Question

${\text{k}}$ for a zero-order reaction is $2 \times {10^{ - 2}}{\text{ }}{{\text{L}}^{ - 1}}{\text{ Se}}{{\text{c}}^{ - 1}}$. If the concentration of the reactant after ${\text{25 sec}}$ is $0 \cdot 5{\text{ M}}$, the initial concentration must have been:
(a) $0 \cdot 5{\text{ M}}$
(b) $1 \cdot 25{\text{ M}}$
(c) $12 \cdot 5{\text{ M}}$
(d) $1 \cdot 0{\text{ M}}$

Answer 1

Hint: In zero order reaction, rate is independent of the concentration of the reactant. This means that increasing or decreasing the concentration would not affect the rate. Hence, rate is directly proportional to the concentration of the reactant raise to the power zero. i.e., ${\text{Rate = k}}{\left[ {\text{A}} \right]^0}$ , where, ${\text{k}}$ is the rate constant, and ${\text{A}}$ is the concentration of the reactant.

Complete step by step answer: Integral rate expression for zero order reaction is given by: ${\text{k = }}\dfrac{{\left[ {{{\text{A}}_ \circ }} \right] - \left[ {\text{A}} \right]}}{{\text{t}}}{\text{ }}.......{\text{(1)}}$,
where, $\left[ {{{\text{A}}_ \circ }} \right]$ is the initial concentration at time, ${\text{t = 0}}$, $\left[ {\text{A}} \right]$ is the concentration of the reactant at any time, ${\text{t = t}}$, ${\text{k}}$ is the rate constant, and ${\text{t}}$ is the time for which reaction occurred.
Value of rate constant ${\text{k}}$ is given $2 \times {10^{ - 2}}{\text{ }}{{\text{L}}^{ - 1}}{\text{ Se}}{{\text{c}}^{ - 1}}$.
Value of concentration of reactant at time, ${\text{t = 25 sec}}$ is given $0 \cdot 5{\text{ M}}$.
Now, we have to calculate the value of $\left[ {{{\text{A}}_ \circ }} \right]$ which is concentration of reactant at ${\text{t = 0}}$ or we can say that, we have to find the initial concentration.
Using equation $\left( {\text{1}} \right)$,
$\left[ {{{\text{A}}_ \circ }} \right]{\text{ = kt}} + \left[ {\text{A}} \right]$
$ \Rightarrow 2 \times {10^{ - 2}} + 0 \cdot 5$
= $0 \cdot 5 + 0 \cdot 5$
= $1{\text{ M}}$
So, initial concentration is $1{\text{ M}}$.

Hence, option (d) is the only correct answer, the initial concentration must have been $1{\text{ M}}$ to satisfy the given conditions in the question.

Note:Half-life of a zero-order reaction is denoted by ${t_{\dfrac{1}{2}}}$,and it is the time required to consume half of the reactant. It is given by ${t_{\dfrac{1}{2}}} = \dfrac{{\left[ {{{\text{A}}_ \circ }} \right]}}{{{\text{2k}}}}$, where, $\left[ {{{\text{A}}_ \circ }} \right]$ is the initial concentration at time, ${\text{t = 0}}$, and ${\text{k}}$ is the rate constant. Unlike the rate constant, half-life is dependent on the concentration of the reactant.