A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
$(i)$ doubled
$(ii)$ reduced to half $?$
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
$(i)$ doubled
$(ii)$ reduced to half $?$
Let the concentration of the reactant be $[ A ]=a$
Rate of reaction, $R=k[A]^{2}$
$=k a^{2}$
$(i)$ If the concentration of the reactant is doubled, i.e. $[ A ]=2 a$, then the rate of the reaction would be
$R ^{\prime}=k(2 a)^{2}$
$=4 ka ^{2}$
$=4 R$
Therefore, the rate of the reaction would increase by $4$ times.
$(ii)$ If the concentration of the reactant is reduced to half, i.e. $[ A ]=\frac{1}{2} a$
the reaction would be
$R ^{\prime \prime}=k\left(\frac{1}{2} a\right)^{2}$
$=\frac{1}{4} k a$
$=\frac{1}{4} R$
Therefore, the rate of the reaction would be reduced to ${\frac{1}{4}^{th}}$
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