Chemical Kinetics: For a first-order reaction, prove that the half-life is independent of the initial concentration

Chapter: Chemical Kinetics

Topic: Half-Life of a First-Order Reaction

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For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant.

Rate ∝ [A]

Therefore:

−d[A]/dt = k[A]

where k is the first-order rate constant.

For a first-order reaction:

−d[A]/dt = k[A]

On integration between the initial concentration [A]₀ and concentration [A] at time t:

ln([A]₀/[A]) = kt

or:

k = (2.303/t) log([A]₀/[A])

At half-life, the concentration of the reactant becomes half of its initial concentration.

[A] = [A]₀/2

Substitute this value into the first-order integrated rate equation:

k = (2.303/t_1/2) log([A]₀/([A]₀/2))

Therefore:

k = (2.303/t_1/2) log 2

Since:

log 2 = 0.3010

we get:

k = (2.303 × 0.3010)/t_1/2

Since 2.303 × 0.3010 ≈ 0.693:

k = 0.693/t_1/2

Hence:

t_1/2 = 0.693/k

The half-life equation obtained for a first-order reaction is:

t_1/2 = 0.693/k

Notice that the expression contains only the rate constant k.

There is no term containing the initial concentration [A]₀.

Therefore, even if the initial concentration is changed, the half-life will remain the same, provided the temperature and other conditions remain unchanged.

Hence proved:

The half-life of a first-order reaction is independent of its initial concentration.

Q. A first-order reaction has a rate constant k = 0.693 min⁻¹. Calculate its half-life.

We know:

t_1/2 = 0.693/k

Substituting:

t_1/2 = 0.693/0.693

t_1/2 = 1 minute

The same half-life would be obtained whether the initial concentration were 1 M, 0.5 M, 0.1 M or any other value, as long as k remains unchanged.

Order	Half-Life	Depends on Initial Concentration?
Zero order	t1/2 = [A]0/2k	Yes
First order	t1/2 = 0.693/k	No
Second order	t1/2 = 1/(k[A]0)	Yes

Special property of first-order reactions:

Half-life is independent of the initial concentration.

Q. Write the expression for the half-life of a first-order reaction.

Answer:

t_1/2 = 0.693/k

Since the expression does not contain the initial concentration, half-life is independent of initial concentration.

Q. Prove that the half-life of a first-order reaction is independent of initial concentration.

For a first-order reaction:

k = (2.303/t) log([A]₀/[A])

At half-life:

[A] = [A]₀/2

Therefore:

k = (2.303/t_1/2) log 2

Since log 2 = 0.301:

t_1/2 = 0.693/k

The initial concentration [A]₀ cancels out. Hence, half-life is independent of initial concentration.

Q. Derive the half-life equation for a first-order reaction and explain its significance.

Integrated first-order rate equation:

k = (2.303/t) log([A]₀/[A])

At t = t_1/2:

[A] = [A]₀/2

Hence:

k = (2.303/t_1/2) log 2

Therefore:

t_1/2 = 0.693/k

The initial concentration is absent from the final expression. Therefore, t_1/2 is independent of initial concentration.

Q. For a first-order reaction, prove mathematically that the half-life is independent of the initial concentration.

Step 1: First-order equation

k = (2.303/t) log([A]₀/[A])

Step 2: Condition at half-life

At t = t_1/2, half of the original reactant remains:

[A] = [A]₀/2

Step 3: Substitute

k = (2.303/t_1/2) log([A]₀/([A]₀/2))

The [A]₀ terms cancel:

k = (2.303/t_1/2) log 2

Step 4: Simplify

k = 0.693/t_1/2

Therefore:

t_1/2 = 0.693/k

Since [A]₀ does not occur in the final equation, changing the initial concentration does not change the half-life.

Q. For a first-order reaction, derive the expression for half-life and prove that it is independent of the initial concentration. Explain the physical significance of this result.

1. Rate Law

For a first-order reaction:

−d[A]/dt = k[A]

2. Integrated Rate Equation

k = (2.303/t) log([A]₀/[A])

3. Apply the Half-Life Condition

At half-life, the concentration becomes half of the initial concentration:

[A] = [A]₀/2

Substituting:

k = (2.303/t_1/2) log([A]₀/([A]₀/2))

4. Cancellation of Initial Concentration

[A]₀/([A]₀/2) = 2

Thus:

k = (2.303/t_1/2) log 2

Using log 2 = 0.3010:

k = (2.303 × 0.3010)/t_1/2

Therefore:

t_1/2 = 0.693/k

5. Proof of Independence

The final equation contains only k. It does not contain [A]₀. Hence, for a fixed temperature and reaction conditions, the half-life remains constant regardless of the initial concentration.

