Reaction 1 has $E_{a} = 40 kJ mo l^{- 1}$ and $A = 1 0^{10} s^{- 1}$ ; reaction 2 has $E_{a} = 80 kJ mo l^{- 1}$ and $A = 1 0^{15} s^{- 1}$ . Use $R = 8.314 J mo l^{- 1} K^{- 1}$ and $ln 1 0^{5} = 11.51$ . Which statements are correct?

Answer: A, B, D

Setting

ln A_{1} - E_{1} / RT = ln A_{2} - E_{2} / RT

gives

T = \frac{E _{2} - E _{1}}{R l n ( A _{2} / A _{1} )} = \frac{40000}{8.314 \times 11.51} = 418

K. Reaction 2 has the steeper line, so it is slower at low T and faster above the crossing. At 300 K

k_{1} / k_{2} = 1 0^{- 5} e^{40000/ (8.314 \times 300)} = 92.2 \neq = 1