Exhaustive Guide: Arrhenius Equation & Collision Theory | Chemical Kinetics

Temperature Dependence & Collision Theory

Module 3 | CBSE Class 12 Chemistry | Kinetics Chapter

Table of Contents

• 1. Temperature Dependence of the Rate of a Reaction
• 2. The Arrhenius Equation
  • 2.1 Activation Energy (E_a)
  • 2.2 Graphical Representation & Two-Temperature Form
• 3. Maxwell-Boltzmann Energy Distribution
• 4. Effect of a Catalyst
• 5. Collision Theory of Chemical Reactions
• 6. NCERT Solved Examples
• 7. Previous Year Questions (PYQs)

1. Temperature Dependence of the Rate of a Reaction

It is a general observation that chemical reactions accelerate when the temperature increases. For example, in a mixture of potassium permanganate (KMnO₄) and oxalic acid, decolorization is much faster at a higher temperature than at a lower temperature.

The "10 Degree" Rule (Temperature Coefficient): For a chemical reaction, with a rise in temperature by 10°C (or 10 K), the rate constant is nearly doubled (and sometimes even trebled).

The Temperature Coefficient is defined as the ratio of the rate constant of the reaction at two temperatures differing by 10°C. Typically, it is measured at 308 K and 298 K.
Temperature Coefficient = k_{(T + 10)} / k_T ≈ 2 to 3

2. The Arrhenius Equation

The precise quantitative relationship between temperature and the rate constant was proposed by Svante Arrhenius. It combines the concepts of activation energy and the Boltzmann distribution.

k = A e^{-E_a / RT}

Where:
k = Rate constant of the reaction.
A = Arrhenius factor, frequency factor, or pre-exponential factor. It represents the frequency of collisions.
E_a = Activation Energy (in Joules/mol).
R = Universal Gas Constant (8.314 J K^-1 mol^-1).
T = Absolute Temperature (in Kelvin).
e^-E_a/RT = The fraction of molecules having kinetic energy equal to or greater than the activation energy (E_a).

2.1 Activation Energy (E_a)

Molecules must possess a certain minimum amount of energy to undergo a chemical reaction. This minimum energy required to form the activated complex (or transition state) is called Activation Energy (E_a).

Threshold Energy: The total minimum energy that colliding molecules must have for the collision to be effective.
Threshold Energy = Initial Kinetic Energy of Reactants + Activation Energy (E_a)

2.2 Graphical Representation & Two-Temperature Form

Taking the natural logarithm (ln) on both sides of the Arrhenius equation:

ln k = - E_a / RT + ln A

Converting to base-10 logarithm (log):

log k = - E_a / (2.303 RT) + log A

Graph of log k vs 1/T:
If we plot log k on the y-axis and 1/T on the x-axis, we obtain a straight line (y = mx + c).

• Slope (m) = - E_a / 2.303 R
• Intercept (c) = log A

Calculating E_a from Two Temperatures:
If the rate constant is k₁ at temperature T₁ and k₂ at temperature T₂, integrating the equation yields the highly tested two-temperature formula:

log (k₂ / k₁) = (E_a / 2.303 R) × [ (T₂ - T₁) / (T₁ T₂) ]

3. Maxwell-Boltzmann Energy Distribution

Why does a mere 10°C rise in temperature double the reaction rate? Does the kinetic energy of molecules double? No.

According to the Maxwell-Boltzmann distribution, kinetic energy is not distributed equally among molecules. As the temperature increases from T to (T + 10), the peak of the curve shifts forward (higher energy) and the curve broadens.

The Real Reason: The fraction of molecules with energy greater than the Activation Energy (E_a) corresponds to the area under the curve beyond E_a. When temperature increases by 10°C, this specific area under the curve almost doubles. Therefore, the number of effective collisions doubles, doubling the rate of reaction.

4. Effect of a Catalyst

A catalyst is a substance which alters the rate of a reaction without itself undergoing any permanent chemical change. (e.g., MnO₂ catalyzes the decomposition of KClO₃).

Mechanism of a Catalyst: A catalyst participates in a chemical reaction by forming temporary bonds with the reactants, creating an intermediate complex. This provides an alternate pathway or reaction mechanism by lowering the activation energy between reactants and products.

By lowering E_a, a larger fraction of molecules cross the energy barrier, significantly increasing the rate.

Vital Characteristics of a Catalyst:

• A catalyst does not alter the Gibbs Free Energy (ΔG) of a reaction.
• It does not alter the equilibrium constant (K_c). It simply helps attain equilibrium faster by catalyzing both the forward and backward reactions equally.
• A small amount of catalyst can catalyze a large amount of reactants.

5. Collision Theory of Chemical Reactions

Arrhenius equation successfully predicted temperature dependence, but it lacked theoretical backing regarding how molecules react. Max Trautz and William Lewis developed the Collision Theory.

According to this theory, reactant molecules are assumed to be hard spheres, and a reaction occurs only when these spheres collide with each other.

Collision Frequency (Z): The number of collisions per second per unit volume of the reaction mixture.

For a bimolecular elementary reaction A + B → Products, Rate = Z_AB e^-E_a/RT. However, experimental rates were often much lower than calculated rates. Why?

