Hess’s law

1. Overview

Hess's law says that the enthalpy change of a reaction does not depend on the route taken from reactants to products. It lets you build energy cycles that link a reaction to data you already have: enthalpy changes of formation, enthalpy changes of combustion or bond energies. With a cycle you can find enthalpy changes that no experiment can measure directly, such as the formation of an organic compound from its elements. This section builds on the definitions, calorimetry and bond-energy method of topic 5.1.

Key Definitions

• Hess's law: the total enthalpy change for a reaction is the same whichever route is taken from the reactants to the products, provided the initial and final conditions are the same.
• Energy cycle (Hess cycle): a diagram that links the same reactants and products by two different routes, so that one unknown enthalpy change can be calculated from the others.
• Standard enthalpy change of reaction (ΔHr): the enthalpy change when the amounts of reactants shown in a balanced equation react completely under standard conditions, with all substances in their standard states.
• Standard enthalpy change of formation (ΔHf): the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.
• Standard enthalpy change of combustion (ΔHc): the enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions, with all reactants and products in their standard states.
• Bond energy: the energy required to break one mole of a particular covalent bond in the gaseous state; it is always positive.

Content

Hess's law

The enthalpy of a set of substances depends only on what they are, how much there is, their physical states, the temperature and the pressure. It does not depend on how they got there. So if reactants can become products by two different routes, $\Delta H$ is the same for both.

This follows from the conservation of energy. Suppose route 1 released 10 kJ more than route 2. You could go forward by route 1 and back by route 2, end with the same chemicals you started with, and have 10 kJ left over. That would be energy made from nothing, which is impossible.

The routes must start and finish in the same conditions: the same substances, the same physical states, the same temperature and pressure.

Constructing an energy cycle

1. Write the reaction you want across the top, balanced and with state symbols. Its arrow is $\Delta H_{\text{r}}$.
2. Choose the third corner to match the data you are given (table below).
3. Draw each arrow in the direction its definition describes, and label it with its symbol or value, multiplied by the number of moles in the equation.
4. Check that every box contains the same atoms. In a combustion cycle, add the oxygen needed for burning to both top boxes.
5. Follow the indirect route: add a value when you travel along its arrow, subtract it when you travel against its arrow.

Data given	Third corner	Direction of the arrows
$\Delta H_{\text{f}}$	the elements in their standard states	from the elements to each side
$\Delta H_{\text{c}}$	the combustion products, such as CO₂(g) and H₂O(l)	from each side to the combustion products
bond energies	gaseous atoms	from reactants to atoms (breaking), from atoms to products (making)

Tables of data give standard values, such as $\Delta H_{\text{f}}^\ominus$. The cycles on this page leave out the $\ominus$ sign to keep them short.

Formation cycle. Formation always starts from the elements, so both arrows point away from the bottom corner.

$$\begin{array}{ccc} \text{reactants} & \xrightarrow{\quad\Delta H_{\text{r}}\quad} & \text{products} \\[4pt] {\scriptstyle\Delta H_1}\;{\Large\nwarrow} & & {\Large\nearrow}\;{\scriptstyle\Delta H_2} \\[4pt] & \text{elements} & \end{array}$$

Going from reactants to products the indirect way, you travel against $\Delta H_1$ and along $\Delta H_2$, so $\Delta H_{\text{r}} = -\Delta H_1 + \Delta H_2$.

Combustion cycle. Burning ends at the combustion products, which are CO₂(g) and H₂O(l) for a compound of carbon, hydrogen and oxygen, so both arrows point towards the bottom corner.

$$\begin{array}{ccc} \text{reactants} & \xrightarrow{\quad\Delta H_{\text{r}}\quad} & \text{products} \\[4pt] {\scriptstyle\Delta H_1}\;{\Large\searrow} & & {\Large\swarrow}\;{\scriptstyle\Delta H_2} \\[4pt] & \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)} & \end{array}$$

This time you travel along $\Delta H_1$ and against $\Delta H_2$, so $\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2$.

