A substance A decomposes by two parallel first-order paths, AtoB (k1, Ea1=90 kJ mol^-1) and AtoC (k2, Ea2=150 kJ mol^-1). At 400 K, k1=3.0×10^-3 s^-1…

For parallel first-order paths the overall rate constant is $k=k_1+k_2=4.0\times10^{-3}\ \mathrm{s^{-1}}$. Differentiating $\ln k$ with respect to $T$ and using $\dfrac{d\ln k_i}{dT}=\dfrac{E_{ai}}{RT^2}$ gives $E_{a,\mathrm{app}}=\dfrac{k_1E_{a1}+k_2E_{a2}}{k_1+k_2}$, i.e. a rate-weighted average, not the simple mean. Numerically $E_{a,\mathrm{app}}=\dfrac{3(90)+1(150)}{4}=105\ \mathrm{kJ\,mol^{-1}}$ (the arithmetic mean $120$ is the trap). The half-life uses the total constant, $t_{1/2}=\dfrac{0.693}{4.0\times10^{-3}}=173\ \mathrm{s}$; using only $k_1$ gives $231\ \mathrm{s}$. Hence $105\ \mathrm{kJ\,mol^{-1}}$ and $173\ \mathrm{s}$. As a check, the fraction of $\mathrm{A}$ going to $\mathrm{C}$ is $k_2/k=25\%$, and this ratio is independent of time because both paths are first order in the same species.