A reaction is second order in A and first order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of A three times?

Given that
Order of reactant A, x = 2
Order of reactant B, y = 1
The differential rate equation will be
-d[R]/dt = k[A]^x[B]^y
Plug the value in this formula
-d[R]/dt = k[A]^2[B]
Given that concentration of A is increased three times
So that
New concentration of A = 3[A]
Put the values we get new rate of reaction
-d[R]/dt = k[3A]^2[B]
Hence, the rate of reaction will increase by 9 times
(iii)Given that
New concentrations of A = 2[A]
New concentration of B = 2[B]
Plug the values we get
-d[R]/dt = k[2A]^2[2B]
Hence the rate of reaction increased by 8 times
(b)k=1/t ln(100/100-x)=1/40 × 0.1548=0.00387 min⁻¹