Enthalpy level diagrams. The same cycle can be drawn with enthalpy on a vertical axis. Each box becomes a level, and each arrow points down for an exothermic change and up for an endothermic one. The diagram below shows the type of cycle used in Worked example 2.

Shortcut equations. Summing the arrows in each cycle gives two general results:

$$\begin{aligned} \Delta H_{\text{r}} = &\sum \Delta H_{\text{f}}(\text{products}) \\ &- \sum \Delta H_{\text{f}}(\text{reactants}) \end{aligned}$$

$$\begin{aligned} \Delta H_{\text{r}} = &\sum \Delta H_{\text{c}}(\text{reactants}) \\ &- \sum \Delta H_{\text{c}}(\text{products}) \end{aligned}$$

The order is opposite because the arrows point the opposite way. If you are unsure which applies, sketch the cycle: it tells you the order.

Three facts save mistakes:

• $\Delta H_{\text{f}}$ of an element in its standard state, such as S(s) or O₂(g), is zero.
• Some values describe the same equation. C(s) + O₂(g) → CO₂(g) is both the combustion of carbon and the formation of CO₂, so $\Delta H_{\text{c}}$(C) $= \Delta H_{\text{f}}$(CO₂), and a question may give you either one. In the same way, $\Delta H_{\text{c}}$(H₂) $= \Delta H_{\text{f}}$(H₂O(l)): the same value, −285.8 kJ mol⁻¹, is given for H₂O(l) in Worked example 1 and for H₂(g) in Worked example 2.
• The oxygen added to a combustion cycle appears on both routes. It is not burned, so it has no $\Delta H_{\text{c}}$ to add.

Adding equations instead of drawing a cycle. You can also combine equations directly:

• Reverse an equation: change the sign of its $\Delta H$.
• Multiply an equation by a number: multiply its $\Delta H$ by the same number.
• Add equations: add their $\Delta H$ values. Cancel a formula only when it appears on both sides in the same state.

Enthalpy changes that cannot be measured directly

Calorimetry (topic 5.1) needs a reaction that goes quickly and cleanly to the products in the equation, at a temperature where you can measure the temperature change of the surroundings. Many reactions fail this test:

• The reaction does not happen as written. Carbon, hydrogen and oxygen cannot be made to combine directly into ethyl ethanoate, so its $\Delta H_{\text{f}}$ must come from a cycle.
• Other products form as well. Carbon burned in a limited supply of oxygen gives a mixture of CO and CO₂, so the heat released is not for forming CO alone.
• The reaction needs heating. Zinc carbonate decomposes only while it is being heated. The flame supplies an unknown amount of energy, so no temperature change of water can tell you $\Delta H$.
• The reaction is too slow. Rusting releases its energy over weeks, too slowly to see a temperature rise.

In each case, you find two or more enthalpy changes you can measure or look up, and join them with a cycle.

Worked example 1: enthalpy change of reaction from formation data

Question: Hydrogen sulfide reacts with sulfur dioxide to form sulfur and water. Calculate the enthalpy change for the equation shown.

2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l)

Substance	$\Delta H_{\text{f}}^\ominus$ / kJ mol⁻¹
H₂S(g)	−20.6
SO₂(g)	−296.8
H₂O(l)	−285.8
H₂O(g)	−241.8

Step 1: draw the cycle. The elements in the bottom corner are 3S(s) + 2H₂(g) + O₂(g): 3 S, 4 H and 2 O, the same atoms as in each top box.

$$\begin{array}{ccc} 2\text{H}_2\text{S(g)} & \xrightarrow{\;\Delta H_{\text{r}}\;} & 3\text{S(s)} \\ +\;\text{SO}_2\text{(g)} & & +\;2\text{H}_2\text{O(l)} \\[4pt] {\scriptstyle\Delta H_1}\;{\Large\nwarrow} & & {\Large\nearrow}\;{\scriptstyle\Delta H_2} \\[4pt] & 3\text{S(s)} + 2\text{H}_2\text{(g)} & \\ & +\;\text{O}_2\text{(g)} & \end{array}$$

