Exhaustive Guide: Integrated Rate Equations & Half-Life | Chemical Kinetics

Integrated Rate Equations & Half-Life

Module 2 | CBSE Class 12 Chemistry | Kinetics Chapter

Table of Contents

• 1. Need for Integrated Rate Equations
• 2. Zero Order Reactions
  • 2.1 Graphical Representation
• 3. First Order Reactions
  • 3.1 Graphical Representation
  • 3.2 First Order Gas Phase Reactions
• 4. Half-Life of a Reaction (t_1/2)
  • 4.1 Half-Life for Zero Order
  • 4.2 Half-Life for First Order
• 5. Pseudo First Order Reactions
• 6. NCERT Solved Examples
• 7. Previous Year Questions (PYQs)

1. Need for Integrated Rate Equations

We already know that the differential rate equation takes the form: Rate = - d[R]/dt = k[R]ⁿ. However, measuring the instantaneous rate by calculating the slope of a tangent on a concentration-time graph is difficult and prone to errors.

To solve this, we integrate the differential rate equation to obtain a relationship directly between the concentration of reactants at different times and the rate constant (k). This integrated equation makes numerical calculations much simpler.

2. Zero Order Reactions

A zero-order reaction means that the rate of the reaction is proportional to the zero power of the concentration of reactants.

Consider the reaction: R → P

Rate = - d[R]/dt = k[R]⁰
Since any quantity to the power zero is unity ([R]⁰ = 1):
- d[R]/dt = k  &implies;  d[R] = - k dt

Integrating both sides, we get: [R] = -kt + I (where I is the constant of integration).

At time t = 0, the concentration of the reactant is the initial concentration, [R]₀. Therefore, I = [R]₀.

Substituting this back gives the integrated rate equation for a zero-order reaction:

[R] = -kt + [R]₀

k = ( [R]₀ - [R] ) / t

NCERT Examples of Zero Order: Some enzyme-catalyzed reactions and reactions occurring on metal surfaces are zero-order. For example, the decomposition of gaseous ammonia (NH₃) on a hot platinum surface at high pressure:
2NH₃(g) → N₂(g) + 3H₂(g)

2.1 Graphical Representation (Zero Order)

The equation [R] = -kt + [R]₀ matches the equation of a straight line, y = mx + c.

• If we plot [R] on the y-axis against time (t) on the x-axis, we get a straight line.
• Slope (m) = -k
• Intercept (c) = [R]₀

3. First Order Reactions

In this class of reactions, the rate of the reaction is proportional to the first power of the concentration of the reactant R.

Rate = - d[R]/dt = k[R]¹

Rearranging: d[R] / [R] = - k dt

Integrating both sides: ln [R] = -kt + I

At t = 0, [R] = [R]₀. Therefore, I = ln [R]₀.

Substituting the value of I: ln [R] = -kt + ln [R]₀

To make calculations easier, we convert natural logarithm (ln) to base-10 logarithm (log) by substituting ln = 2.303 log:

k = ( 2.303 / t ) × log ( [R]₀ / [R] )

Where:
[R]₀ = Initial concentration of the reactant.
[R] = Concentration of the reactant remaining at time 't'.

Crucial Fact for Boards: All natural and artificial radioactive decay of unstable nuclei take place by first-order kinetics. (e.g., Radium-226 decaying into Radon-222).

3.1 Graphical Representation (First Order)

From the equation: log [R] = (-k / 2.303)t + log [R]₀

• If we plot log [R] on the y-axis against time (t) on the x-axis, we get a straight line.
• Slope = -k / 2.303
• Intercept = log [R]₀

3.2 First Order Gas Phase Reactions

For a typical first-order gas phase reaction A(g) → B(g) + C(g), we measure pressure instead of concentration. Let p_i be the initial pressure of A, and p_t be the total pressure at time 't'.

The integrated rate equation becomes:

k = ( 2.303 / t ) × log [ p_i / (2p_i - p_t) ]

Note: This specific formula applies when 1 mole of reactant gas decomposes into 2 moles of product gases.

4. Half-Life of a Reaction (t_1/2)

Half-Life (t_1/2): The time in which the concentration of a reactant is reduced to one-half of its initial concentration.

4.1 Half-Life for a Zero Order Reaction

We know that for zero order, k = ( [R]₀ - [R] ) / t.
At half-life, t = t_1/2 and [R] = [R]₀ / 2.

Substituting these values:

t_1/2 = [R]₀ / 2k

Conclusion: For a zero-order reaction, t_1/2 is directly proportional to the initial concentration of the reactant.

4.2 Half-Life for a First Order Reaction

We know that for first order, k = ( 2.303 / t ) log ( [R]₀ / [R] ).
At half-life, t = t_1/2 and [R] = [R]₀ / 2.

Substituting these values:
t_1/2 = ( 2.303 / k ) log ( [R]₀ / ([R]₀ / 2) )
t_1/2 = ( 2.303 / k ) log (2)
Since log(2) = 0.301:

t_1/2 = 0.693 / k

Highly Tested Concept: For a first-order reaction, the half-life (t_1/2) is a constant (0.693/k) and is completely independent of the initial concentration of the reacting species.

