Question

A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?

Answer 1

Consider,

The concentration of reactant $\left[A\right]=a$

and its order is 2

So, rate of reaction is $K[A{]}^2$...............................................[1]

Rate $=K{a}^2$

i) If the concentration is doubled

i.e $\left[A\right]=2a$

Write the values in equation (1)

Rate of reaction $R=K[2a{]}^2=K4a^2$

Initial rate $=K{a}^2$

Final rate $R_1=K4a^2$

so, $R_1=4R$

hence, the rate of reaction increased 4 times

ii) If the concentration of the reactant is reduced to half

so, the concentration of reactant $\left[A\right]=\left[\frac{1}{2}\right]a$

Final rate $R_2=K\left(\right(\frac{1}{2})a{)}^2$ $=\left(\frac{1}{4}\right)K{a}^2$

we know $R=K{a}^2$

$R_1=\left(\frac{1}{4}\right)R$

Hence rate of reaction reduce to $\left(\frac{1}{4}\right)$.