Enthalphy of Reactions Hess's Law

Question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g)2FeO(s)...... ΔH° = -544.0 kJ

(2) 2Hg(l) + O₂(g)2HgO(s)......ΔH° = -181.6 kJ

what is the standard enthalpy change for the reaction:

(3) FeO(s) + Hg(l)Fe(s) + HgO(s)......ΔH° = ?

 kJ

Answer 1

FeO(s) + Hg(l) ==> Fe(s) + HgO(s) TARGET EQUATION

Given:

(1) 2Fe(s) + O₂(g) ==> 2FeO(s) ....  ΔH° = -544.0 kJ

(2) 2Hg(l) + O₂(g)2HgO(s) .... ΔH° = -181.6 kJ

Procedure:

Reverse (1)÷2: FeO(s) ==> Fe(s) + 1/2O₂ (g) ... ∆H = +272 kJ (change sign and divide by 2)

Copy (2)÷2: Hg(l) + 1/2O₂(g) ==> HgO(s) ... ∆H = -90.8 (divide by 2)

Add together to get...

FeO(s) + Hg(l) + 1/2O₂(g) ==> Fe(s) + 1/2)O₂(g) + HgO(s)

Combine and/or cancel like items to get ...

FeO(s) + Hg(l) ==> Fe(s) + HgO(s) ... TARGET EQUATION

∆H = 272 kJ + -90.8 kJ

∆H = 181.2 kJ