Question

Calculate enthalpy change for the change $8S(g) \longrightarrow S_8(g)$, given that
$H_2S_2(g) \rightarrow 2H(g) + 2S(g), \Delta H = 239.0 kcalmol^{-1}$
$H_2S(g) \rightarrow 2H(g) + S(g), \Delta H = 175.0 kcalmol^{-1}$
(a) + 512.0 k cal
(b) – 512.0 k cal
(c) 508.0 k cal
(d) – 508.0 k cal

Answer 1

(b) – 512.0 k cal

ΔH1 - ΔH2 = 1 S-S Bond Energy = 239-175 = 64

So for 8S → S8

(-64)*8 = -512.0 kcal