The activation energy of the reaction: `A + B rarr` Products is `105.73 kJ mol^(-1)`. At `40^(@)C`, the Products are found at the rate of `0.133 mol L^(-1) min^(-1)`. What will be the rate of formation of Products at `80^(@)C`?

Explanation

Let the rate law be difined as
At `T_(1) : r_(1) = k_(1)[A]^(x)[B]^(y)`
At `T_(2) : r_(2) = k_(2)[A]^(x)[B]^(y)`
`rArr r_(2) = r_(1) ((k_(2))/(k_(1)))`
Uisng Arrhenius equation, find `k` at `40^(@)C`.
`log.(k_(2))/(k_(1)) = (E_(a))/(2.303R) ((T_(2) - T_(1))/(T_(1)T_(2)))`
`rArr log.(k_(2))/(k_(1)) = (105.73 xx 10^(3))/(2.303 xx 8.31) ((40)/(313 xx 353))`
`rArr log.(k_(2))/(k_(1)) = 2.0`
`rArr (k_(2))/(k_(1)) = 100`
`rArr r_(2) = 0.133 xx 100 = 13.3 mol L^(-1) min^(-1)`