Question
For a reaction A + B → P, the rate is given by Rate = k[A] [B]2
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
Solution
A+B
Rate = k [A][B]2
(i) Since the given reaction has order two with respect to reactant B, thus if the concentration of B is double in the given reaction, then the rate of reaction will become four times.
(ii) If A is present in large excess then the reaction will be independent of the concentration of A and will be independent only on the concentration of B.as [B]2 will be the only determining factor in the order of the reaction will be two.
