Question

When initial concentration of the reactant is doubled, the half life period of a zero order reaction?
(A) Is tripled
(B) Is halved
(C) Remains unchanged
(D) Is doubled

Answer 1

Hint: For a zero order reaction the rate is proportional to zero power of the reactant concentration. Half life is the time in which the concentration of the reactant gets reduced to half. For a zero order reaction the half life is: ${{t}_{{1}/{2}\;}} = \dfrac{\left[ {{R}_{0}} \right]}{2k}$.

Complete Solution:
-First of all let us see what order of a reaction is.
The order of a reaction is basically the sum of the powers of the reactant concentration in the rate law expression. The order of a reaction can be 0, 1, 2, 3 and also a fraction.
Also for a zero order reaction the reaction rate is proportional to zero power of the reactant concentration.
\[R\to P\]
\[Rate=-\dfrac{d[R]}{dt}k{{[R]}^{0}}\]
Example of a zero order reaction is: decomposition of gaseous ammonia on hot platinum
\[2N{{H}_{3}}(g)\xrightarrow[Pt]{1130K}{{N}_{2}}(g)+3{{H}_{2}}(g)\]
\[\Rightarrow Rate = k{{[N{{H}_{3}}]}^{0}}=k\]

- Now let us talk about what half life is $({{t}_{{1}/{2}\;}})$.
The time in which the concentration of the reactant gets reduced to half of its concentration is known as half life and is represented as.

- For a zero order reaction the half life will be:
${{t}_{{1}/{2}\;}}=\dfrac{\left[ {{R}_{0}} \right]}{2k}$ ………(1)
 where, =$\left[ {{R}_{0}} \right]$ is the initial concentration of the reactant

- According to the question, the initial concentration of the reactant is doubled and we have to find out the new half life of the zero order reaction.
Initially when concentration of reactant is$\left[ {{R}_{0}} \right]$, half life will be:
${{t}_{{1}/{2}\;}}=\dfrac{\left[ {{R}_{0}} \right]}{2k}$ ……… (2)
After the concentration of the reactant is doubled:
\[{{[{{R}_{0}}]}^{'}}=2[{{R}_{0}}]\] ………. (3)

Now substitute equation (3) in equation (1)
${{t}_{{1}/{2}\;}}^{'}=\dfrac{\left[ {{R}_{0}} \right]}{2k}$
\[\Rightarrow {{t}_{{1}/{2}\;}}^{'} = 2\times \dfrac{\left[ {{R}_{0}} \right]}{2k}\]
From (2): ${{t}_{{1}/{2}\;}}^{'} = 2\times {{t}_{{1}/{2}\;}}$
Hence we can say that after doubling the concentration of the reactant, the half life period of this zero order reaction will be doubled.

So, the correct answer is “Option D”.

Note: The formula for half life is different for different order reactions. For a first order reaction: $k=\dfrac{2.303}{t}\log \dfrac{[{{R}_{0}}]}{[R]}$
At ${{t}_{{1}/{2}\;}}$: $[R]=\dfrac{[{{R}_{0}}]}{2}$; and above equation becomes:
$k=\dfrac{2.303}{t}\log \dfrac{[{{R}_{0}}]}{[R]/2}$
\[k=\dfrac{2.303}{{{t}_{{1}/{2}\;}}}\log 2\]
This will finally lead to:\[{{t}_{{1}/{2}\;}}=\dfrac{0.693}{k}\]
This shows us that half life for a first order reaction is independent of the initial concentration of the reactant.