Discussion Forum

Calculate enthalpy change for the change

8S_(g) → S_8(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

Answer:

Explanation:

$\triangle H_{S-S}+2\triangle H_{H-S} = 239$

$2\triangle H_{H-S} = 175$

Hence,

$\triangle H_{S-S}=239-175=64kcal mol^{-1}$

Then, ΔH for 8S_(g) → S_8(g)  is 8 x (-64) = -512kcal