hess's law

Question

Given: C(s) + O₂(g) → CO₂(g), ΔH = −394 kJ; 2CO(g) + O₂(g) → 2CO₂(g), ΔH = −566 kJ. The enthalpy of formation of CO(g) is:

RankGuru · Accepted Answer

Target reaction for enthalpy of formation of $CO(g)$: $C(s) + \frac{1}{2}O_2(g) 	o CO(g)$
Given Eq 1: $C(s) + O_2(g) 	o CO_2(g), \Delta H_1 = -394 	ext{ kJ}$
Given Eq 2: $2CO(g) + O_2(g) 	o 2CO_2(g), \Delta H_2 = -566 	ext{ kJ}$
Divide Eq 2 by 2 and reverse it:
$CO_2(g) 	o CO(g) + \frac{1}{2}O_2(g), \Delta H_3 = +\frac{566}{2} = +283 	ext{ kJ}$
Add Eq 1 and Eq 3:
$C(s) + O_2(g) + CO_2(g) 	o CO_2(g) + CO(g) + \frac{1}{2}O_2(g)$
Simplifying: $C(s) + \frac{1}{2}O_2(g) 	o CO(g)$
$\Delta H_f = \Delta H_1 + \Delta H_3 = -394 + 283 = -111 	ext{ kJ}$ (Exact value is around $-110.5 	ext{ kJ}$).

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