  Question

# In a reaction, A+B→ Product, rate is doubled when the concentration of B is doubled and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as :

A

Rate =k[A][B]2

B

Rate =k[A]2[B]2

C

Rate =k[A][B]

D

Rate =k[A]2[B]

Solution

## The correct option is D Rate =k[A]2[B] Let the order of reaction with respect to A and B is x and y respectively. So, the rate law can be given as R=k[A]x[B]y  …(i) When the concentration of only B is doubled, the rate is doubled, so R′=k[A]x[2B]y=2R ….(ii) If concentrations of both the reactants A and B are doubled, the rate increases by a factor of 8, so R′′=k[2A]x[2B]y=8R ….(iii) ⇒k2x2y[A]x[B]y=8R ....(iv) Form Eqs. (i) and (ii), we get ⇒2RR=([A]x[2B]y)([A]x[B]y)2=2y∴y=1 From Eqs.(i) and (iv), we get ⇒8RR=(2x2y[A]x[B]y)([A]x[B]y)or8=2x2y Substitution of the value of y gives, 8=2x214=2x(2)2=(2)x∴X=2 Substitution of the value of x and y in Eq. (i) gives, R=k[A]2[B]  Suggest corrections   