Question

Question #edde8

Answer 1

Reaction rate differs with temperature. Here, we're looking to relate those two variables with the equation,

`ln(k_1/k_2) = (E_a)/R(1/T_2-1/T_1)`

We can deduce each rate,

`0.0112s^-1 = k_1`
`k_2 = 2k_1 = 0.0224s^-1`

And solve for the other temperature,

`ln((0.0112s^-1)/(0.0224s^-1)) = ((41.1*10^3J)/(mol))/((8.314J)/(mol*K))(1/T_2-1/(298K))`

`thereforeT_2 approx 311K`