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

Explanation

Since Rate = K `[A]^(2)`
Let [A] = a `" " therefore " " Rate = Ka^(2)…. (1)`
(i) `" "` If [A] = 2a `" " therefore " "` Rate = K `(2a^(2)) = 4 Ka^(2) = 4 ` times
(ii) `" "` If [A] = `(a)/(2) " " therefore " " Rate = K ((a)/(2))^(2) = (1)/(4)`
`Ka^(2) = (1)/(4)` th