5.2 Hess’s Law

Syllabus

What does this mean?

  • Learn the definition!

But what does it mean?

It is basically a re-statement of the Law of Conservation of Energy – energy can’t be created or destroyed. So, if a reaction can happen in two different ways the overall energy change for the first way must be equal to the overall energy change for the second.

You already used this without knowing when you used: ΔH⊖reaction=∑ΔH⊖f(products)− ∑ΔH⊖f(Reactants)

However, if the reaction is more complex it is easier to use the triangle method above and apply Hess’s Law.

In a calculation you will generally be given:

a) ∆Hcombustion for all substances in the reaction

Or

b) ∆Hformation for all substances in the reaction (except elements)

Or

c) A mixture of both

a) Only Combustion Data

Note:

1. The arrows point away from the substances that were burned.

2. The sign was changed on the arrow “facing the wrong way”

b) Only Formation Data

Note:

The arrows point to the substances that are formed.

Either

-109 = +52.2 -92.3 + ∆H

Or

∆H = -109 -52.2 + 92.3 (sign changed for wrong way arrows)

=-68.9 kJmol-1

Questions using Formation Data

1 Calculate the overall enthalpy change for this reaction:

CH4(g) + 2 O2(g) --> CO2(g) + 2 H2O(l)

∆Hf (CH4(g)) = -75, (CO2(g)) = -393, (H2O(l)) = -286 kJ/mol

2 The enthalpy change for the following reaction is -2877 kJ/mol:

C4H10(g) + 6 ½ O2(g) → 4 CO2(g) + 5 H2O(l)

Calculate the enthalpy change of formation of butane (C4H10(g)) given the following data:

∆Hf (CO2(g)) = -393, (H2O(l)) = -286 kJ/mol

3 Use the enthalpies of formation below to calculate the enthalpy change for the following reaction.

3 Fe(s) + 4 H2O(g) → 4 H2(g) + Fe3O4(s)

∆Hf: H2O(g) -242; Fe3O4(s)-1117 kJ mol-1

4 ∆Hreaction for the following reaction is shown.

Use it and the ∆Hf values below to calculate the ∆Hf of Pb(NO3)2(s).

Pb(NO3)2(s) → PbO(s) + 2 NO2(g) + ½ O2(g) ∆Hreaction = +301 kJ mol-1

∆Hf: PbO(S) -217; NO2(g) +33 kJ mol-1


Questions using Formation Data

1 Calculate the enthalpy of formation of ethanol (C2H5OH) given the following enthalpies of combustion. Write an equation for the formation first!

∆Hc C(s) = -393, H2(g) = -286, C2H5OH (l) = -1371 kJ/mol

2 Calculate the enthalpy change for this reaction given the following data.

C(s) + 2 H2(g) → CH4(g)

∆Hc C(s) = -393, H2(g) = -286, CH4(g) = -890 kJ/mol

3) Calculate the enthalpy of combustion of propane, C3H8(g), given the following enthalpy changes.

∆Hc: C(s) -393; H2(g) -286 kJ/mol, ∆Hf: C3H8(l) -103 kJ/mol

4) Calculate the enthalpy change for the following reaction using the enthalpies of combustion given.

C(graphite) → C(diamond)

∆Hc: C(graphite) -393; C(diamond) -395 kJ/mol

5) Calculate the enthalpy of formation of pentane, C5H12(l), given the following enthalpies of combustion.

∆Hc: H2(g) -286; C(s) -393; C5H12(l) -3509 kJ/mol

Questions using mixed data

(write your own equations first)

1) Calculate the enthalpy of combustion of propanone, CH3COCH3(l), given

∆Hc : H2(g) -286; C(s) -393 ∆Hf : CH3COCH3(l) -217 kJ/mol

2) Calculate the enthalpy of combustion of CS2(l) given the data.

∆Hc: C(s) -393; S(s) -297 kJ/mol, ∆Hf: CS2(l) +88 kJ/mol

3) Calculate the enthalpy change for SO2(g) + 2 H2S(g) → 3 S(s) + 2 H2O(l)

∆Hc: S(s) -297 kJ/mol ∆Hf :H2O(l) -286; H2S(g) -20 kJ/mol