Step 2: value of each arrow. Sulfur is an element in its standard state, so its $\Delta H_{\text{f}}$ is zero. The equation has liquid water, so use the value for H₂O(l). $$\begin{aligned} \Delta H_1 &= 2(-20.6) + (-296.8) \\ &= -338.0\ \text{kJ mol}^{-1} \end{aligned}$$ $$\begin{aligned} \Delta H_2 &= 3(0) + 2(-285.8) \\ &= -571.6\ \text{kJ mol}^{-1} \end{aligned}$$

Step 3: go round the cycle. Against $\Delta H_1$, then along $\Delta H_2$: $$\begin{aligned} \Delta H_{\text{r}} &= -(-338.0) + (-571.6) \\ &= -233.6\ \text{kJ mol}^{-1} \end{aligned}$$

Answer: $\Delta H_{\text{r}} = -233.6$ kJ mol⁻¹ for the equation as written. The data are given to 0.1 kJ mol⁻¹, so the answer is too.

States matter. With the table's value for steam, −241.8 kJ mol⁻¹, the answer would be −145.6 kJ mol⁻¹. The 88.0 kJ difference is the energy released when 2 mol of steam condenses to liquid water.

Worked example 2: enthalpy change of formation from combustion data

Question: Use the data to calculate the standard enthalpy change of formation of ethyl ethanoate, CH₃COOCH₂CH₃(l).

4C(s) + 4H₂(g) + O₂(g) → CH₃COOCH₂CH₃(l)

Substance	$\Delta H_{\text{c}}^\ominus$ / kJ mol⁻¹
C(s)	−393.5
H₂(g)	−285.8
CH₃COOCH₂CH₃(l)	−2238

Step 1: draw the cycle. Burning the ester, CH₃COOCH₂CH₃(l) + 5O₂(g) → 4CO₂(g) + 4H₂O(l), needs 5O₂, so add 5O₂(g) to both top boxes. The left box then holds 6O₂ in total (1 from the formation equation and 5 added): 4 to burn the carbon and 2 to burn the hydrogen. The cycle shows the ester by its molecular formula, C₄H₈O₂, which makes the atoms easy to count.

$$\begin{array}{ccc} 4\text{C(s)} & \xrightarrow{\;\Delta H_{\text{f}}\;} & \text{C}_4\text{H}_8\text{O}_2\text{(l)} \\ +\;4\text{H}_2\text{(g)} & & +\;5\text{O}_2\text{(g)} \\ +\;6\text{O}_2\text{(g)} & & \\[4pt] {\scriptstyle\Delta H_1}\;{\Large\searrow} & & {\Large\swarrow}\;{\scriptstyle\Delta H_2} \\[4pt] & 4\text{CO}_2\text{(g)} & \\ & +\;4\text{H}_2\text{O(l)} & \end{array}$$

Step 2: value of each arrow. $$\begin{aligned} \Delta H_1 &= 4(-393.5) + 4(-285.8) \\ &= -2717.2\ \text{kJ mol}^{-1} \end{aligned}$$ $$\Delta H_2 = \Delta H_{\text{c}}(\text{ester}) = -2238\ \text{kJ mol}^{-1}$$

Step 3: go round the cycle. Along $\Delta H_1$, then against $\Delta H_2$: $$\begin{aligned} \Delta H_{\text{f}} &= \Delta H_1 - \Delta H_2 \\ &= -2717.2 - (-2238) \\ &= -479.2\ \text{kJ mol}^{-1} \end{aligned}$$

Answer: $\Delta H_{\text{f}}$ of ethyl ethanoate $= -479$ kJ mol⁻¹ (3 s.f., since the ester's $\Delta H_{\text{c}}$ is given to the nearest kJ).

