1) Calculate enthalpy change for the change 8S(g) → S8(g), given that H2S2(g) → 2H(g) + 2S(g) ΔH = 239.0 kcal mol-1 H2S(g) → 2H(g) + S(g) ΔH = 175.0 kcal mol-1 A) + 512.0 k cal B) - 512.0 k cal C) +508.0 k cal D) - 508.0 k cal Answer: Option BExplanation:$\triangle H_{S-S}+2\triangle H_{H-S} = 239$ $2\triangle H_{H-S} = 175$ Hence, $\triangle H_{S-S}=239-175=64kcal mol^{-1}$ Then, ΔH for 8S(g) → S8(g) is 8 x (-64) = -512kcal