Hess's Law Calculations Revision Notes | A Level AQA Chemistry

1. Why Hess's Law Matters

Many reactions cannot be measured directly in a calorimeter. Combustion of carbon to carbon monoxide, for instance, always produces some carbon dioxide as well, making a clean direct measurement impossible. Hess's law gives you a way around this: if you cannot measure a reaction directly, you can calculate its enthalpy change by combining reactions you can measure.

The underlying reason this works is that enthalpy is a state function. Think of it like altitude on a map: it does not matter which path you take from London to Edinburgh, the difference in altitude between the two cities is always the same. In the same way, the enthalpy difference between reactants and products is fixed, regardless of the chemical route you take between them.

There are two main types of Hess's law calculation at A Level: those using enthalpies of formation and those using enthalpies of combustion. The cycle you draw looks different in each case, and the direction of the arrows matters enormously. Getting this right is the single most important skill in this topic.

2. Enthalpies of Formation: The Downward Cycle

The standard enthalpy of formation, $\Delta_f H^\circ$, is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states. By definition, the enthalpy of formation of any element in its standard state is zero — carbon as graphite, oxygen as $O_2(g)$, hydrogen as $H_2(g)$, and so on.

When you build a Hess's law cycle using formation data, the elements in their standard states sit at the bottom of the cycle. Arrows point upward from those elements to both the reactants and the products. This gives the cycle its characteristic shape: a triangle with the reaction arrow across the top and two formation arrows pointing upward from the elements.

The logic behind this formula comes directly from the cycle. Route 1 is the direct reaction (top arrow). Route 2 goes down-left from reactants to elements (reversing the formation of reactants, so the sign flips) and then up-right from elements to products (using the formation of products as written, following the arrows as drawn). Because both routes must give the same total enthalpy change, the formula above follows automatically.

Worked example: Formation Cycle: Enthalpy of Reaction for Methanol Combustion

Use the following standard enthalpies of formation to calculate $\Delta H^\circ$ for the reaction: $CH_3OH(l) + \frac{3}{2}O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$ $\Delta_f H^\circ [CH_3OH(l)] = -239 \text{ kJ mol}^{-1}$ $\Delta_f H^\circ [CO_2(g)] = -394 \text{ kJ mol}^{-1}$ $\Delta_f H^\circ [H_2O(l)] = -286 \text{ kJ mol}^{-1}$ $\Delta_f H^\circ [O_2(g)] = 0 \text{ kJ mol}^{-1}$ (element in standard state)

1. Products: $1 \times CO_2(g)$ and $2 \times H_2O(l)$ Reactants: $1 \times CH_3OH(l)$ and $\frac{3}{2} \times O_2(g)$ — Always check stoichiometric coefficients first — they multiply the formation enthalpies.
2. $\sum \Delta_f H^\circ (\text{products}) = (1 \times -394) + (2 \times -286)$ $= -394 + (-572) = -966 \text{ kJ mol}^{-1}$ — Multiply each $\Delta_f H^\circ$ by its stoichiometric coefficient before adding.
3. $\sum \Delta_f H^\circ (\text{reactants}) = (1 \times -239) + (\frac{3}{2} \times 0)$ $= -239 + 0 = -239 \text{ kJ mol}^{-1}$ — $O_2(g)$ is an element in its standard state, so its formation enthalpy is zero — do not forget to include it, but it contributes nothing numerically.
4. $\Delta H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants})$ $= -966 - (-239)$ $= -966 + 239$ $= -727 \text{ kJ mol}^{-1}$ — Subtracting a negative number means adding its magnitude. This is where sign errors most commonly occur — write out the subtraction explicitly.
5. Combustion reactions are exothermic, so $\Delta H^\circ$ should be negative. $-727 \text{ kJ mol}^{-1}$ is negative. [M1] formula applied correctly; [A1] correct final value. — Always check the sign makes physical sense. An exothermic combustion giving a positive value would signal an arithmetic error.

Answer: $\Delta H^\circ = -727 \text{ kJ mol}^{-1}$

3. Enthalpies of Combustion: The Upward Cycle

When you are given combustion data rather than formation data, the cycle is drawn differently. Combustion products — typically $CO_2(g)$ and $H_2O(l)$ — sit at the bottom of the cycle. The reactants and products of your target reaction both connect downward to those combustion products via their combustion enthalpies.

This means the formula is the reverse of the formation version: you subtract products from reactants, not reactants from products. The reason is that in the combustion cycle, the arrows from reactants and products both point down to the combustion products. To travel from reactants to products via the bottom of the cycle, you go down on the left (using $\Delta_c H^\circ$ of reactants) and then back up on the right (reversing $\Delta_c H^\circ$ of products, so the sign flips).

A reliable way to avoid confusing the two formulas is to draw the cycle every time and trace the arrows. If you follow the arrows in the direction they point, you add the value. If you travel against an arrow, you reverse the sign. The formula is just a shortcut for that arrow-tracing process.

