A reaction has $E_{a} = 100 kJ mo l^{- 1}$ without a catalyst and $75 kJ mo l^{- 1}$ with a catalyst, with the same pre-exponential factor. At $300 K$ , 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 0.01.

Answer: 4.35

From

k = A e^{- E_{a} / RT}

with equal

A

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

. This is

25000/ (2.303 \times 8.314 \times 300) = 4.352

, i.e. 4.35.