Question

For a reaction A + B → P  the rate is given by Rate = k[A] [B]²

(a) How is the rate of reaction affected if the concentration of B is doubled ? 

(b) What is the overall order of reaction if A is present in large excess ?

Answer 1

A + B → P

Rate = k[A] [B]² 

(a) When concentration of B is doubled it means conc. of B becomes 2 times.

Thus, Rate = k[A]¹ [2B]²  

= = k[A] [4B]²

(b) Order of reaction is the no. of molecules whose conc. alters after the reaction. If A is present in excess i.e., its conc. is unaffected. So, rate is depending only on the conc. of B. As

k = [B]²

Thus the reaction is of second order.