹

0.500 mol L⁻¹

Correct Answer: B

Solution:

For a first order reaction, the concentration at time

t

is given by

[A]_t = [A]_0 e^{-kt}

. Substituting the values,

[A]_t = 1.0 \times e^{-2.0 \times 10^{-2} \times 100} = 0.818

mol L⁻¹.

mol⁻¹ L s⁻¹

mol⁻1/2 L¹/2 s⁻¹

mol⁻3/2 L³/2 s⁻¹

mol L⁻¹ s⁻¹

Correct Answer: B

Solution:

For a rate law of the form Rate =

k[CH₃OCH₃]^{3/2}

, the units of

k

must be such that they cancel out the units of concentration raised to the power of 3/2, resulting in the units of rate. Thus, the units of

k

are mol⁻1/2 L¹/2 s⁻¹.

60 min

80 min

100 min

120 min

Correct Answer: B

Solution:

For a first order reaction,

t_{1/2} = \frac{0.693}{k}

. Using the formula

k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}

, where

t = 40

min and

\frac{[A]}{[A]_0} = 0.7

, we find

k

. Then,

t_{1/2} = \frac{0.693}{k}

, which gives approximately 80 min.

First order

Second order

Third order

Zero order

Correct Answer: C

Solution:

The order of the reaction is the sum of the powers of the concentration terms in the rate law, which is 1 (for

A

) + 2 (for

B

) = 3.

120 minutes

180 minutes

240 minutes

90 minutes

Correct Answer: B

Solution:

For a first order reaction, the time taken to reduce to

\frac{1}{8}

th is three half-lives. Since the half-life is 60 minutes, the time taken is

3 \times 60 = 180

minutes.

2.5 g

5.0 g

1.25 g

0.625 g

Correct Answer: C

Solution:

After 11460 years, which is two half-lives, the amount remaining is

\frac{10}{2^2} = 2.5 \text{ g}

231 s

462 s

693 s

924 s

Correct Answer: B

Solution:

For a first-order reaction, the time

t

required for the concentration to decrease to a fraction

f

of its initial value is given by

t = \frac{1}{k} \ln \frac{1}{f}

. Here,

f = 0.25

and

k = 3.0 \times 10^{-3} \text{s}^{-1}

. Therefore,

t = \frac{1}{3.0 \times 10^{-3}} \ln 4 = 462 \text{s}

0.036 mol L⁻¹

0.046 mol L⁻¹

0.056 mol L⁻¹

0.066 mol L⁻¹

Correct Answer: A

Solution:

For a pseudo first-order reaction, the concentration at time

t

is given by

[A] = [A]_0 e^{-kt}

. Here,

[A]_0 = 0.1 \text{ mol L}^{-1}

k = 1.0 \times 10^{-4} \text{s}^{-1}

, and

t = 3600 \text{s}

. Therefore,

[A] = 0.1 e^{-1.0 \times 10^{-4} \times 3600} = 0.036 \text{ mol L}^{-1}

It is independent of the initial concentration.

It is directly proportional to the initial concentration.

It is inversely proportional to the initial concentration.

It is exponentially related to the initial concentration.

Correct Answer: B

Solution:

For a zero order reaction, the half-life is directly proportional to the initial concentration of the reactants.

52.897 kJ/mol

43.897 kJ/mol

62.897 kJ/mol

32.897 kJ/mol

Correct Answer: A

Solution:

The activation energy can be calculated using the Arrhenius equation. Given that the rate doubles,

E_a

is approximately 52.897 kJ/mol.

Zero order

First order

Second order

Third order

Correct Answer: A

Solution:

The units of the rate constant for a zero order reaction are mol L⁻¹ s⁻¹. Here, the given rate constant is 2.5 x 10⁻⁴ mol⁻¹ Ls⁻¹, which matches the units for a zero order reaction.

52.897 kJ/mol

104.5 kJ/mol

209 kJ/mol

418 kJ/mol

Correct Answer: A

Solution:

Using the Arrhenius equation, the activation energy is calculated to be 52.897 kJ/mol.

