A substance A decomposes by two parallel first-order paths, A→ B (activation energy 60 kJ mol^-1) and A→ C (activation energy 80 kJ mol^-1). At 300 K…

For parallel first-order paths the ratio of products formed equals the ratio of the rate constants at all times, so $\dfrac{[\mathrm{B}]}{[\mathrm{C}]}=\dfrac{k_1}{k_2}=\dfrac{A_1}{A_2}e^{-(E_1-E_2)/RT}$. The pre-exponential factors are treated as temperature independent, so taking the ratio at the two temperatures gives $\ln\dfrac{(k_1/k_2)_{350}}{(k_1/k_2)_{300}}=-\dfrac{E_1-E_2}{R}\left(\dfrac1{350}-\dfrac1{300}\right)$. With $E_1-E_2=-20000\ \mathrm{J\,mol^{-1}}$ the right side is $\dfrac{20000}{8.314}\times(0.0028571-0.0033333)=2405.6\times(-4.762\times10^{-4})=-1.1456$. Hence the ratio at $350\ \mathrm{K}$ is $2.0\times e^{-1.1456}=2.0\times0.3181=0.64$. The path with the higher activation energy gains ground as the temperature rises: $k_1/k_2$ at $350\ \mathrm{K}$ is $e^{-1.1456}\approx0.32$ times its $300\ \mathrm{K}$ value of $2.0$, so the product $\mathrm{C}$, which needs the larger barrier, is favoured more strongly at higher temperature.