5. Pseudo First Order Reactions

Sometimes, a reaction which is not truly first order appears to follow first-order kinetics under certain conditions.

Pseudo First Order Reaction: A chemical reaction in which the molecularity is 2 or more but the order of the reaction is 1. This happens when one of the reacting species is present in such a large excess that its concentration remains almost constant during the reaction.

Example: Acid-catalyzed hydrolysis of Ethyl Acetate
CH₃COOC₂H₅ + H₂O (excess) → CH₃COOH + C₂H₅OH

The rate law should theoretically be: Rate = k'[CH₃COOC₂H₅][H₂O].
However, since water is in a massive excess, its concentration does not change significantly. The term k'[H₂O] becomes a new constant, 'k'.
So, Rate = k[CH₃COOC₂H₅]. The reaction behaves as a first-order reaction.

Another common example: Inversion of cane sugar (sucrose) in the presence of an acid.

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

NCERT Example 4.5: The initial concentration of N₂O₅ in the following first order reaction: N₂O₅(g) → 2NO₂(g) + 1/2 O₂(g) was 1.24 × 10^-2 mol L^-1 at 318 K. The concentration of N₂O₅ after 60 minutes was 0.20 × 10^-2 mol L^-1. Calculate the rate constant of the reaction at 318 K.

Solution:
Given:
[R]₀ = 1.24 × 10^-2 mol L^-1
[R] = 0.20 × 10^-2 mol L^-1
t = 60 min

Formula for First Order:
k = ( 2.303 / t ) log ( [R]₀ / [R] )
k = ( 2.303 / 60 ) log ( 1.24 × 10^-2 / 0.20 × 10^-2 )
k = ( 2.303 / 60 ) log ( 6.2 )
k = ( 2.303 / 60 ) × 0.7924
k = 0.0304 min^-1

NCERT Example 4.7: A first order reaction is found to have a rate constant, k = 5.5 × 10^-14 s^-1. Find the half-life of the reaction.

Solution:
For a first order reaction, the half-life is independent of concentration.
t_1/2 = 0.693 / k
t_1/2 = 0.693 / (5.5 × 10^-14 s^-1)
t_1/2 = 1.26 × 10¹³ s

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

Part A: Conceptual (1-2 Marks)

[CBSE 2018, 2021]

Q1. A reaction is first order in A and second order in B. How is the rate affected when the concentrations of both A and B are doubled?

Answer:
Initial Rate (r₁) = k[A]¹[B]².
When concentrations are doubled: New Rate (r₂) = k[2A]¹[2B]² = k × 2 × 4 [A][B]² = 8 k[A][B]².
The rate increases by a factor of 8 times.

[CBSE 2017, 2019]

Q2. For a zero order reaction, will the half-life period remain the same if the initial concentration of the reactant is doubled? Give reason.

Answer: No. For a zero order reaction, t_1/2 = [R]₀ / 2k. Thus, the half-life is directly proportional to the initial concentration. If the initial concentration is doubled, the half-life period will also be doubled.

Part B: Assertion-Reason Type (1 Mark)

[CBSE Sample Paper 2023]

Q3. Assertion (A): The half-life of a radioactive substance is independent of its initial mass.
Reason (R): Radioactive decay follows first-order kinetics.

Answer: Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion. Since radioactive decay is a first-order process, its half-life (t_1/2 = 0.693/k) depends only on the decay constant (k) and is totally independent of the initial concentration/mass.

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

[CBSE 2016, 2018, NCERT Intext]

Q4. A first order reaction takes 40 minutes for 30% decomposition. Calculate t_1/2.

Answer:
Step 1: Understand the data
Let initial concentration [R]₀ = 100.
Since 30% has decomposed, the amount remaining [R] = 100 - 30 = 70.
t = 40 minutes.

Step 2: Calculate rate constant (k)
k = (2.303 / t) log ([R]₀ / [R])
k = (2.303 / 40) log (100 / 70) = (2.303 / 40) log (1.428)
k = (2.303 / 40) × 0.1548 = 0.00891 min^-1

Step 3: Calculate half-life (t_1/2)
t_1/2 = 0.693 / k
t_1/2 = 0.693 / 0.00891 = 77.77 minutes.

[CBSE 2015]

Q5. Show that for a first order reaction, the time required for 99% completion of a reaction is twice the time required to complete 90% of the reaction.

Answer:
Case 1: 99% completion
[R]₀ = 100. [R] remaining = 100 - 99 = 1.
t_99% = (2.303 / k) log (100 / 1) = (2.303 / k) log(10²) = (2.303 / k) × 2 --- (Equation 1)

Case 2: 90% completion
[R]₀ = 100. [R] remaining = 100 - 90 = 10.
t_90% = (2.303 / k) log (100 / 10) = (2.303 / k) log(10¹) = (2.303 / k) × 1 --- (Equation 2)

Dividing Equation 1 by Equation 2:
t_99% / t_90% = 2 / 1
t_99% = 2 × t_90% (Hence Proved).

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This module is strictly mapped to the latest rationalised NCERT syllabus for Class 12 Chemistry.
Coming up in Module 3: Temperature Dependence of Reaction Rates (Arrhenius Equation) & Collision Theory.

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