Finding an enthalpy change by experiment and a cycle

When a reaction cannot be measured directly, you can often measure two other reactions that can, then join them with a cycle. A common plan is to add each of two solids to the same dilute acid in a polystyrene cup, measure each temperature change, and work out each $\Delta H$ with $q = mc\Delta T$ and $\Delta H = -mc\Delta T/n$, as in topic 5.1.

• Use the same acid, in excess, for both reactions, so that both end at exactly the same products: the shared corner of the cycle.
• Turn each result into kJ per mole of the solid before you use the cycle.
• Both reactions must be fast at room temperature: use powders and stir.

Worked example 3: decomposition of zinc carbonate

Question: Zinc carbonate decomposes only while it is being heated, so the enthalpy change of ZnCO₃(s) → ZnO(s) + CO₂(g) cannot be measured directly. A student reacts each solid separately with excess dilute hydrochloric acid and obtains these results:

Reaction	$\Delta H$ / kJ mol⁻¹
ZnCO₃(s) + 2HCl(aq) → ZnCl₂(aq) + H₂O(l) + CO₂(g)	$\Delta H_1 = -20.4$
ZnO(s) + 2HCl(aq) → ZnCl₂(aq) + H₂O(l)	$\Delta H_2 = -89.2$

Calculate the enthalpy change of decomposition, $\Delta H_{\text{r}}$.

Step 1: draw the cycle. Add 2HCl(aq) to both top boxes. The shared corner is what both reactions make.

$$\begin{array}{ccc} \text{ZnCO}_3\text{(s)} & \xrightarrow{\;\Delta H_{\text{r}}\;} & \text{ZnO(s)} \\ +\;2\text{HCl(aq)} & & +\;\text{CO}_2\text{(g)} \\ & & +\;2\text{HCl(aq)} \\[4pt] {\scriptstyle\Delta H_1}\;{\Large\searrow} & & {\Large\swarrow}\;{\scriptstyle\Delta H_2} \\[4pt] & \text{ZnCl}_2\text{(aq)} & \\ & +\;\text{H}_2\text{O(l)} & \\ & +\;\text{CO}_2\text{(g)} & \end{array}$$

The CO₂ in the right-hand top box takes no part in the ZnO reaction; it simply carries through to the corner.

Step 2: go round the cycle. Along $\Delta H_1$, then against $\Delta H_2$: $$\begin{aligned} \Delta H_{\text{r}} &= \Delta H_1 - \Delta H_2 \\ &= -20.4 - (-89.2) \\ &= +68.8\ \text{kJ mol}^{-1} \end{aligned}$$

Check by adding equations. Keep equation 1 and reverse equation 2, which changes its sign to +89.2 kJ mol⁻¹. On adding, ZnCl₂(aq), H₂O(l) and 2HCl(aq) appear on both sides and cancel, leaving ZnCO₃(s) → ZnO(s) + CO₂(g) with $\Delta H = -20.4 + 89.2 = +68.8$ kJ mol⁻¹.

Answer: $\Delta H_{\text{r}} = +68.8$ kJ mol⁻¹. The positive sign shows that the decomposition is endothermic.

Bond energies in energy cycles

The bond-energy rule from topic 5.1 is itself a Hess cycle. The indirect route breaks every bond in the gaseous reactants to make gaseous atoms, then makes every bond in the gaseous products. Here $E$ is a bond energy in kJ mol⁻¹.

$$\begin{array}{ccc} \text{reactants (g)} & \xrightarrow{\;\Delta H_{\text{r}}\;} & \text{products (g)} \\[4pt] {\scriptstyle\Delta H_1}\;{\Large\searrow} & & {\Large\nearrow}\;{\scriptstyle\Delta H_2} \\[4pt] & \text{gaseous} & \\ & \text{atoms} & \end{array}$$

Breaking bonds takes in energy and making bonds releases it, so:

$$\begin{aligned} \Delta H_1 &= +\textstyle\sum E(\text{bonds broken}) \\ \Delta H_2 &= -\textstyle\sum E(\text{bonds formed}) \end{aligned}$$

Going along both arrows, $\Delta H_{\text{r}} = \Delta H_1 + \Delta H_2$, which is the topic 5.1 rule.

