Hess’s Law Practice Problems Answers
Determine ∆Ho for each of the following problems. Use a separate piece of paper to show
your work. You can always check your answer using molar enthalpies of formation (∆Hof)
1. Find the standard molar enthalpy for the reaction C(s) + ½ O2(g) → CO(g)
Given that C(s) + O2(g) → CO2(g) ∆Ho = -394 kJ
and CO2(g) → CO(g) + ½ O2(g) ∆Ho = +283 kJ
1. Cancel
C(s) + O2(g) → CO2(g) ∆Ho = -394 kJ
CO2(g) → CO(g) + ½ O2(g) ∆Ho = +283 kJ
2. Add
C(s) + ½ O2(g) → CO(g) ∆Ho = -111 kJ
2. The standard enthalpy changes for the formation of aluminium oxide and iron (III) oxide
are 2 Al(s) + 3/2 O2(g) → Al2O3(s) ∆Ho = -1676 kJ
2 Fe(s) + 3/2 O2(g) → Fe2O3(s) ∆Ho = -824 kJ
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Calculate ∆H for the reaction: Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)
1. Inverse B
2 Al(s) + 3/2 O2(g) → Al2O3(s) ∆Ho = -1676 kJ
Fe2O3(s) → 2 Fe(s) + 3/2 O2(g) ∆Ho = +824 kJ
2. Cancel
2 Al(s) + 3/2 O2(g) → Al2O3(s) ∆Ho = -1676 kJ
Fe2O3(s) → 2 Fe(s) + 3/2 O2(g) ∆Ho = +824 kJ
3. Add
Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) ∆Ho = -852 kJ
3. Coal gasification converts coal into a combustible mixture of carbon monoxide and
hydrogen, called coal gas, in a gasifier: H2O(l) + C(s) → CO(g) + H2(g) ∆Ho = ?
Calculate the standard enthalpy change for this reaction from the following chemical
equations: 2 C(s) + O2(g) → 2 CO(g) ∆Ho = -222 kJ
2 H2(g) + O2(g) → 2H2O(g) ∆Ho = -484 kJ
H2O(l) → H2O(g) ∆Ho = +44 kJ
1. Divide equation A and B by 2 and inverse equation B
C(s) + ½O2(g) → CO(g) ∆Ho = -111 kJ
H2O(g) → H2(g) + ½O2(g) ∆Ho = +242 kJ
H2O(l) → H2O(g) ∆Ho = +44 kJ
2. Cancel
C(s) + ½O2(g) → CO(g) ∆Ho = -111 kJ
H2O(g) → H2(g) + ½O2(g) ∆Ho = +242 kJ
H2O(l) → H2O(g) ∆Ho = +44 kJ
3. Add
H2O(l) + C(s) → CO(g) + H2(g) ∆Ho = +175 kJ
4. This coal gas can be used a fuel: CO(g) + H2(g) + O2(g) → CO2(g) + H2O(g)
Predict the change in enthalpy for this combustion reaction from the following equations:
2 C(s) + O2(g) → 2 CO(g) ∆Ho = -222 kJ
C(s) + O2(g) → CO2(g) ∆Ho = -394 kJ
2 H2(g) + O2(g) → 2H2O(g) ∆Ho = -484 kJ
1. Divide equation A and C by 2 and inverse equation A
CO(g) → C(s) + ½O2(g) ∆Ho = +111 kJ
C(s) + O2(g) → CO2(g) ∆Ho = -394 kJ
H2(g) + ½O2(g) → H2O(g) ∆Ho = -242 kJ
2. Cancel
CO(g) → C(s) + ½O2(g) ∆Ho = +111 kJ
C(s) + O2(g) → CO2(g) ∆Ho = -394 kJ
H2(g) + ½O2(g) → H2O(g) ∆Ho = -242 kJ
3. Add
CO(g) + H2(g) + O2(g) → CO2(g) + H2O(g) ∆Ho = -525 kJ
5. Use the following calorimetrically determined enthalpy changes to predict the standard
enthalpy change for the reaction of ethene with chlorine gas.
C2H4(g) + Cl2(g) → C2H3Cl(g) + HCl(g) ∆Ho = ?
H2(g) + Cl2(g) → 2HCl(g) ∆Ho = -185 kJ
C2H4(g) + HCl(g) → C2H5Cl(l) ∆Ho = -65 kJ
C2H3Cl(g) + H2(g) → C2H5Cl(l) ∆Ho = -140 kJ
1. Inverse equation C
H2(g) + Cl2(g) → 2HCl(g) ∆Ho = -185 kJ
C2H4(g) + HCl(g) → C2H5Cl(l) ∆Ho = -65 kJ
C2H5Cl(l) → C2H3Cl(g) + H2(g) ∆Ho = +140 kJ
2. Cancel
H2(g) + Cl2(g) → 2HCl(g) ∆Ho = -185 kJ
C2H4(g) + HCl(g) → C2H5Cl(l) ∆Ho = -65 kJ
C2H5Cl(l) → C2H3Cl(g) + H2(g) ∆Ho = +140 kJ
3. Add
C2H4(g) + Cl2(g) → C2H3Cl(g) + HCl(g) ∆Ho = -110 kJ
6. Given: CaCO3(s) → CaO(s) + CO2(g) ∆Ho = 175 kJ
Ca(OH)2(s) → H2O(l) + CaO(s) ∆Ho = 67 kJ
Ca(OH)2(s) + 2 HCl(g) → CaCl2(s) + 2 H2O(l) ∆Ho = -198 kJ
Predict ∆Ho for CaCO3(s) + 2 HCl(g) → CaCl2(s) + H2O(l) + CO2(g)
1. Inverse B
CaCO3(s) → CaO(s) + CO2(g) ∆Ho = 175 kJ
CaO(s) + H2O(l) → Ca(OH)2(s) ∆Ho = -67 kJ
Ca(OH)2(s) + 2 HCl(g) → CaCl2(s) + 2 H2O(l) ∆Ho = -198 kJ
2. Cancel
CaCO3(s) → CaO(s) + CO2(g) ∆Ho = 175 kJ
CaO(s) + H2O(l) → Ca(OH)2(s) ∆Ho = -67 kJ
Ca(OH)2(s) + 2 HCl(g) → CaCl2(s) + 2 H2O(l) ∆Ho = -198 kJ
3. Add
CaCO3(s) + 2 HCl(g) → CaCl2(s) + H2O(l) + CO2(g) ∆Ho = -90 kJ
7. Find 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) ∆Ho = ?
if C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O (g) ∆Ho = -1323 kJ
and C2H4(g) + H2(g) → C2H6(g) ∆Ho = -137 kJ
and H2(g) + ½ O2(g) → H2O(g) ∆Ho = -242 kJ
1. Inverse equation B and multiply equation A, B and C by 2
2 C2H4(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(g) ∆Ho = -2646 kJ
2 C2H6(g) → 2C2H4(g) + 2 H2(g) ∆Ho = +274 kJ
2 H2(g) + 1 O2(g) → 2 H2O(g) ∆Ho = -484 kJ
2. Cancel
2 C2H4(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(g) ∆Ho = -2646 kJ
2 C2H6(g) → 2C2H4(g) + 2 H2(g) ∆Ho = +274 kJ
2 H2(g) + 1 O2(g) → 2 H2O(g) ∆Ho = -484 kJ
3. Add
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) ∆H = -2856 kJ
I need to divided –2856 by 2 because the question is to find the molar
enthalpy of formation.
The final answer is ∆Ho = -1428 kJ/mol
