# Question #edde8

Dec 1, 2017

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

$\ln \left({k}_{1} / {k}_{2}\right) = \frac{{E}_{a}}{R} \left(\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right)$

We can deduce each rate,

$0.0112 {s}^{-} 1 = {k}_{1}$
${k}_{2} = 2 {k}_{1} = 0.0224 {s}^{-} 1$

And solve for the other temperature,

$\ln \left(\frac{0.0112 {s}^{-} 1}{0.0224 {s}^{-} 1}\right) = \frac{\frac{41.1 \cdot {10}^{3} J}{m o l}}{\frac{8.314 J}{m o l \cdot K}} \left(\frac{1}{T} _ 2 - \frac{1}{298 K}\right)$

$\therefore {T}_{2} \approx 311 K$