Worked example: Combustion Cycle: Enthalpy of Hydrogenation of Ethene

Calculate $\Delta H^\circ$ for: $C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)$ $\Delta_c H^\circ [C_2H_4(g)] = -1411 \text{ kJ mol}^{-1}$ $\Delta_c H^\circ [H_2(g)] = -286 \text{ kJ mol}^{-1}$ $\Delta_c H^\circ [C_2H_6(g)] = -1560 \text{ kJ mol}^{-1}$

1. Reactants: $C_2H_4(g)$ (coefficient 1) and $H_2(g)$ (coefficient 1) Product: $C_2H_6(g)$ (coefficient 1) — Check coefficients carefully — they multiply the combustion enthalpies.
2. $\sum \Delta_c H^\circ (\text{reactants}) = (1 \times -1411) + (1 \times -286)$ $= -1411 + (-286) = -1697 \text{ kJ mol}^{-1}$ — Both reactants are combusted, so both values are included.
3. $\sum \Delta_c H^\circ (\text{products}) = 1 \times -1560 = -1560 \text{ kJ mol}^{-1}$ — Only $C_2H_6$ is a product here, so only one value is needed.
4. $\Delta H^\circ = \sum \Delta_c H^\circ (\text{reactants}) - \sum \Delta_c H^\circ (\text{products})$ $= -1697 - (-1560)$ $= -1697 + 1560$ $= -137 \text{ kJ mol}^{-1}$ — Again, subtracting a negative — write it out explicitly to avoid sign errors. [M1] correct formula; [M1] correct substitution; [A1] correct answer.
5. Hydrogenation of alkenes is exothermic, so a negative value is expected. $-137 \text{ kJ mol}^{-1}$ is consistent with known hydrogenation enthalpies. — A positive answer here would indicate a sign error in step 4 — the most common place to lose marks.

Answer: $\Delta H^\circ = -137 \text{ kJ mol}^{-1}$

4. Drawing the Cycles: A Visual Framework

Both types of cycle follow the same underlying logic, but the position of the 'anchor' species at the bottom changes. For formation cycles, the anchor is the elements in standard states. For combustion cycles, the anchor is the combustion products ($CO_2$ and $H_2O$). Drawing the cycle before applying any formula is strongly recommended — it makes the arrow directions explicit and prevents the most common errors.

Notice that the two formulas are mirror images of each other in terms of which set of values is subtracted. This is not a coincidence — it reflects the fact that the arrows in the two cycles point in opposite directions relative to the target reaction. Keeping the table above in mind, or redrawing the cycle each time, is the most reliable way to avoid mixing them up under exam pressure.

5. Multi-Step Worked Example with Stoichiometric Coefficients

Coefficients greater than one are where many students drop marks. Each formation or combustion enthalpy value is defined per mole of the substance, so if a species appears with a coefficient of 2 in the balanced equation, you must multiply its $\Delta H^\circ$ value by 2 before summing. The following example involves coefficients on both sides and uses formation data.

Worked example: Formation Cycle with Non-Unity Coefficients: Enthalpy of Reaction for Ammonia Synthesis

Calculate $\Delta H^\circ$ for: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ $\Delta_f H^\circ [NH_3(g)] = -46 \text{ kJ mol}^{-1}$ $\Delta_f H^\circ [N_2(g)] = 0 \text{ kJ mol}^{-1}$ $\Delta_f H^\circ [H_2(g)] = 0 \text{ kJ mol}^{-1}$

1. Products: $2 \times NH_3(g)$ Reactants: $1 \times N_2(g)$ and $3 \times H_2(g)$ — Both $N_2$ and $H_2$ are elements in their standard states, so their formation enthalpies are zero.
2. $\sum \Delta_f H^\circ (\text{products}) = 2 \times (-46) = -92 \text{ kJ mol}^{-1}$ — The coefficient of 2 must multiply the formation enthalpy of $NH_3$.
3. $\sum \Delta_f H^\circ (\text{reactants}) = (1 \times 0) + (3 \times 0) = 0 \text{ kJ mol}^{-1}$ — Both reactants are elements in standard states — their formation enthalpies are zero by definition.
4. $\Delta H^\circ = -92 - 0 = -92 \text{ kJ mol}^{-1}$ — [M1] correct identification of zero values for elements; [A1] correct final answer. The negative sign confirms the reaction is exothermic, consistent with the industrial Haber process.

Answer: $\Delta H^\circ = -92 \text{ kJ mol}^{-1}$

6. Exam Technique and Common Errors

The most reliable exam strategy for Hess's law questions is to draw the cycle before writing any numbers. Label each arrow with the relevant $\Delta H^\circ$ value and its sign, then trace the route from reactants to products. If you travel with an arrow, use the value as given. If you travel against an arrow, reverse the sign. This arrow-tracing method works for any cycle, however complex, and removes the need to memorise which formula applies.

AQA mark schemes award method marks for a correctly drawn cycle with arrows in the right directions, even if the final arithmetic is wrong. Drawing the cycle is therefore worth marks in its own right — do not skip it under time pressure.

Hess's law calculations underpin a significant portion of the AQA energetics topic. Mastering the two cycle types — formation and combustion — and applying the correct formula with careful attention to stoichiometric coefficients and sign conventions gives you a robust, transferable method for any enthalpy calculation the exam presents.