A reaction is second order with respected to a reactant. How is the rate of reaction affected if the concentration of the reactant is
reduced to half

Explanation

Let the concentration of the reactant be [A] = a
Rate of reaction `R = k[A]^(2)`
`R = ka^(2)`
If the concentration of the reactant is reduced to half `[A]=(1)/(2)` a than the rate of reaction would be
`R=k((1)/(2)a)2`
`=(1)/(4)ka^(2)`
`=(1)/(4)kR`
`therefore` The rate of the reaction would reduced by `(1^(th))/(4)`.