First Order Reaction | Half-life, Numericals

In a first order reaction, the sum of components of the concentration of reactants in a rate expression is equal to one. Let us take a substance (reactant) “A” which decomposes into products. Its initial concentration is “A_°” mol/dm³. Let after time “t” the amount of substance left behind is “A” mol/dm³.

A → Product

Rate of reaction (ROR) = r = k [A]¹

The rate of reaction depends upon the concentration of A.

Integrated Rate Law

A → Product

Rate = r = k [A]¹ ———- (1)

Rate = r = – dA / dt ———-(2)

Compare eq (1) and (2)

– dA / dt = k [A]¹

dA / dt = – K dt

∫dA / dt = – ∫K dt

At t=0 the initial concentration of A is A_° and after time t the concentration of A is A.

_A°^A∫dA / dt = – ₀^t∫K dt

|log_e A|^A_A° = -K |t|^t₀ ∵ ∫dA / dt = log_e A

log_e A – log_e A_° = -K (t-0)

log_e A – log_e A_° = -Kt

log_e A = log_e A_° – Kt (or) ln A = ln A_° – Kt

Now convert log_e to log₁₀

2.303 log₁₀ A = 2.303 log₁₀ A_° – Kt

Divided both sides by 2.303 and we get

log₁₀ A = log₁₀ A_° – Kt/2.303

This is the integrated rate law of first order kinetics.

Other forms of Rate Law

log₁₀ A = log₁₀A_° – Kt/2.303

log₁₀ A_° – log₁₀A = Kt/2.303

log₁₀ (A_° / A) = Kt/2.303

t = 2.303/K log₁₀ (A_° / A)

As we know that

log_e A = log_e A_° – Kt

log_e A – log_e A_° = – Kt

log_e (A / A_°) = – Kt

A / A_° = e^-kt

A = A_° e^-kt

Half-Life of First Order Reaction

Half-life is a time in which the reaction completes 50%. The initial concentration is A_° but when t=t_1/2 the final concentration becomes half, A_°/2.

Initial concentration = A_°

at t=t_1/2, A = A_°/2

t = 2.303/K log₁₀ (A_° / A)

By putting the value of “t” and “A” we get,

t_1/2 = 2.303/K log₁₀ (A_° / A_°/2)

t_1/2 = 2.303 log₁₀ (2) / K

t_1/2 = 2.303 (0.3) / K

t_1/2 = 0.693 / K

Lifetime of Reaction (t_lf)

The lifetime of the reaction is a time in which 100% reaction is complete.

At t = t_lf , A = 0

t_lf = 2.303/K log₁₀ (A_° / 0)

t_lf = ∞

The first-order reaction can never complete 100%.

Graphical Representation of First Order Reaction

Graph: Conc. of Reactant vs time

Graph: ln [A] vs time

Numericals

In a first order reaction, the concentration of the reactant decrease from 800 mol/dm³ to 50 mol/dm³ in 2×10⁴ s. Find the rate constant.

t = 2.303/K log₁₀ (A_° / A)

K = 2.303/t log₁₀ (A_° / A)

K = 2.303/2×10⁴ log₁₀ (800 / 50)

K = 2.303/2×10⁴ log₁₀ 16

K = 2.303/2×10⁴ log₁₀ 2⁴

K = 2.303/2×10⁴ 4 (log₁₀ 2)

K = 2.303/2×10⁴ 4 (0.3)

K = 2.303/2×10⁴ 4 (03/10)

K = 13.818 / 10⁵

K = 13.8×10^-5

K = 1.38×10^-4

A first order kinetics is 20% complete in 5 min. Find the time in which it is 60% complete.

