At $298 K$ a catalyst makes a reaction exactly $1 0^{4}$ times faster, with no change in the pre-exponential factor. By how many $kJ mo l^{- 1}$ does it lower the activation energy? Answer to two decimals. Use $R = 8.314 J mo l^{- 1} K^{- 1}$ and $2.303$ for $ln 10$ .

Numerical answer type, answer in kJ/mol, tolerance 0.02.

Answer: 22.82 kJ/mol

Δ E_{a} = 2.303 RT lo g_{10} (k_{c a t} / k_{u n c a t}) = 2.303 \times 8.314 \times 298 \times 4 = 22823 J mo l^{- 1}

. This is 22.82

kJ mo l^{- 1}