Video Coming Soon
Before we learn about Hess Cycles, let's revisit some key concepts from thermochemistry.
Thermochemistry is the study of heat changes during chemical reactions. The heat change depends on the amount (number of moles) of reactants involved, and is measured in kJ mol−1.
When a heat change is measured at constant pressure, it is called an enthalpy change, given the symbol ΔH.
When an enthalpy change is measured under standard conditions (pressure = 100 kPa, temperature = 298 K, with all substances in their standard states), it is written as:
ΔH°
The standard enthalpy of combustion (ΔHc°) is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions. The standard enthalpy of formation (ΔHf°) is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.
Germain Henri Hess was a 19th-century chemist who discovered that energy changes in chemical reactions do not depend on the route taken. His law states:
The enthalpy change (ΔH) of a reaction is independent of the route taken, and depends only on the initial and final states.
This is enormously useful. If you have a reaction whose enthalpy change is difficult to measure directly, you can calculate it indirectly by using a different route, provided you know the enthalpy changes for the individual steps in that route. This is where Hess Cycles come in.
If the exam question provides standard enthalpies of combustion, you should draw a combustion Hess Cycle. In this cycle, both reactants and products are combusted to form the same combustion products (usually CO2 and H2O), and the arrows point downwards.
The general layout looks like this:
The formula to calculate the unknown enthalpy change is:
ΔH = ∑(ΔHc reactants) − ∑(ΔHc products)
Combustion = Burn Down! So the arrows in a combustion Hess Cycle point down.
Use the combustion data below to calculate the standard enthalpy of formation of butane (C4H10).
| Substance | Standard enthalpy of combustion / kJ mol−1 |
|---|---|
| C | −394 |
| H2 | −286 |
| C4H10 | −2877 |
The formation reaction for butane is:
4C + 5H2 → C4H10
Draw the Hess Cycle with the combustion products (4CO2 + 5H2O) at the bottom, then apply the formula:
ΔH = ∑(ΔHc reactants) − ∑(ΔHc products)
ΔH = [4(−394) + 5(−286)] − [−2877]
ΔH = [−1576 + (−1430)] − [−2877]
ΔH = −3006 + 2877
ΔH = −129 kJ mol−1
Notice that the balanced equation requires 4 moles of C and 5 moles of H2, so we multiply their combustion values accordingly. Always balance your cycle first.
If the exam question provides standard enthalpies of formation, you should draw a formation Hess Cycle. In this cycle, the elements in their standard states sit at the bottom, and the arrows point upwards.
The general layout looks like this:
The formula to calculate the unknown enthalpy change is:
ΔH = ∑(ΔHf products) − ∑(ΔHf reactants)
Formation = Grow Up! So the arrows in a formation Hess Cycle point up.
Use the formation data below to calculate the standard enthalpy of combustion of benzene (C6H6).
| Substance | Standard enthalpy of formation / kJ mol−1 |
|---|---|
| C6H6 | +49.2 |
| CO2 | −394 |
| H2O | −286 |
The combustion reaction for benzene is:
C6H6 + 7.5O2 → 6CO2 + 3H2O
Draw the Hess Cycle with the elements (6C + 3H2 + 7.5O2) at the bottom, then apply the formula:
ΔH = ∑(ΔHf products) − ∑(ΔHf reactants)
ΔH = [6(−394) + 3(−286)] − [+49.2]
ΔH = [−2364 + (−858)] − [+49.2]
ΔH = −3222 − 49.2
ΔH = −3271.2 kJ mol−1
Note that oxygen is an element in its standard state, so its enthalpy of formation is zero. We only need the formation values for C6H6, CO2, and H2O.
This is straightforward: look at the data you have been given in the question.
| Data given | Cycle to draw | Arrow direction | Formula |
|---|---|---|---|
| Enthalpies of combustion | Combustion cycle | Down | ΔH = ∑ΔHc(reactants) − ∑ΔHc(products) |
| Enthalpies of formation | Formation cycle | Up | ΔH = ∑ΔHf(products) − ∑ΔHf(reactants) |