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

Solution.

delta H reaction = -(1)/2+(2)/2

FeO - Fe - 0.5O2 + Hg + 0.5O2 - HgO

FeO + Hg = HgO + Fe

delta H(Hg) = 0

delta H(Fe) = 0

delta H reaction = delta H(products) - delta H(reactants)

delta H reaction = -90.8 - (-272) = 181.2 kJ

Answer:

delta H reaction = -90.8 - (-272) = 181.2 kJ