JEE MAIN - Chemistry (2022 - 26th July Morning Shift - No. 22)
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
2
Explanation
$t_{1 / 2} \propto \frac{1}{\left(a_{0}\right)^{n-1}}$
$$ \begin{array}{ll} \mathrm{t}_{1 / 2}=100 \,\mathrm{sec} & \mathrm{a}_{0}=0.5 \\ \mathrm{t}_{1 / 2}=50 \,\mathrm{sec} & \mathrm{a}_{0}=1 \end{array} $$
$$\frac{100}{50}=\left(\frac{1}{0 \cdot 5}\right)^{n-1}$$
$$(2)=(2)^{n-1}$$
$$\mathrm{n}-1=1$$
$$n=2$$
$$ \begin{array}{ll} \mathrm{t}_{1 / 2}=100 \,\mathrm{sec} & \mathrm{a}_{0}=0.5 \\ \mathrm{t}_{1 / 2}=50 \,\mathrm{sec} & \mathrm{a}_{0}=1 \end{array} $$
$$\frac{100}{50}=\left(\frac{1}{0 \cdot 5}\right)^{n-1}$$
$$(2)=(2)^{n-1}$$
$$\mathrm{n}-1=1$$
$$n=2$$
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