The rate of reaction:
`H_(2)O + 2S_(2)O_(4)^(2-)(aq) rarr 2HSO_(3)^(ɵ)(aq) + S_(2)O_(3)^(2-)(aq)`
was studied as follows. In an experiment, the concentration of `S_(2)O_(4)^(2-)(0.4 M)` half changed in `3.73 xx 10^(6) s`. In another experiment, `0.25 M S_(2)O_(4)^(2-)` half changed in `6.0 xx 10^(6) s`. Find the order of the reaction.

Explanation

`(t_(2))/(t_(1)) = ((a_(1))/(a_(2)))^(n-1)`
`log.(t_(2))/(t_(1)) = (n-1)log((a_(1))/(a_(2)))`
`log = (6.0 xx 10^(6)s)/(3.73 xx 10^(6)s) = (n-1) log.(0.4 M)/(0.25 M)`
`0.2065 = (n-1)(0.2041)`
`(n-1) = (0.2041)/(0.2065) = 1 rArr n = 1 + 1 = 2`
It is a second orde reaction.