Seeing it as a cycle lets you use bond energies in more ways than in topic 5.1:

• Find an unknown bond energy. If $\Delta H$ for a reaction is known, the cycle has only one unknown, so you can solve for it (Worked example 4).
• Combine with formation data. For $\Delta H_{\text{f}}$ the reactants are elements, so you also break the bonds in molecules such as H₂, N₂, O₂ and Cl₂. Count them for the amounts in the formation equation: one mole of NH₃(g) needs only ½ mol of N≡N broken.
• Add a step for any state that is not gas. Bond energies are defined for gases. If a substance in the cycle is a liquid or a solid, the cycle needs one more arrow for turning it into a gas. That value is not in the Data Booklet, so the question must give it to you or give data to work it out (see the end of Worked example 4).

Worked example 4: a bond energy from an enthalpy change of formation

Question: Hydrazine has the structure H₂N-NH₂. Its standard enthalpy change of formation is +95.4 kJ mol⁻¹ for N₂H₄(g) and +50.6 kJ mol⁻¹ for N₂H₄(l). Use the appropriate value and the bond energies N≡N 944, H-H 436 and N-H 390 kJ mol⁻¹ to calculate the N-N bond energy in hydrazine.

Step 1: draw the cycle through gaseous atoms. Bond energies are for gases, so use the value for N₂H₄(g), +95.4 kJ mol⁻¹.

$$\begin{array}{ccc} \text{N}_2\text{(g)} + 2\text{H}_2\text{(g)} & \xrightarrow{\;\Delta H_{\text{f}}\;} & \text{N}_2\text{H}_4\text{(g)} \\[4pt] {\scriptstyle\Delta H_1}\;{\Large\searrow} & & {\Large\nearrow}\;{\scriptstyle\Delta H_2} \\[4pt] & 2\text{N(g)} & \\ & +\;4\text{H(g)} & \end{array}$$

Step 2: value of each arrow. Breaking one N≡N and two H-H bonds: $$\Delta H_1 = 944 + 2(436) = +1816\ \text{kJ mol}^{-1}$$ Making one N-N bond (energy $E$) and four N-H bonds releases energy, so the arrow's value is negative: $$\Delta H_2 = -(E + 4 \times 390) = -(E + 1560)$$

Step 3: go round the cycle and solve. $$\begin{aligned} \Delta H_{\text{f}} &= \Delta H_1 + \Delta H_2 \\ +95.4 &= 1816 - E - 1560 \end{aligned}$$ $$E = 1816 - 1560 - 95.4 = 160.6\ \text{kJ mol}^{-1}$$

Answer: the N-N bond energy in hydrazine is 161 kJ mol⁻¹ (3 s.f.). This is close to the Data Booklet's average value for N-N, 160 kJ mol⁻¹. It is only an estimate, because the N-H value used is itself an average.

The state matters here too. The two values given differ by 95.4 − 50.6 = 44.8 kJ mol⁻¹, the energy taken in when 1 mol of liquid hydrazine turns into gas. Putting the liquid's value into the gas cycle would give $E = 205.4$ kJ mol⁻¹, far from the 161 kJ mol⁻¹ found above.

Key Equations

Hess's law around a cycle (kJ mol⁻¹): the direct change equals the indirect route, adding each step travelled along its arrow and subtracting each step travelled against it. For example, for a formation cycle: $$\Delta H_{\text{r}} = -\Delta H_1 + \Delta H_2$$

From enthalpy changes of formation (each $\Delta H_{\text{f}}$ in kJ mol⁻¹, times the moles in the equation; zero for elements in their standard states): $$\begin{aligned} \Delta H_{\text{r}} = &\sum \Delta H_{\text{f}}(\text{products}) \\ &- \sum \Delta H_{\text{f}}(\text{reactants}) \end{aligned}$$

