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 (ii) reduced to half?

Question 4.6: 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 (ii) reduced to half?
Answer
Let the concentration of the reactant,[A] = a
Order of reaction = 2 so that
Rate of reaction will Rate = k [A]² … (1)
Plug the values we get
Rate = ka²
(i)Given that concentration of the reactant is doubled
So that [A] = 2a,
Plug the values in equation (1) we get
New Rate of reaction, R₁ = k (2a)² = 4ka²
We have already calculate that initial rate R = ka²
Plug the values we get R₁ = 4 R
Hence rate of reaction will increased to 4 times
(ii) Given that concentration of the reactant is reduced to half
So that [A] = (1/2)a,
Plug the values in equation (1) we get
New Rate of reaction, R₂ = k ((1/2)a)² = (1/4)ka²
We have already calculate that initial rate R = ka²
Plug the values we get R₁ = (1/4) R
Hence rate of reaction will reduced to ¼