The rate will increase by a factor of 3.

The rate will increase by a factor of 6.

The rate will increase by a factor of 9.

The rate will remain unchanged.

Correct Answer: C

Solution:

For the rate law

\text{Rate} = k[A]^2

, if the concentration of A is tripled, the rate increases by a factor of

3^2 = 9

1150 years

2300 years

5730 years

4300 years

Correct Answer: D

Solution:

Using the first-order decay equation

N = N_0 e^{-kt}

and the given half-life, the decay constant

k

is calculated. Substituting

N/N_0 = 0.8

and solving for

t

, the age of the sample is found to be approximately 4300 years.

1.0 \times 10^{12} \text{s}^{-1}

2.0 \times 10^{11} \text{s}^{-1}

4.5 \times 10^{11} \text{s}^{-1}

3.0 \times 10^{12} \text{s}^{-1}

Correct Answer: C

Solution:

Using the Arrhenius equation

k = A e^{-E_a/RT}

, rearrange to find

A

. Given

k = 2.418 \times 10^{-5} \text{s}^{-1}

E_a = 179.9 \text{kJ/mol} = 179900 \text{J/mol}

R = 8.314 \text{J/mol K}

, and

T = 546 \text{K}

, solve for

A

to get

4.5 \times 10^{11} \text{s}^{-1}

52.8 kJ/mol

44.0 kJ/mol

58.0 kJ/mol

60.0 kJ/mol

Correct Answer: A

Solution:

Using the Arrhenius equation,

k = Ae^{-E_a/RT}

, and the fact that the rate doubles, we have

E_a = 2.303 \times R \times \log(2) \times \frac{298 \times 308}{10} \approx 52.8 \text{ kJ/mol}

3.6 x 10^-8 mol L^-1 s^-1

1.8 x 10^-8 mol L^-1 s^-1

7.2 x 10^-8 mol L^-1 s^-1

4.5 x 10^-8 mol L^-1 s^-1

Correct Answer: A

Solution:

Using the rate law: rate = k[A][B]^2, substitute the given values:

rate = 2.0 \times 10^{-6} \times 0.2 \times (0.3)^2 = 3.6 \times 10^{-8} \text{ mol L}^{-1} \text{ s}^{-1}

0.5 minutes

1 minute

2 minutes

10 minutes

Correct Answer: B

Solution:

For a first order reaction, the half-life

t_{1/2}

is given by

t_{1/2} = \frac{0.693}{k}

. Here,

k = 0.693 \text{ min}^{-1}

, so

t_{1/2} = \frac{0.693}{0.693} = 1 \text{ minute}

1.23 \times 10^{-5}

3.67 \times 10^{-4}

2.89 \times 10^{-6}

5.12 \times 10^{-3}

Correct Answer: A

Solution:

The fraction of molecules with energy equal to or greater than

E_a

is given by the Boltzmann factor

e^{-E_a/RT}

. Substituting

E_a = 209.5 \times 10^3 \text{ J mol}^{-1}

R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}

, and

T = 581 \text{ K}

, the fraction is

1.23 \times 10^{-5}

0.5

Correct Answer: B

Solution:

The rate law is

\text{rate} = k[A]^2[B]

. Doubling

[A]

increases the rate by a factor of 4 (since

[A]^2

becomes

(2[A])^2 = 4[A]^2

), and halving

[B]

decreases the rate by a factor of 0.5. Therefore, the overall change is

4 \times 0.5 = 2

1 with respect to

A

and 2 with respect to

B

2 with respect to

A

and 1 with respect to

B

2 with respect to

A

and 2 with respect to

B

1 with respect to

A

and 1 with respect to

B

Correct Answer: A

Solution:

The order of a reaction with respect to a reactant is the exponent of its concentration term in the rate law. Here, the rate law is

\text{rate} = k[A][B]^2

, indicating the reaction is first order with respect to

A

and second order with respect to

B

20°C

30°C

40°C

50°C

Correct Answer: B

Solution:

Using the Arrhenius equation,

k = Ae^{-E_a/RT}

, and solving for the temperature when

k = 1.5 \times 10^4 \text{s}^{-1}

gives approximately 30°C.

