Enthalpy cycles, Hess cycles, thermochemical cycles

What is enthalpy exactly?

An enthalpy change is the heat transferred at constant pressure by a chemical reaction

Heat changes in chemical reactions are the result of bond making and bond breaking.

Zn_(s) + CuSO_4(aq) ⇾ ZnSO_4(aq) + Cu_(s)

Hess’s Law states that ΔH (the change in enthalpy) depends only on the finial and initial conditions of the system and is independent of the route taken to get from A to B

We can use Hess’s Law to indirectly determine an enthalpy change for a chemical reaction, which is nice.

Our first example is the oxidation of carbon (graphite) to from carbon dioxide. We can accomplish this in two ways (via two different routes).

1. burn carbon in a plentiful supply of oxygen to produce carbon dioxide
2. burn carbon in a limited supply of oxygen so that we have incomplete combustion – carbon monoxide is produced, which we can then burn in more oxygen to give us carbon dioxide

Either way, we start off with carbon and oxygen and we end up with carbon dioxide.

Δ_rH^⦵ is the same either way as the initial and final conditions are the same (elements and compounds in their standard states under standard conditions, 298K, 100 kPa).

At A level, two enthalpy changes are routinely used as an ‘alternative route’ to indirectly find Δ_rH^⦵ for a reaction that cannot be carried out at 298K and 100kPa.

e.g. 2C_(s) + 3H_2(g) ⇾ C₂H_6(g)

Graphite and hydrogen don’t react at 298K (activation energy is far too high) and in any case we can’t guarantee the product would be ethane as there are lots of possibilities.

Time for a couple more definitions. LEARN THEM 😳

The standard enthalpy change of formation, Δ_fH^⦵, is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions.
½N_2(g) + 1½H_2(g) ⇾ NH_3(g) Δ_fH^⦵ = -46.1 kJ mol^-1

Note that the equation is written for the formation of 1 mole of ammonia because Δ_fH^⦵ is for the formation of 1 mole of a compound, as per the definition.

The standard enthalpy change of combustion, Δ_cH^⦵, is the enthalpy change when 1 mole of a substance reacts completely with oxygen gas at 298K and 100kPa
CH_4(g) + 2O_2(g) ⇾ CO_2(g) + 2H₂O_(l) Δ_cH^⦵ = -890.3 kJ mol^-1

Note that the state symbol for water is (l) not (g) as water is indeed a liquid at 298K.

Using Δ_fH^⦵ data in an enthalpy cycle

Given the following data, calculate the enthalpy change of reaction at 298K for:

3Fe₃O_3(s) + 2NH_3(g) ⇾ 6FeO_(s) + 3H₂O_(l) + N_2(g)

Substance	ΔfH⦵ / kJ mol-1
Fe2O3	-824
FeO	-266
NH3	-46
H2O	-286

Note that there is no data for nitrogen as Δ_fH^⦵ is for the formation of compounds and nitrogen is an element!

The data is for the formation of each of these compounds in the equation from its elements, so we can draw a Hess cycle where the alternative route is to make and unmake compounds from their elements …

Once we have drawn our Hess cycle we can add the data to each arrow / reaction, then draw in our alternative route …

In order to get from reactants to products via route 2, our alternative route, we have to UNMAKE iron (III) oxide and ammonia to form elements which can then be reassembled to make iron (II) oxide, water and nitrogen.

If forming 1 mole of Fe₂O₃ from its elements releases -824 kJ mol^-1, then unmaking it must use +824 kJ mol^-1. Essentially, if we are travelling in the opposite direction to the arrow as drawn on our Hess cycle, we reverse the sign on the data.

Now to calculate a value for the enthalpy change of reaction, Δ_rH^⦵, via the alternative route, all we need to do is add everything together and remember to put in the correct sign for an endothermic or exothermic reaction …

(3 x +824) + (2 x +46) + (6 x -266) + (3 x -286) =. +110 kJ mol^-1

Using Δ_cH^⦵ data in an enthalpy cycle

Given the following data, calculate the enthalpy change of reaction at 298K for:

CH₃CHO_(l) + H_2(g) ⇾ CH₃CH₂OH_(l)

Substance	ΔcH⦵ / kJ mol-1
CH3CHO	-1167
H2	-286
CH3CH2OH	-1367

We tackle this in exactly the same way as above – step 1 is to draw out a Hess cycle. In this case the Δ_cH^⦵ data is for the combustion of the substances in the equation to form carbon dioxide and water … our alternative route must involve burning ethanal and hydrogen to form CO₂ and H₂O, and then using that CO₂ and H₂O to make ethanol.

Step 2 is to add the data to the arrows, then draw in our alternative route.

Step 3 is to check the direction of travel – if our alternative route is going in the opposite direction to the arrows of the Hess cycle, we know we are going to have to reverse the sign on the data. In this case, we are told that burning ethanol releases -1367 kJ mol^-1, so unburning it (converting carbon dioxide and water into ethanol) must use up + 1367 kJ mol^-1.

Δ_rH^⦵ = (-1167) + (-286) + (+1367) = -86 kJ mol^-1

Practice questions

1. Calculate the enthalpy change of formation for butane, given that Δ_cH^⦵ (H_2(g)) is -286 kJ mol^-1, Δ_cH^⦵ (C_(s)) is -394 kJ mol^-1 and Δ_cH^⦵ (C₄H_10(g)) is -2877 kJ mol^-1.

4C_(s)  +  5H_2(g)  ⇾  C₄H_10(g)

2. (a) Write an equation to represent Δ_cH^⦵ for cyclohexane.

      (b) State the conditions under which this reaction occurs.

      (c) Calculate Δ_cH^⦵ for the cyclohexane using the following data

Substance	Cyclohexane	Carbon dioxide	Water
ΔfH⦵ / kJ mol-1	-156.3	-393.5	-285.8

3. A liquid hydrocarbon with a density of 0.692 g cm^-3 releases 47.8 kJ g-¹ when burnt in excess oxygen. Calculate the heat energy released in kJ when 1.00 dm³ of the hydrocarbon is completely combusted.  

Answers

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