For a reaction $\mathrm{A} \rightarrow 2 \mathrm{~B}+\mathrm{C}$ the half lives are $100 \mathrm{~s}$ and $50 \mathrm{~s}$ when the concentration of reactant $\mathrm{A}$ is $0.5$ and $1.0 \mathrm{~mol} \mathrm{~L}^{-1}$ respectively. The order of the reaction is ______________ . (Nearest Integer)
Answer (integer)
2
Solution
$t_{1 / 2} \propto \frac{1}{\left(a_{0}\right)^{n-1}}$
<br/><br/>
$$
\begin{array}{ll}
\mathrm{t}_{1 / 2}=100 \,\mathrm{sec} & \mathrm{a}_{0}=0.5 \\
\mathrm{t}_{1 / 2}=50 \,\mathrm{sec} & \mathrm{a}_{0}=1
\end{array}
$$
<br/><br/>
$\frac{100}{50}=\left(\frac{1}{0 \cdot 5}\right)^{n-1}$
<br/><br/>
$(2)=(2)^{n-1}$
<br/><br/>
$\mathrm{n}-1=1$
<br/><br/>
$n=2$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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