For two firstorder reactions having same concentration of `A` and `B` at `t = 0`, use the given data to calculate the tempretature at which both occur with same rate.
Reaction I: `ArarrZ`, `k_(1) = 10^(16)e^((-(3000)/(T)))`
Reaction II: `BrarrY`, `k_(2) = 10^(15) e^((-(2000)/(T)))`

Explanation

`r_(1) = k_(1)[A]`
`r_(2) = k_(2)[B]`
Since `[A] = [B], r_(1) = r_(2)` when `k_(1) = k_(2)` Thus, `10^(16)e^((-(3000)/(T))) = 10^(15)e^((-(2000)/(T)))`
or `16 - (3000)/(T xx 2.303) = 15 - (2000)/(T xx 2.303)`
`1 = (1000)/(T xx 2.303)` or `T = (1000)/(1 xx 2.303) = 434.22 K`