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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.

$P_{A_0} \text{ (atm)}$	$0.1$	$0.025$
$t_{1/2} \text{ (sec)}$	$100$	$50$

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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