A catalyst lowers the activation energy of a reaction from $96 kJ mo l^{- 1}$ to $62 kJ mo l^{- 1}$ but also lowers the pre-exponential factor to $0.010$ times its original value (the site is partly blocked by the catalyst surface). At $310 K$ calculate $lo g_{10} (k_{c a t} / k_{u n c a t})$ to two decimals. Use $R = 8.314 J mo l^{- 1} K^{- 1}$ and $2.303$ for $ln 10$ .

Numerical answer type, tolerance 0.02.

Answer: 3.73

lo g_{10} (k_{c a t} / k_{u n c a t}) = lo g_{10} (A_{c} / A_{u}) + Δ E_{a} / (2.303 RT)

. The first term is

- 2

, the second is

34000/ (2.303 \times 8.314 \times 310) = 5.728

. The total is 3.73.