[A]_0 / text{mol } L^{-1}
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Question

(C) For a reaction of order $n$,$t_{1/2} \propto [A]_0^{1-n}$.
Using the data: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{([A]_0)_1}{([A]_0)_2}

Vedclass · Accepted Answer

$A, B$ and $D$ only

Vedclass · Answer

(C) For a reaction of order $n$,$t_{1/2} \propto [A]_0^{1-n}$.
Using the data: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{([A]_0)_1}{([A]_0)_2} ight)^{1-n}$.
$\frac{200}{100} = \left( \frac{0.100}{0.025} ight)^{1-n} \implies 2 = (4)^{1-n}$.
$2 = (2^2)^{1-n} = 2^{2-2n}$.
$1 = 2-2n \implies 2n = 1 \implies n = 1/2$.
Thus,statement $A$ is correct.
Since $n = 1/2$,$t_{1/2} \propto [A]_0^{1-1/2} = [A]_0^{1/2}$.
For $[A]_0 = 1 	ext{ M}$: $\frac{200}{t_{1/2}} = \left( \frac{0.1}{1} ight)^{1/2} = \sqrt{0.1} = \frac{1}{\sqrt{10}}$.
$t_{1/2} = 200 \sqrt{10} 	ext{ min}$. Thus,statement $B$ is correct.
Statement $C$ is incorrect because the order of a reaction is a constant property for a given reaction under specific conditions.
For $[A]_0 = 1.6 	ext{ M}$: $\frac{200}{t_{1/2}} = \left( \frac{0.1}{1.6} ight)^{1/2} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
$t_{1/2} = 200 	imes 4 = 800 	ext{ min}$. Thus,statement $D$ is correct.
Therefore,$A, B$ and $D$ are correct.

$[A]_0 / \text{mol } L^{-1}$	$t_{1/2} / \text{min}$
$0.100$	$200$
$0.025$	$100$

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