A reaction is of 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 ?
i) If the concentration of reactant A is doubled, i.e.
[A] = 2 a
Then, rate = k (${{2a}^{2}}$)= 4 ${{ka}^{2}}$
Rate of reaction becomes 4 times.
(ii) When concentration of A is reduced to i.e.
[A] = 1/2 a
Then, rate = K[${{a/2}^{2}}$] = 1/4 ${{ka}^{2}}$
Rate of reaction becomes 1/4 times i.e. reduced to one-fourth.