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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 o 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) Fe2O3(s) → 2 Fe(s) + 3/2 O2(g)

∆Ho = -1676 kJ ∆Ho = +824 kJ

2 Al(s) + 3/2 O2(g) → Al2O3(s) Fe2O3(s) → 2 Fe(s) + 3/2 O2(g)

∆Ho = -1676 kJ ∆Ho = +824 kJ

2. Cancel

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) C2H4(g) + HCl(g) → C2H5Cl(l) C2H3Cl(g) + H2(g) → C2H5Cl(l) 1. Inverse equation C H2(g) + Cl2(g) → 2HCl(g) C2H4(g) + HCl(g) → C2H5Cl(l) C2H5Cl(l) → C2H3Cl(g) + H2(g) 2. Cancel H2(g) + Cl2(g) → 2HCl(g) C2H4(g) + HCl(g) → C2H5Cl(l) C2H5Cl(l) → C2H3Cl(g) + H2(g) 3. Add C2H4(g) + Cl2(g) → C2H3Cl(g) + HCl(g)

∆Ho = -185 kJ ∆Ho = -65 kJ ∆Ho = -140 kJ ∆Ho = -185 kJ ∆Ho = -65 kJ ∆Ho = +140 kJ ∆Ho = -185 kJ ∆Ho = -65 kJ ∆Ho = +140 kJ ∆Ho = -110 kJ

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

6. Given:

Predict ∆Ho for 1. Inverse B

2. Cancel

CaCO3(s) + 2 HCl(g) → CaCl2(s) + H2O(l) + CO2(g) 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

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 if and and

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) ∆Ho = ? C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O (g) ∆Ho = -1323 kJ C2H4(g) + H2(g) → C2H6(g) ∆Ho = -137 kJ 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...