A first-order reaction is 20% complete in 30 min at 300 K and 20% complete in 5.0 min at 330 K. With $R = 8.314 J mo l^{- 1} K^{- 1}$ and $ln 6 = 1.792$ , the activation energy in $kJ mo l^{- 1}$ is

Answer: C

For equal extent of a first-order reaction,

k \propto 1/ t

, so

k_{330} / k_{300} = 30/5.0 = 6

E_{a} = \frac{R l n 6}{1/300 - 1/330} = \frac{8.314 \times 1.792}{0.0003030} J mo l^{- 1} = 49.2 kJ mo l^{- 1}