A reaction has $E_{a} = 55 kJ mo l^{- 1}$ . At what temperature (in K, one decimal place) will its rate constant be double the value at 300 K? Use $R = 8.314 J mo l^{- 1} K^{- 1}$ and $ln 2 = 0.693$ .

Numerical answer type, answer in K, tolerance 0.3.

Answer: 309.7 K

ln 2 = \frac{E _{a}}{R} (\frac{1}{300} - \frac{1}{T _{2}})

, so

\frac{1}{T _{2}} = \frac{1}{300} - \frac{8.314 \times 0.693}{55000} = 0.0032286

and

T_{2} = 309.7