25°C

35°C

45°C

55°C

Correct Answer: B

Solution:

Using the Arrhenius equation

\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)

, where

k_1 = 4.5 \times 10^3 \text{ s}^{-1}

k_2 = 1.5 \times 10^4 \text{ s}^{-1}

E_a = 60 \times 10^3 \text{ J mol}^{-1}

, and

R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}

. Solving gives

T_2 = 308 \text{ K}

35°C

t_{1/2} = \frac{0.693}{k}

t_{1/2} = \frac{1}{2k}

t_{1/2} = \frac{1}{k}

t_{1/2} = \frac{0.5}{k}

Correct Answer: A

Solution:

For a first-order reaction, the half-life is given by the equation

t_{1/2} = \frac{0.693}{k}

. This is a standard result derived from the integrated rate law for first-order reactions.

True or False

Correct Answer: True

Solution:

The order of a reaction is determined experimentally and can be zero, a whole number, or even a fraction.

Correct Answer: False

Solution:

For a first order reaction, the time required to reduce the concentration to 1/16th of its initial value can be calculated using the formula:

t = \frac{4 \times 0.693}{k}

. Given

k = 60 \text{s}^{-1}

t = \frac{4 \times 0.693}{60} \approx 0.0462 \text{s}

Correct Answer: True

Solution:

Rate of reaction depends on the concentration of reactants, and generally, higher concentrations lead to higher reaction rates.

Correct Answer: True

Solution:

For a first order reaction, the half-life is given by the equation

t_{1/2} = \frac{0.693}{k}

, which shows that it is constant and independent of the initial concentration.

Correct Answer: True

Solution:

The decomposition of H₂O₂ follows first order kinetics as indicated by the rate constant expression provided.

Correct Answer: True

Solution:

The reaction is second order with respect to

B

because the exponent of

Solution:

The given equation is used to calculate the rate constant for the first order decomposition of

Solution:

For zero order reactions, the half-life is directly proportional to the initial concentration of the reactants, unlike first order reactions where it is independent.

Chemical Kinetics

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Learning Objectives

Learning Objectives for Chemical Kinetics

Detailed Notes

Chemical Kinetics Notes

Overview

Key Concepts

Rate of Reaction

Factors Affecting Reaction Rates

Rate Laws

Orders of Reaction

Examples

Important Equations

First Order Reaction:

Experimental Data

Conclusion

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.A reaction is first order with respect to A and zero order with respect to B. If the concentration of A is doubled, how will the rate change?

Correct Answer: B

Q2.The half-life of a first order reaction is 60 minutes. What is the rate constant kkk for this reaction?

Correct Answer: B

Q3.A first order reaction has a rate constant k=0.693 min−1k = 0.693 \text{ min}^{-1}k=0.693 min−1. How long will it take for the concentration of the reactant to reduce to 25% of its initial value?

Correct Answer: B

Q4.The rate constant for a first order reaction is given by the equation: log⁡k=14.34−1.25×104T\log k = 14.34 - \frac{1.25 \times 10^4}{T}logk=14.34−T1.25×104​. What is the energy of activation EaE_aEa​ for this reaction?

Correct Answer: C

Correct Answer: A

Q6.For the decomposition of a substance A, the rate constant is given as 4.5 X 10³ s⁻¹ at 10°C. If the activation energy is 60 kJ/mol, at what temperature would the rate constant be 1.5 X 10⁴ s⁻¹?

Correct Answer: B

Q7.In a reaction where the rate law is given by rate = k[A]^2[B], if the concentration of A is doubled and B is halved, by what factor does the rate of reaction change?

Correct Answer: C

Q8.The decomposition of a hydrocarbon follows the equation k=(4.5×1011s−1)e−28000K/Tk = (4.5 \times 10^{11} \text{s}^{-1}) e^{-28000 \text{K}/T}k=(4.5×1011s−1)e−28000K/T. What is the activation energy EaE_aEa​?

