A decomposes by two parallel first-order paths with activation energies $E_{1} = 50.0$ and $E_{2} = 90.0 kJ mo l^{- 1}$ . At the temperature of interest the products from the two paths form in the ratio $3 : 1$ . Treating the overall rate constant as $k = k_{1} + k_{2}$ , find the apparent activation energy of the disappearance of A in $kJ mo l^{- 1}$ (one decimal place).

Numerical answer type, answer in kJ/mol, tolerance 0.1.

Answer: 60 kJ/mol

From

ln k

E_{a} = R T^{2} d ln k / d T = R T^{2} (d k_{1} / d T + d k_{2} / d T) / k = (k_{1} E_{1} + k_{2} E_{2}) / (k_{1} + k_{2})

, a weighted average. The product ratio equals

k_{1} : k_{2} = 3 : 1

, so

E_{a} = 0.75 \times 50.0 + 0.25 \times 90.0 = 60.0 kJ mo l^{- 1}