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

Explanation

(a) A reaction is second order in A and first order in B. Differential rate equation:
`"Rate"=(-d[R])/(dt)=K[A]^2[B]`
(ii) On invreasing the concentration A three times i.e., 3A.
`"Rate"=k[3A]^2[B]`
`=9k[A]^2[B]`
9 (Rate), 8 times the initial rate.
(iii) On increasing the concertration of A and B as 2A and 2B :
`"Rate,"=k[2A]^2[2B]`
`=k(4xx2)[A]^2[B]`
`=8k[A]^2[B]=8`(Rate), 8 times the initial rate.
(b) `AtoP`
`t=0" "a" "0`
`t=0" "(a-x)" "x`
Now, it takes 40 minutes for 30% decoposition i.e.,reactant left after 40 minutes is 70% of its initial concertration.
So, `(a-x)=70/100xxa=7/10a`
`k=2.303/tlog.(a)/((7//10)a)=2.303/40log1.428`
`:. k=0.00891 "min"^(-1)`
`:. t_(1//2)=0.693/k=0.693/0.008913=77.78 min`