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(C) The reaction $X ightarrow Y$ follows second order kinetics.The rate law for this reaction is given by: $	ext{Rate} = k[X]^2$.Let the initial co

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It will increase by $9$ times.

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(C) The reaction $X ightarrow Y$ follows second order kinetics.The rate law for this reaction is given by: $	ext{Rate} = k[X]^2$.Let the initial concentration of $X$ be $[X]_1 = a \ 	ext{mol L}^{-1}$.Then,$	ext{Rate}_1 = k(a)^2 = ka^2$.If the concentration of $X$ is increased to three times,the new concentration is $[X]_2 = 3a \ 	ext{mol L}^{-1}$.The new rate of reaction is: $	ext{Rate}_2 = k(3a)^2 = k(9a^2) = 9(ka^2)$.Comparing the two rates: $	ext{Rate}_2 = 9 	imes 	ext{Rate}_1$.Therefore,the rate of formation of $Y$ will increase by $9$ times.

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

The conversion of molecules $X$ to $Y$ follows second order kinetics. If the concentration of $X$ is increased to three times,how will it affect the rate of formation of $Y$?

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