A forms B (activation energy $E_{1}$ ) and C (activation energy $E_{2}$ ) by parallel first-order paths, with $E_{1} - E_{2} = 20.0 kJ mo l^{- 1}$ and a temperature-independent ratio of the pre-exponential factors. The ratio $k_{B} / k_{C}$ is $0.10$ at $300 K$ . Find $k_{B} / k_{C}$ at $400 K$ ( $R = 8.314 J mo l^{- 1} K^{- 1}$ ).

Answer: A

ln \frac{( k _{B} / k _{C} ) _{400}}{( k _{B} / k _{C} ) _{300}} = \frac{E _{1} - E _{2}}{R} (\frac{1}{300} - \frac{1}{400}) = \frac{20000}{8.314} (8.333 \times 1 0^{- 4}) = 2.0047

. So the ratio becomes

0.10 \times e^{2.0047} = 0.74

; the path with higher

E_{a}

is favoured by raising

T