ThermochemistryHard

Hess's law

Calculate ΔH° for: C(s) + ½O₂(g) → CO(g) using:
(1) C(s) + O₂(g) → CO₂(g) ΔH° = -393.5 kJ
(2) CO(g) + ½O₂(g) → CO₂(g) ΔH° = -283.0 kJ

Given data

Rxn (1): C(s) + O₂(g) → CO₂(g) ΔH° = -393.5 kJRxn (2): CO(g) + ½O₂(g) → CO₂(g) ΔH° = -283.0 kJ

Steps

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Apply Hess: target = rxn(1) - rxn(2). ΔH° = ?
Write your working — the numerical result must appear in the text kJ

Full worked solution

Apply Hess: target = rxn(1) - rxn(2). ΔH° = ?
$-393.5 - (-283.0)$
$-393.5 - (-283.0) = -393.5+283.0 = -110.5 = \mathbf{-110.5\,kJ}$ . CO formation enthalpy via Hess's law.

Result:ΔH°_f(CO) = -110.5 kJ/mol.

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