6. Physical Significance

Suppose two samples of the same first-order reactant have initial concentrations of 1.0 M and 0.5 M. Both will take the same time to reduce their respective concentrations by half.

1.0 M → 0.5 M → 0.25 M → 0.125 M

0.5 M → 0.25 M → 0.125 M → 0.0625 M

Each successive half-life is equal because the half-life depends only on the rate constant.

Final Conclusion:

For a first-order reaction, t_1/2 = 0.693/k.

Therefore, the half-life is independent of the initial concentration.

1. The half-life of a first-order reaction is:

A) [A]₀/2k

B) 0.693/k

C) 1/k[A]₀

D) k/0.693

✅ Answer

B) 0.693/k

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2. The half-life of a first-order reaction depends on:

A) Initial concentration only

B) Rate constant

C) Initial volume only

D) Initial mass only

✅ Answer

B) Rate constant

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3. The half-life of a first-order reaction is independent of:

A) Rate constant

B) Temperature

C) Initial concentration

D) Reaction conditions

✅ Answer

C) Initial concentration

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4. At half-life, [A] is equal to:

A) 2[A]₀

B) [A]₀

C) [A]₀/2

D) Zero

✅ Answer

C) [A]₀/2

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5. The value of log 2 is approximately:

A) 0.3010

B) 0.693

C) 2.303

D) 1.301

✅ Answer

A) 0.3010

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6. The numerical value 0.693 is approximately equal to:

A) log 2

B) 2.303 log 2

C) log 10

D) 1/log 2

✅ Answer

B) 2.303 log 2

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7. If k = 0.693 min⁻¹, t_1/2 is:

A) 0.693 min

B) 1 min

C) 2 min

D) 0.5 min

✅ Answer

B) 1 min

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8. If the initial concentration is doubled, the first-order half-life:

A) Doubles

B) Becomes half

C) Remains unchanged

D) Becomes zero

✅ Answer

C) Remains unchanged

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9. For a first-order reaction, the rate law is:

A) Rate = k

B) Rate = k[A]

C) Rate = k[A]²

D) Rate = [A]/k

✅ Answer

B) Rate = k[A]

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10. The unit of k for a first-order reaction is:

A) mol L⁻¹ s⁻¹

B) s⁻¹

C) L mol⁻¹ s⁻¹

D) mol² L⁻² s⁻¹

✅ Answer

B) s⁻¹

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11. Which order has a half-life independent of initial concentration?

A) Zero order

B) First order

C) Second order

D) Third order

✅ Answer

B) First order

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12. If k increases, the half-life:

A) Increases

B) Decreases

C) Remains unchanged

D) Becomes infinite

✅ Answer

B) Decreases

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13. If k is doubled, t_1/2 becomes:

A) Double

B) Half

C) Four times

D) Unchanged

✅ Answer

B) Half

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14. The integrated rate equation for a first-order reaction is:

A) [A] = [A]₀ − kt

B) k = (2.303/t) log([A]₀/[A])

C) 1/[A] = kt

D) Rate = k[A]²

✅ Answer

B

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15. After two half-lives, the fraction remaining is:

A) 1/2

B) 1/3

C) 1/4

D) 1/8

✅ Answer

C) 1/4

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16. After three half-lives, the fraction remaining is:

A) 1/2

B) 1/4

C) 1/6

D) 1/8

✅ Answer

D) 1/8

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17. The first-order half-life equation contains:

A) Initial concentration

B) Final concentration

C) Rate constant

D) Both A and B

✅ Answer

C) Rate constant

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18. A first-order reaction is characterised by a constant:

A) t_1/2 only for all temperatures

B) Fraction decomposed per unit time

C) Initial concentration

D) Volume

✅ Answer

B

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19. If t_1/2 = 10 s, the rate constant is approximately:

A) 0.0693 s⁻¹

B) 6.93 s⁻¹

C) 0.693 s⁻¹

D) 10 s⁻¹

✅ Answer

A) 0.0693 s⁻¹

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20. The statement "half-life is independent of initial concentration" is true for:

A) Zero-order reactions

B) First-order reactions

C) Second-order reactions

D) All reactions

✅ Answer

B) First-order reactions

🎯 Quick Revision

First-order rate equation: k = (2.303/t) log([A]₀/[A])
At half-life: [A] = [A]₀/2
Half-life: t_1/2 = 0.693/k
Key point: [A]₀ cancels out → Half-life is independent of initial concentration.