Chemical Kinetics

The rate of reaction $k$ is given by Arrhenius equation $k = Ae^{-\frac{E_a}{RT}}$ , where $A$ is a frequency factor of the reaction, $E_a$ the activation energy, $R$ the universal gas constant and $T$ temperature in K. Therefore we have:

$\begin{aligned} \frac{k_2}{k_1} & = \frac{Ae^{\frac{E_a}{RT_1}}}{Ae^{\frac{E_a}{RT_2}}} \\ 2 & = e^{\frac{E_a}{8.314}\left(\frac{1}{300}-\frac{1}{310}\right)} \\ \ln 2 & = \frac{E_a}{8.314}\left(\frac{1}{300}-\frac{1}{310}\right) \\ \Rightarrow E_a & = \frac{300\times 310 \times 8.314 \ln 2}{310-300} = 53594.3 = \boxed{53.6} \space kJ/mol \end{aligned}$

From arrhenius equation,logk2/k1= -Ea/2.303R (1/T2-1/T1) Given, k2/k1=2; T2=310K T1=300K On putting values, log2=-Ea/2.303*8.314(1/310-1/300) Ea=53598.6 J/mol=53.6 kJ/mol

$ln\dfrac{k_1}{k_2}=\dfrac{E_a}{R}\left( \dfrac{1}{T_1}-\dfrac{1}{T_2} \right)\\ ln\dfrac{1}{2}=\dfrac{E_a}{8.314}\left(\dfrac{1}{300}-\dfrac{1}{310}\right)\\ \text{Calculating, } E_a=53594.28J/mol\approx\boxed{53.6}kJ/mol$

For proof, see this wiki Rate of chemical reaction