From enthalpy changes of combustion (each $\Delta H_{\text{c}}$ in kJ mol⁻¹, times the moles in the equation): $$\begin{aligned} \Delta H_{\text{r}} = &\sum \Delta H_{\text{c}}(\text{reactants}) \\ &- \sum \Delta H_{\text{c}}(\text{products}) \end{aligned}$$

From bond energies ($E$ = bond energy in kJ mol⁻¹, gases, always positive): $$\begin{aligned} \Delta H_{\text{r}} = &\sum E(\text{bonds broken}) \\ &- \sum E(\text{bonds formed}) \end{aligned}$$

Changing an equation (kJ mol⁻¹): reversing it changes the sign, and multiplying it by $n$ multiplies $\Delta H$ by $n$. $$\Delta H(\text{reverse}) = -\Delta H(\text{forward})$$

Common Mistakes to Avoid

• Mistake: adding a value when you travel against its arrow. Fix: going against an arrow reverses that change, so change its sign. In Worked example 3, $\Delta H_{\text{r}} = -20.4 - (-89.2) = +68.8$ kJ mol⁻¹; adding the two values instead gives −109.6 kJ mol⁻¹.
• Mistake: ignoring the numbers in front of formulae. Fix: multiply each $\Delta H_{\text{f}}$ or $\Delta H_{\text{c}}$ by its number of moles. In Worked example 1, 2H₂O(l) contributes 2 × (−285.8) = −571.6; using −285.8 once gives +52.2 kJ mol⁻¹, which even has the wrong sign.
• Mistake: giving an element its own enthalpy change of formation. Fix: $\Delta H_{\text{f}}$ is zero for an element in its standard state. The 3S(s) in Worked example 1 contributes 3 × 0 = 0.
• Mistake: using the formation shortcut with combustion data. Fix: formation data give products minus reactants; combustion data give reactants minus products. In Worked example 2, −2717.2 − (−2238) = −479.2 kJ mol⁻¹; the other way round gives +479.2.
• Mistake: using data for the wrong state. Fix: match each value to the state symbol in the equation. With the value for H₂O(g) (−241.8) in place of H₂O(l), Worked example 1 gives −145.6 instead of −233.6 kJ mol⁻¹.
• Mistake: cycle arrows unlabelled or pointing the wrong way. Fix: formation arrows start at the elements; combustion arrows end at the combustion products; label every arrow. In Worked example 2 both arrows point down to 4CO₂(g) + 4H₂O(l).
• Mistake: breaking a whole mole of N≡N for one mole of NH₃. Fix: count bonds for the amounts in the equation. For ½N₂(g) + 1½H₂(g) → NH₃(g), the bonds broken are ½ × 944 = 472 kJ and 1½ × 436 = 654 kJ.
• Mistake: combining experimental results before turning them into molar values. Fix: find each $\Delta H$ per mole first. If 0.0200 mol of zinc carbonate releases 408 J, then $\Delta H_1 = -408 \div 0.0200 = -20\,400$ J mol⁻¹ $= -20.4$ kJ mol⁻¹; only then use the cycle.

Exam Tips

• Construct or draw a labelled energy cycle: put a balanced formula with state symbols in every box, add any extra reactant (O₂, an acid) to both top boxes, point every arrow as its definition says, and label it with its symbol or value.
• Calculate: show the cycle, or the sum, before the arithmetic. Write each value in brackets with its sign and multiplier, for example 2(−285.8) − [2(−20.6) + (−296.8)], then give the answer with its sign, kJ mol⁻¹ and sensible significant figures.
• Deduce an expression: when a question names the steps $\Delta H_1$, $\Delta H_2$ and so on, write the relationship in those symbols first, for example $\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2$, then substitute numbers if asked.
• Use the data given: take values from the question; go to the Data Booklet only for values the question does not supply, such as bond energies.
• Explain why an enthalpy change cannot be measured directly with the reason for that reaction: it does not happen as written, it gives other products too, it needs continuous heating, or it is too slow.
• Suggest why a bond-energy answer differs from a data value: bond energies are averages from many compounds, and they are for gases, while the data value may be for a liquid or a solid.