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/dm^{3}. Let after time “t” the amount of substance left behind is “A” mol/dm^{3}.

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|^{t}_{0} ∵ ∫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**

_{°}– KtNow convert log_{e} to log_{10}

2.303 log_{10} A = 2.303 log_{10} A_{°} – Kt

Divided both sides by 2.303 and we get

**log _{10} A = log_{10} A_{°} – Kt/2.303**

This is the integrated rate law of first order kinetics.

#### Other forms of Rate Law

log_{10} A = log_{10}A_{°} – Kt/2.303

log_{10} A_{°} – log_{10}A = Kt/2.303

log_{10} (A_{°} / A) = Kt/2.303

**t = 2.303/K log _{10} (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 log_{10} (A_{°} / A)

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

t_{1/2} = 2.303/K log_{10} (A_{°} / A_{°}/2)

t_{1/2} = 2.303 log_{10} (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 log_{10} (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/dm^{3} to 50 mol/dm^{3} in 2×10^{4} s. Find the rate constant.

t = 2.303/K log_{10} (A_{°} / A)

K = 2.303/t log_{10} (A_{°} / A)

K = 2.303/2×10^{4} log_{10} (800 / 50)

K = 2.303/2×10^{4} log_{10} 16

K = 2.303/2×10^{4} log_{10} 2^{4}

K = 2.303/2×10^{4} 4 (log_{10} 2)

K = 2.303/2×10^{4} 4 (0.3)

K = 2.303/2×10^{4} 4 (03/10)

K = 13.818 / 10^{5}

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 log_{10} (A_{°} / A)

5 = 2.303/K log_{10} (100 / 80) ———- (1)

A_{°} = 100

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

t = ?

t = 2.303/K log_{10} (100 / 40) ———- (2)

Divide equation (2) by (1) we get

t/5 = log_{10} 10 – log 4 / log_{10} 10 – log 8

t/5 = log_{10} 10 – log 2^{2} / log_{10} 10 – log 2^{3}

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 log_{10} (A_{°} / A)

K = 2.303/40 log_{10} (0.1 / 0.025)

K = 2.303/40 log_{10} 4

K = 2.303/40 2 log_{10} 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

- 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%}) - 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%}) - 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%}) - 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 log_{10} (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 log_{10} (100 / 50) ———- (1)

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

t_{75%} = 2.303/K log_{10} (100 / 25) ———- (2)

Divide (2) by (1)

t_{75%} / t_{50%} = (2.303/K log_{10} (100 / 25)) / (2.303/K log_{10} (100 / 50))

t_{75%} / t_{50%} = 2 log_{10} 2 / log_{10} 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 log_{10} (100 / 0.1)

t_{99.9%} = 2.303/K log_{10} (1000 / 1)

t_{99.9%} = 2.303/K log_{10} 10^{3} ———- (2)

Divide (2) by (1)

t_{99.9%} / t_{50%} = (2.303/K log_{10} 10^{3}) / (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

- Nuclear physics reaction (Radioactive decay)
- Growth and decay of bacteria
- Inversion of cane sugar in an acidic medium
- Acidic hydrolysis of ester (esterification)
- Dissociation of N
_{2}O_{5} - Dissociation of H
_{2}O_{2}