The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. [R = 8.314 JK^-1 mol^-1]

52.905 kJ mol^-1

log\(\frac{K_2}{K_1}\) = \(\frac{E_a}{2.303R}\)[\(\frac{1}{T_1}\) - \(\frac{1}{T_2}\)]

let K₁ = x and K₂ = 2x

log\(\frac{2x}{x}\) = \(\frac{E_a}{19.15}\)[\(\frac{1}{298}\) - \(\frac{1}{308}\)]

log2 = \(\frac{E_a}{19.15}\)[\(\frac{1}{298}\) - \(\frac{1}{308}\)]

0.30 = \(\frac{E_a}{19.15}\)[\(\frac{308 - 298}{298×308}\)]

E_a = 52905 J mol^-1 = 52.905 kJ mol^-1