## Half Life (t1/2)

Half Life of a reaction is time required to reduce the initial concentration to half. The half-life of a reaction depends on the order of reaction. The variation of half-life with order is given as

t_{1/2 } ∝ [A_{0}]^{1-n}

Where, [A_{0}] = initial concentration

n = Order of reaction

for, Zeroth order reaction

t_{1/2} ∝ [A_{0}]

for first order reaction,

t_{1/2} is independent of initial concentration

for second order reaction,

t_{1/2} ∝ [1/A_{0}]

**Integrated rate law expression**** **

For Zeroth order reaction

A → product

Initially a o

at t = t_{1 }(a -x) x

The rate of reaction at time ‘t’

Integrating on both sides, we get,

when t = 0, x = 0

c = 0

x = kt

k = x/t

When t = t_{1/2}, x = a/2

k = a/2t_{1/2}

t_{1/2} = a/2k

For first Order Reaction

A → product

Initially a o

At, t = t (a -x) x_{1}

The rate of reaction at time ‘t’

Integrating on both sides, we get,

when t = 0, x = 0

-ln a = c

-ln (a-x) = kt – ln a

When t = t_{1/2},

x = a/2

i.e. (a-x) = a/2

So, half-life of first order reaction is independent of initial concentration.

**Numerical**

The half-life of a first order reaction is 50 mins. Calculate the time required to complete 75% of the reaction.

Given,

Half Life (t_{1/2}) = 50 mins

Then,

Rate constant (k)

= 0.01386 min^{-1}

Again,

initial concentration (a) = 100 (let) then

at time t,

Concentration left (a-x) = 100-75 = 25

We have,

**Calculate the half period of first order reaction when rate constant is 5 year ^{-1}.**

We have,

For 1

^{st}order reaction

t

_{1/2}= 0.693/5

= 0.1386 year

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