A catalyst lowers $E_{a}$ from 100 to 80 kJ/mol but also lowers the pre-exponential factor by a factor of 50 (the catalysed reaction has $A_{c a t} = A /50$ ). Using $R = 8.314 J mo l^{- 1} K^{- 1}$ and $ln 50 = 3.912$ , find the temperature (in K, one decimal) at which the uncatalysed reaction has the same rate constant as the catalysed reaction at 300 K.

Numerical answer type, answer in K, tolerance 0.4.

Answer: 334.2 K

Equal

k

ln A - \frac{E _{1}}{R T _{2}} = ln A - ln 50 - \frac{E _{2}}{R T _{1}}

, so

\frac{E _{1}}{R T _{2}} = \frac{E _{2}}{R T _{1}} + ln 50 = 32.074 + 3.912 = 35.986

. Hence

T_{2} = \frac{100000}{8.314 \times 35.986} = 334.2