For the hypothetical reaction
`2A + B rarr` Products
following data obtained:
`|{:("Experiment number","Initial conc of (A)" (mol L^(-1)),"Initial conc of (B)" (mol L^(-1)),"Initial rate mol" L^(-1) s^(-1),),(1,0.10,0.20,3 xx 10^(2),),(2,0.30,0.40,3.6 xx 10^(10^(3)),),(3,0.30,0.80,1.44 xx 10^(4),),(4,0.10,0.40,...,),(5,0.20,0.60,...,),(6,0.30,1.20,...,):}|`
Find out how the rate of the reaction depends upon the concentration of `A` and `B` and fill in the blanks given in the table.

Explanation

form experiments (2) and (3), it is clear that when the concentration of `A` is kept constant and that of `B` is doubled, the rate increases four times. This shows that the reaction is of second order with respect to `B`.
ismilarly, form experiments (1) and (2), it is observed that when the concentration of `A` is increased theee times and that of `B` two times, the rate becomes `12` times. Hence, the reaction is of first order with respect to `A`.
Thus, the rate law for the reaction is
Rate `= k[A][B]^(2)`
Substituting the values of experiment (1) in the rate equation,
`3 xx 10^(2) = k[0.10][0.20]^(2)`
or `k = (3 xx 10^(2))/([0.10][0.20]^(2)) = 7.5 xx 10^(4) L^(2) mol^(-2) s^(-1)`
Experiment (4): Rate `= k[0.10][0.40]^(2)`
`= 7.5 xx 10^(4) xx 0.10 xx 0.40 xx 0.40`
`= 1.2 xx 10^(3) mol L^(-1) s^(-1)`
Experiment (5): Rate `= k[0.20][0.60]^(2)`
`= 7.5 xx 10^(4) xx 0.20 xx 0.60 xx 0.60`
`= 5.4 xx 10^(3) mol L^(-1) s^(-1)`
Experiment (6): Rate `= k[0.30][1.20]^(2)`
`= 7.5 xx 10^(4) xx 0.30 xx 1.20 xx 1.20`
`= 3.24 xx 10^(4) mol L^(-1) s^(-1)`