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) If A is present in large excess, then how will the rate of reaction depend on the concentrations of A and B?

a (i) For a reaction, A + B → P
R1 = k[A][B]² ————— (i)
If the concentration of B is doubled,
R2 = k[A][2B]² ————— (ii)
On dividing (i) and (ii)
R1/R2 = k[A][B]²/k[A][2B]²
R1/R2 = B²/4B²
R2 = 4R1
The rate of reaction will be four times the initial rate.
(ii) If A is present in large excess, then the rate of the reaction will be independent of A and will depend only on the concentration of B. The overall rate of the reaction will be 2.
(b) K = 0.693/30 = 0.0231
The time required to complete 90%