Substance A decomposes by two parallel first-order paths, Axrightarrowk1B and Axrightarrowk2C. At 300 K, k1/k2=4. The activation energies are E1=60…

The overall rate constant is $k=k_1+k_2$, so $E_{\mathrm{eff}}=RT^2\,\mathrm{d}\ln k/\mathrm{d}T=(k_1E_1+k_2E_2)/(k_1+k_2)$. With $k_1=4k_2$: $E_{\mathrm{eff}}=(4\times60+100)/5=68\ \mathrm{kJ\,mol^{-1}}$ (the simple average, $80$, is wrong because it ignores weighting). The product ratio equals $k_1/k_2$ at the given temperature: $\ln\frac{(k_1/k_2)_{330}}{(k_1/k_2)_{300}}=-\frac{E_1-E_2}{R}\left(\frac1{330}-\frac1{300}\right)=\frac{40000}{8.314}\times(-3.03\times10^{-4})=-1.458$. Thus $(k_1/k_2)_{330}=4e^{-1.458}=0.93$. The pathway with the lower barrier loses ground as the temperature rises, so the ratio falls below $4$. Option B is the inverted ratio and D has the wrong sign of the exponent. Answer: A. Since the ratio $[\mathrm{B}]/[\mathrm{C}]$ in parallel first-order reactions equals $k_1/k_2$ at all times, no concentration data are needed beyond the rate-constant ratio.