A reaction is of second order with respect to a reactant. How will the rate of reaction be affected if the concentration of this reactant is doubled?

It becomes \(\frac{1}{4}\) It becomes four times It becomes \(\frac{1}{2}\) It becomes two times

It becomes four times

Since Rate = K[A]2 For second order reaction Let [A] = a then Rate = Ka2 If [A] = 2a then Rate = K (2a)2 = 4 Ka2 ∴Rate of reaction becomes 4 times.