A reacton is first order in A and second order in B. (i)Wite the differential rate equation. (ii)How is the rate affected on increasing the concentration of B three times? (iii)How Is the rate affected when the concentration of both A and B are doubled?

A reacton is first order in A and second order in B. (i)Wite the diff

1.	A reacton is first order in A and second order in B. (i)Wite the differential rate equation. (ii)How is the rate affected on increasing the concentration of B three times? (iii)How Is the rate affected when the concentration of both A and B are doubled?
Answer» Solution :Reaction:`A+BtoP` Order with respect to A=1 Order with respect to B=2 Total order of reaction =1+2=3 (i)Rate law of REACTON is as follow: `(d[R])/(DT)=k[A]^(1)[B]^(2)` In short r=`k[A]^(1)[B]^(2)` (ii)Effect on rate if the concentration increase by three times: `r_(2)=k[A]^(1)[3B]^(2)` `r_(2)=9 r_(1)` Rate increases by 9 times. (iii)If concentration A and B both double then the rate: Concentration of A=2A Concentration of B=2B So,`r_(3)=k[2A]^(1)[2B]^(2)` `=k2[A]^(2)xx4[B]^(2)` `therefore` The rate becomes 8 times (Note:If only concentration of B is double then rate become double)