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(A) The half-life of a reaction of order $n$ is related to the initial concentration $[A_0]$ as $t_{\frac{1}{2}} \propto \frac{1}{[A_0]^{n-1}}$.Given:

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$2$

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(A) The half-life of a reaction of order $n$ is related to the initial concentration $[A_0]$ as $t_{\frac{1}{2}} \propto \frac{1}{[A_0]^{n-1}}$.Given:$t_{\frac{1}{2}, 1} = 100 \ s$ at $[A_0]_1 = 0.5 \ mol \ L^{-1}$$t_{\frac{1}{2}, 2} = 50 \ s$ at $[A_0]_2 = 1.0 \ mol \ L^{-1}$Taking the ratio:$\frac{t_{\frac{1}{2}, 1}}{t_{\frac{1}{2}, 2}} = \left( \frac{[A_0]_2}{[A_0]_1} \right)^{n-1}$$\frac{100}{50} = \left( \frac{1.0}{0.5} \right)^{n-1}$$2 = (2)^{n-1}$$2^1 = 2^{n-1}$Comparing the exponents:$n - 1 = 1$$n = 2$The order of the reaction is $2$.

Observation	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Rate $(mol \ L^{-1} \ sec^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.4$	$0.1$	$3.2 \times 10^{-3}$
$3$	$0.1$	$0.2$	$8 \times 10^{-3}$

For a reaction $A \rightarrow 2 B + C$ the half-lives are $100 \ s$ and $50 \ s$ when the concentration of reactant $A$ is $0.5 \ mol \ L^{-1}$ and $1.0 \ mol \ L^{-1}$ respectively. The order of the reaction is (Nearest Integer).

Similar Questions

What is the order of reaction $A + B \to C$?

Observation $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ Rate $(mol \ L^{-1} \ sec^{-1})$

$1$ $0.1$ $0.1$ $2 \times 10^{-3}$

$2$ $0.4$ $0.1$ $3.2 \times 10^{-3}$

$3$ $0.1$ $0.2$ $8 \times 10^{-3}$

If the reaction between $A$ and $B$ to give $C$ shows first-order kinetics in $A$ and second-order in $B$,the rate equation can be written as:

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