Let A_° = 100

20% reaction is complete so the remaining concentration is A = 80

t = 5 min

t = 2.303/K log₁₀ (A_° / A)

5 = 2.303/K log₁₀ (100 / 80) ———- (1)

A_° = 100

60% reaction is complete so the remaining concentration is A = 40

t = ?

t = 2.303/K log₁₀ (100 / 40) ———- (2)

Divide equation (2) by (1) we get

t/5 = log₁₀ 10 – log 4 / log₁₀ 10 – log 8

t/5 = log₁₀ 10 – log 2² / log₁₀ 10 – log 2³

t/5 = 1 – 2 log 2 / 1 – 3 log 2

t/5 = 1 – 2 (0.3) / 1 – 3 (0.3)

t/5 = 0.4 / 0.1

t = 4 × 5

t = 20 min

In first order reaction, “A” gives a product. The concentration of “A” changes from 0.1 M to 0.025 M in 40 mins. Find the rate of reaction when the concentration of “A” is 0.01 M.

Rate of reaction =?

Rate of reaction = r = K [A]¹

First to find the rate constant, to find the K

t = 2.303/K log₁₀ (A_° / A)

K = 2.303/40 log₁₀ (0.1 / 0.025)

K = 2.303/40 log₁₀ 4

K = 2.303/40 2 log₁₀ 2

K = (2.303) (2) (0.3) / 40

K = (2.303) (2) (3) / 400

K = (2.303) (6) / 400

K = 13.818 / 4 × 100

K = 3.454/100

Rate = K [A]¹

Rate = 3.454/100 × 0.01 M

Rate = 3.454/100 × 1/100

Rate = 3.454/10000

Rate = 3.4 × 10^-4

Important Points to Remember

1. The time taken for 75% completion of the first-order reaction is equal to 2 times taken for 50% completion. (t_75% = 2 t_50%)
2. The time taken for 99.9% completion of the first-order reaction is equal to 10 times taken for 50% completion. (t_99.9% = 10 t_50%)
3. The time taken for 25% completion of the first-order reaction is equal to 0.415 times taken for 50% completion. (t_25% = 0.415 t_50%)
4. The time taken for 87.5% completion of the first-order reaction is equal to 3 times taken for 50% completion. (t_87.5% = 3 t_50%)

Prove that t_75% = 2 t_50%.

t = 2.303/K log₁₀ (A_° / A)

Let the initial concentration (A_°) of reactant is 100, when 50% reaction is goes to completion the remaining concentration of reactant is 50.

t_50% = 2.303/K log₁₀ (100 / 50) ———- (1)

When the reaction is 75%complete, the remaining concentration is 25.

t_75% = 2.303/K log₁₀ (100 / 25) ———- (2)

Divide (2) by (1)

t_75% / t_50% = (2.303/K log₁₀ (100 / 25)) / (2.303/K log₁₀ (100 / 50))

t_75% / t_50% = 2 log₁₀ 2 / log₁₀ 2

t_75% / t_50% = 2

t_75% = 2 t_50%

Prove that t_99.9% = 10 t_50%.

Let the initial concentration (A_°) of reactant is 100, when 50% reaction goes to completion the remaining concentration of reactant is 50.

t_50% = 0.693 / K ——— (1)

when 99.9% reaction goes to completion, the remaining concentration of the reactant is 0.1.

t_99.9% = 2.303/K log₁₀ (100 / 0.1)

t_99.9% = 2.303/K log₁₀ (1000 / 1)

t_99.9% = 2.303/K log₁₀ 10³ ———- (2)

Divide (2) by (1)

t_99.9% / t_50% = (2.303/K log₁₀ 10³) / (0.693 / K)

t_99.9% / t_50% = 2.303 × 3 / 0.693

t_99.9% / t_50% = 6.909 / 0.693

t_99.9% / t_50% =10

t_99.9% = 10 t_50%

Examples of First Order Reaction

1. Nuclear physics reaction (Radioactive decay)
2. Growth and decay of bacteria
3. Inversion of cane sugar in an acidic medium
4. Acidic hydrolysis of ester (esterification)
5. Dissociation of N₂O₅
6. Dissociation of H₂O₂