Compound A disappears by two parallel first-order paths, A→ B (rate constant k1, Ea1=60 kJ mol^-1) and A→ C (rate constant k2, Ea2=100 kJ mol^-1). At…

The overall rate constant is $k=k_1+k_2$, and $E_{a}=\dfrac{d\ln k}{d(1/T)}(-R)=\dfrac{k_1E_{a1}+k_2E_{a2}}{k_1+k_2}$. At $300\ \mathrm{K}$, $k_1/(k_1+k_2)=0.80$, so $E_a=0.8(60)+0.2(100)=68\ \mathrm{kJ\,mol^{-1}}$. A simple average, $80\ \mathrm{kJ\,mol^{-1}}$, ignores the weighting. For the product ratio, $\dfrac{k_1}{k_2}\propto\exp\!\left[\dfrac{E_{a2}-E_{a1}}{R}\left(-\dfrac1T\right)\right]$, hence $\left(\dfrac{k_1}{k_2}\right)_{330}=4\exp\!\left[-\dfrac{40000}{8.314}\left(\dfrac1{300}-\dfrac1{330}\right)\right]=4e^{-1.458}=0.931$. The fraction ending as B equals $k_1/(k_1+k_2)=0.931/1.931=0.482$, that is $48\%$. The value $80\%$ is the yield at $300\ \mathrm{K}$, and $52\%$ is the yield of C at $330\ \mathrm{K}$. The yield of the higher-energy product C increases with temperature, because the path with larger activation energy is more sensitive to heating, so the share of B falls from $80\%$ to $48\%$.