Medium
$A$ 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?
(N/A) Let the initial concentration of the reactant be $[A] = a$.
The rate of reaction is given by $R = k[A]^2 = ka^2$.
$(i)$ If the concentration is doubled,$[A] = 2a$:
$R' = k(2a)^2 = 4ka^2 = 4R$.
Thus,the rate of reaction increases by $4$ times.
$(ii)$ If the concentration is reduced to half,$[A] = \frac{1}{2}a$:
$R'' = k(\frac{1}{2}a)^2 = \frac{1}{4}ka^2 = \frac{1}{4}R$.
Thus,the rate of reaction is reduced to $\frac{1}{4}$ of its initial value.
The rate of reaction is given by $R = k[A]^2 = ka^2$.
$(i)$ If the concentration is doubled,$[A] = 2a$:
$R' = k(2a)^2 = 4ka^2 = 4R$.
Thus,the rate of reaction increases by $4$ times.
$(ii)$ If the concentration is reduced to half,$[A] = \frac{1}{2}a$:
$R'' = k(\frac{1}{2}a)^2 = \frac{1}{4}ka^2 = \frac{1}{4}R$.
Thus,the rate of reaction is reduced to $\frac{1}{4}$ of its initial value.
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