A reaction is first order in A and second order in B.Write the differential rate equation for the reaction.

Explanation

i) `((d) (x))((dt))=(k)[(A)]^(*)[(B)]^(2)` or rate=(k)[(A)][(B)]^(2)`
ii) When the concentration of `(B)` is tripled rate =`(k)[(A)][3 (B)]^(2)=9 (k)[(A)][(B)]^(2)`
Rate becomes 9 xx.
iii) If the conc. of the both `(A)` and `(B)` is doubled, then rate `=(k)[2 (A)][2 (B)]^(2)=(k)[2 (A)] xx 4[(B)]^(2)=8 (k)[(A)][(B)]^(2)`
therefore Rate becomes 8 xx.