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## Half Life Period of a Reaction

The half life of a reaction is defined as the time required for the reactant concentration to reach one half its initial value. For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration.

The rate constant for a first order reaction is given by

Let us calculate the half life period for a zero order reaction.

Hence, in contrast to the half life of a first order reaction, the half life of a zero order reaction is directly proportional to the initial concentration of the reactant.

**Example 1**

A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989; log 10 = 1)

Solution:

For a first order reaction

k = \(\frac{2.303}{t}\)log[latex]\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}[/latex] …………. (1)

Let [A_{°}] = 100M

When t = t_{90%}; [A]=10M (given that t_{90%} = 8hours)

t = t_{90%}; [A]=20M

Find the value of k using the given data

Substitute the value of k in equation (2)

t_{80%} = \(\frac{2.303}{2.303/8hours}\) log(5)

t_{80%} = 8hours × 0.6989

t_{80%} = 5.59hours

**Example 2**

The half life of a first order reaction x → products is 6.932 × 10^{4}s at 500k. What percentage of x would be decomposed on heating at 500K for 100 min.(e^{0.06}=1.06).

Solution:

Given t_{1/2} = 0.6932 × 10^{4}s

\(\frac{\left[\mathrm{A}_{0}\right]-[\mathrm{A}]}{\left[\mathrm{A}_{0}\right]}\) × 100

We know that

For a first order reaction, t_{1/2} = \(\frac{0.6932}{k}\)

**Example 3**

Show that case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.

Solution:

Let [A_{°}] = 100

When t = t_{99.9%}; [A] = (100 – 99.9) = 0.1