KCET · Chemistry · Chemical Kinetics
Half-life of a reaction is found to be inversely proportional to the fifth power of its initial concentration, the order of reaction is
- A 4
- B 5
- C 6
- D 3
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
The half-life period of \(n\)th order reaction is
\[
t_{1 / 2}=\frac{k}{a^{n-1}} \text {. }
\]
\(a\) is the initial concentration of reactant \(t_{1 / 2} \propto \frac{1}{a^5}\)
\[
\begin{aligned}
\therefore \quad t_{1 / 2} & =\frac{k}{a^5} \text { and } \frac{k}{a^{n-1}}=\frac{k}{a^5} \\
\Rightarrow \quad a^5 & =a^{n-1} \\
n-1 & =5=6
\end{aligned}
\]
\(\therefore\) The order of the reaction is 6 .
\[
t_{1 / 2}=\frac{k}{a^{n-1}} \text {. }
\]
\(a\) is the initial concentration of reactant \(t_{1 / 2} \propto \frac{1}{a^5}\)
\[
\begin{aligned}
\therefore \quad t_{1 / 2} & =\frac{k}{a^5} \text { and } \frac{k}{a^{n-1}}=\frac{k}{a^5} \\
\Rightarrow \quad a^5 & =a^{n-1} \\
n-1 & =5=6
\end{aligned}
\]
\(\therefore\) The order of the reaction is 6 .
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