Correct Answer: B

Q9.The rate constant for the first-order decomposition of H₂O₂ is given by the equation log⁡k=14.34−1.25×104T\log k = 14.34 - \frac{1.25 \times 10^4}{T}logk=14.34−T1.25×104​. Calculate the activation energy EaE_aEa​ for this reaction.

Correct Answer: A

Correct Answer: C

Correct Answer: A

Q12.In a reaction where the rate is given by Rate = k[CH₃OCH₃]^{3/2}, what are the units of the rate constant k if the rate is measured in bar/min?

Correct Answer: A

Q13.For a reaction 2A+B→C+D2A + B \rightarrow C + D2A+B→C+D, the rate law is given by rate=k[A][B]2\text{rate} = k[A][B]^2rate=k[A][B]2. If the concentration of AAA is reduced to 0.06 mol L⁻¹ from 0.1 mol L⁻¹, what happens to the rate of reaction?

Correct Answer: B

Q14.The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of AAA is 4×10104 \times 10^{10}4×1010 s⁻¹, what is the rate constant kkk at 318K?

Correct Answer: A

Correct Answer: A

Q16.The rate constant for a first order reaction is 60 s⁻¹. How much time will it take for the concentration of the reactant to reduce to 1/16th of its initial value?

Correct Answer: B

Q17.The rate constant for a reaction is given by the Arrhenius equation. What does the pre-exponential factor represent?

Correct Answer: B

Q18.Consider a reaction with the rate law rate=k[A]3\text{rate} = k[A]^3rate=k[A]3. If the concentration of AAA is increased by 50%, by what factor does the rate of reaction increase?

Correct Answer: C

Q19.In a reaction between A and B, the initial rate of reaction was measured for different initial concentrations of A and B. If the concentration of A is doubled, how does the rate change?

Correct Answer: A

Q20.The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

Correct Answer: A

Q21.In the reaction 2A+B→C+D2A + B \rightarrow C + D2A+B→C+D, the rate law is given by Rate = k[A][B]2k[A][B]^2k[A][B]2. If the concentration of BBB is tripled, how does the rate change?

Correct Answer: C

Q22.The rate constant for the first order decomposition of a substance is given by the equation log⁡k=14.34−1.25×104T\log k = 14.34 - \frac{1.25 \times 10^4}{T}logk=14.34−T1.25×104​. If the activation energy EaE_aEa​ is to be calculated, which of the following expressions is correct?

Correct Answer: B

Q23.In a reaction between A and B, the initial rate of reaction r0r_0r0​ was measured for different initial concentrations of A and B. If doubling the concentration of A quadruples the rate and doubling the concentration of B doubles the rate, what is the overall order of the reaction?

Correct Answer: B

Q24.In a reaction where the rate law is given by Rate = k[A][B]², what is the overall order of the reaction?

Correct Answer: C

Q25.For a zero-order reaction, the concentration of a reactant decreases from 0.8 M to 0.4 M in 20 minutes. What is the rate constant kkk of the reaction?

Correct Answer: A

Q26.In a reaction between A and B, the initial rate of reaction r0r_0r0​ was measured for different initial concentrations of A and B. What is the order of the reaction with respect to A and B?

Correct Answer: A

Q27.In a reaction where the rate constant is given by the equation log⁡k=14.34−1.25×104T\log k = 14.34 - \frac{1.25 \times 10^4}{T}logk=14.34−T1.25×104​, what is the activation energy EaE_aEa​ for this reaction?

Correct Answer: D

Q28.A first order reaction has a half-life of 60 minutes. How much time will it take for the concentration of the reactant to reduce to 1/16th of its initial value?

Correct Answer: C

Q29.The rate constant for a first order reaction is 60 s⁻¹. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Correct Answer: B

Q30.The activation energy for a certain reaction is 50 kJ/mol. By how much will the rate constant increase if the temperature is raised from 300 K to 310 K, assuming the Arrhenius equation holds?

Correct Answer: A

Q31.For a reaction where the rate quadruples when the temperature increases from 293 K to 313 K, calculate the activation energy EaE_aEa​ assuming it does not change with temperature.