The rate constant of a reaction at three temperatures is tabulated as $lo g_{10} k$ : -0.45 at 300 K, 1.05 at 350 K and 2.17 at 400 K ( $k$ in $s^{- 1}$ ). Using the end points, $R = 8.314 J mo l^{- 1} K^{- 1}$ and 2.303 for $ln 10$ , the activation energy in $kJ mo l^{- 1}$ is

Answer: D

lo g_{10} k = lo g_{10} A - \frac{E _{a}}{2.303 R} \frac{1}{T}

, so slope

= - E_{a} /2.303 R

. Slope from the end points

= \frac{( 2.17 ) - ( - 0.45 )}{1/400 - 1/300} = - 3144

K, so

E_{a} = - (- 3144) \times 2.303 \times 8.314 = 60.2 kJ mo l^{- 1}