Question

If the initial concentration is reduced to 1/4th in a zero-order reaction, then the time taken for half the reaction to complete:
(A)- remains the same
(B)- becomes 4 times
(C)- becomes one-fourth
(D)- doubles

Answer 1

The time taken for the concentration to reach 50% of its initial concentration is known as the half-life of a chemical reaction. In other words, the half-life of a chemical reaction is the time taken for the reactant concentration to reach half of its initial value.

Complete Step by step solution:
-Let us now derive the formula for the half-life for a zero-order reaction.
The integral form for the zero-order reaction is given as-
$[A]=[{{A}_{0}}]-kt$
Replacing the concentration and time terms for half-life as $[A]=\dfrac{1}{2}[A];t={{t}_{1/2}}$
$\begin{aligned} & \dfrac{1}{2}[A]=[{{A}_{0}}]-k{{t}_{1/2}} \\\ & \Rightarrow k{{t}_{1/2}}=\dfrac{1}{2}{{[A]}_{0}} \\\ \end{aligned}$
Therefore, the final half-life for the zero-order reaction is-
${{t}_{1/2}}=\dfrac{1}{2k}{{[A]}_{0}}$
where ${{t}_{1/2}}$is the time taken for the reaction to become half.
${{[A]}_{0}}$is the initial or starting concentration of a reactant
k is the rate constant of the reaction.
-Let ${{[A]}_{0}}=a$ , then ${{t}_{1/2}}=\dfrac{a}{2k}...(1)$
When the initial concentration is reduced to one fourth, then $[{{A}_{0}}]=\dfrac{1}{4}\times a$
$t{{'}_{1/2}}=\dfrac{\dfrac{1}{4}\times a}{2k}...(2)$
Now dividing the equation (2) by (1), we get
$\dfrac{t{{'}_{1/2}}}{{{t}_{1/2}}}=\dfrac{\dfrac{1}{4}\times a}{\dfrac{a}{2k}}=\dfrac{1}{4}$

So, the correct answer is option C.

Note: If the rate of reaction is independent of the concentration of the reactants, the reaction is said to be of zero-order reaction. Characteristics of zero-order reaction are-
(i) The concentration of reactant decreases linearly with time, and whose reaction is given ${{[A]}_{t}}={{[A]}_{0}}-kt$.
(ii) The time required for a zero-order reaction to complete is given as ${{t}_{completion}}=\dfrac{{{[A]}_{0}}}{k}=\dfrac{\text{Initial concentration}}{\text{Rate constant}}$
(iii) The unit of k for a zero-order reaction is given as $\text{mol }{{\text{L}}^{-1}}\text{ tim}{{\text{e}}^{-1}}$.
Some examples of zero-order reactions are-
(i) Photochemical reaction between hydrogen and chlorine - ${{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow{h\nu }2HCl$
(ii) Decomposition of ${{N}_{2}}O$ on hot platinum surface - ${{N}_{2}}O\to {{N}_{2}}+\dfrac{1}{2}{{O}_{2}}$
(iii) Decomposition of $N{{H}_{3}}$in presence of molybdenum or tungsten - $2N{{H}_{3}}\xrightarrow{Mo}{{N}_{2}}+3{{H}_{2}}$