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

Explanation

Ans: Rate `=(k)[(A)]^(2)=(ka)^(2)`
if `[A]=2 a, rate =k[2 a]^(2)=4 ka^(2)=4xx`
`[(A)]=(1)(2) (a) , rate =(k)(a)(2)^(2)=(1)(4) (ka)^(2)=(1)(4) (th)`
Thus the rate of 2nd order reaction becomes 4 xx when the concentration of the reactant is doubled and it becomes `(1)(4) th` when the conc. of reactant is halved.