Because not all collisions lead to product formation. Only Effective Collisions yield products. For a collision to be effective, it must satisfy two barriers:

1. Energy Barrier: The colliding molecules must possess kinetic energy greater than or equal to the threshold energy (Activation Energy).
2. Orientation Barrier (Steric Factor, P): The molecules must be oriented in a specific geometry during the collision to allow old bonds to break and new bonds to form.

Thus, the modified Collision Theory equation is:

Rate = P × Z_AB × e^-E_a/RT

Where P is the Probability factor or Steric factor.

6. NCERT Solved Examples (Step-by-Step)

NCERT Example 4.9: The rate constants of a reaction at 500 K and 700 K are 0.02 s^-1 and 0.07 s^-1 respectively. Calculate the values of E_a and A.

Solution:
Given:
T₁ = 500 K, k₁ = 0.02 s^-1
T₂ = 700 K, k₂ = 0.07 s^-1
R = 8.314 J K^-1 mol^-1

Step 1: Calculate Activation Energy (E_a)
Using the formula: log (k₂ / k₁) = (E_a / 2.303 R) × [ (T₂ - T₁) / (T₁ T₂) ]
log (0.07 / 0.02) = [ E_a / (2.303 × 8.314) ] × [ (700 - 500) / (700 × 500) ]
log (3.5) = (E_a / 19.147) × [ 200 / 350000 ]
0.544 = (E_a / 19.147) × 5.71 × 10^-4
E_a = (0.544 × 19.147) / (5.71 × 10^-4) = 18230 J mol^-1 (or 18.23 kJ mol^-1)

Step 2: Calculate Arrhenius Factor (A)
Using log k = log A - E_a / 2.303 RT (for T₁ = 500 K)
log (0.02) = log A - 18230 / (2.303 × 8.314 × 500)
-1.699 = log A - (18230 / 9573.5)
-1.699 = log A - 1.904
log A = 1.904 - 1.699 = 0.205
A = Antilog (0.205) = 1.6 s^-1

7. Previous Year Questions (PYQs) & Exhaustive Question Bank

Part A: Conceptual & Graphs (1-2 Marks)

[CBSE 2017, 2020]

Q1. A plot of ln k vs 1/T for a reaction is a straight line. What does the slope of this line represent?

Answer: According to the Arrhenius equation (ln k = -E_a/RT + ln A), the plot of ln k against 1/T yields a straight line where the slope is equal to -E_a / R. (Note: If the plot was log k vs 1/T, the slope would be -E_a / 2.303R).

[CBSE 2018, 2021]

Q2. Why do most chemical reactions require an initial input of energy (activation energy) to proceed, even if they are highly exothermic?

Answer: Even in exothermic reactions, reactant molecules must overcome an initial energy barrier to break old bonds and form an unstable intermediate called the activated complex (transition state). The energy required to reach this transition state is the activation energy. Without it, the effective collisions necessary for the reaction will not occur.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2023]

Q3. Assertion (A): A catalyst does not alter the equilibrium constant of a reaction.
Reason (R): A catalyst lowers the activation energy of both the forward and backward reactions by the exact same amount.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. Since the activation energies of both the forward and backward pathways are lowered equally, the rates of both reactions increase by the same factor. Hence, the equilibrium state (K_c) remains entirely unchanged; it is simply reached faster.

Part C: Numerical Problems (3 Marks - Highly Repeated)

[CBSE 2016, NCERT Intext]

Q4. The rate of a chemical reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. (R = 8.314 J K^-1 mol^-1)

Answer:
Step 1: Understand the Data
T₁ = 293 K, T₂ = 313 K.
The rate quadruples, which means k₂ / k₁ = 4.

Step 2: Apply the Two-Temperature Arrhenius Equation
log (k₂ / k₁) = (E_a / 2.303 R) × [ (T₂ - T₁) / (T₁ T₂) ]
log (4) = (E_a / (2.303 × 8.314)) × [ (313 - 293) / (293 × 313) ]
0.602 = (E_a / 19.147) × [ 20 / 91709 ]
0.602 = E_a × (20 / 1755951)
E_a = (0.602 × 1755951) / 20
E_a = 52854 J mol^-1 = 52.85 kJ mol^-1.

[CBSE 2019]

Q5. For a reaction A → B, the activation energy (E_a) is zero. If the rate constant at 280 K is 1.6 × 10⁶ s^-1, what will be the rate constant at 300 K?

Answer:
According to the Arrhenius equation: log (k₂ / k₁) = (E_a / 2.303 R) × [ (T₂ - T₁) / (T₁ T₂) ]
Since Activation Energy (E_a) = 0, the entire right side of the equation becomes 0.
log (k₂ / k₁) = 0
k₂ / k₁ = Antilog(0) = 1
Therefore, k₂ = k₁.
The rate constant at 300 K remains exactly the same: 1.6 × 10⁶ s^-1. (Reactions with zero activation energy are independent of temperature).

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This concludes the exhaustive series on the Chemical Kinetics Chapter for CBSE Class 12. Optimized for Board and Competitive Exam Excellence.

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