A plot of $ln k$ ( $k$ in $s^{- 1}$ ) against $1/ T$ ( $T$ in K) for a reaction is a straight line of slope $- 6000 K$ and intercept 20. Taking $R = 8.314 J mo l^{- 1} K^{- 1}$ and $ln 10 = 2.303$ , the activation energy in $kJ mo l^{- 1}$ is

Answer: C

Slope

= - E_{a} / R

, so

E_{a} = 6000 \times 8.314 = 49884 J mo l^{- 1} = 49.9 kJ mo l^{- 1}

. The natural-log plot needs no factor of 2.303.