For a second-order reaction $2 A products$ with $[A]_{0} = 0.50 M$ in both runs, the half-life is 100 s at 300 K and 20 s at 320 K. Using $R = 8.314 J mo l^{- 1} K^{- 1}$ and $ln 5 = 1.609$ , find $E_{a}$ in kJ/mol (one decimal).

Numerical answer type, answer in kJ/mol, tolerance 0.3.

Answer: 64.2 kJ/mol

For second order

t_{1/2} = 1/ (k [A]_{0})

, so

k \propto 1/ t_{1/2}

and

k_{320} / k_{300} = 5.0

E_{a} = \frac{R l n 5}{1/300 - 1/320} = \frac{8.314 \times 1.609}{0.0002083} J mo l^{- 1} = 64.2 kJ mo l^{- 1}