Question

Given the molecular formula: 2 H2S (g) + 3 O2(g) --> 2 H2O (l) +

2 SO2 (g). Theenthalpy change for the reaction of gaseous hydrogen
sulfideburning is what?

Answer 1

ΔHrxn = - 1123.66 KJ/mol

Step-by-step explanation:

• 2 H2S(g) + 3 O2(g) → 2 H2O(l) + 2 SO2(g)

⇒ ΔHrxn(298K) = 2 ΔHfH2O(l) + 2 ΔHfSO2(g) - 2 ΔHfH2S(g) - 3 ΔHfO2(g)

∴ ΔHf H2O(l) = - 285.83 KJ/mol

∴ ΔHf SO2(g) = - 296.83 KJ/mol

∴ ΔHf H2S(g) = - 20.83 KJ/mol

∴ ΔHf O2(g) = 0 KJ/mol

⇒ ΔHrxn = 2(- 285.83) + 2(- 296.83) - 2(- 20.83)

⇒ ΔHrxn = - 1123.66 KJ/mol