Enthalpy - Chemistry

Question

Calculate ΔH^o for the process Zn(s) + S(s) + 2 O₂(g) → ZnSO₄(s)

from the following information

Zn(s) + S(s) → ZnS(s).

ΔHo = -158.9 kJ/mol

ZnS(s) + 2 O2(g) → ZnSO4(s)

ΔHo = -247.4 kJ/mol

Answer 1

Hess' Law

Given:

eq1: Zn(s) + S(s) ==>ZnS(s) ... ∆Hº = -158.9 kJ/mol

eq2: ZnS(s) + 2 O2(g) ==> ZnSO4(s) ... ∆Hº = -247.4 kJ/mol

Target equation: Zn(s) + S(s) + 2 O2(g) ==> ZnSO4(s)

Procedure:

Copy eq1: Zn(s) + S(s) ==>ZnS(s) ... ∆Hº = -158.9 kJ/mol

Copy eq2: ZnS(s) + 2 O2(g) ==> ZnSO4(s) ... ∆Hº = -247.4 kJ/mol

Add them together to get...

Zn(s) + S(s) + ZnS(s) + 2 O2(g) => ZnS(s) + ZnSO4(s)

Cancel the ZnS(s) from both sides and you are left with the target equation...

Zn(s) + S(s) + 2 O2(g) ==> ZnSO4(s)

∆Hº for the target reaction is -158.9 + -247.4 = -406.3 kJ