Hess's Law and Energy Cycles

AS · 11 min

Many enthalpy changes cannot be measured directly. You cannot make methane by mixing carbon and hydrogen, and you cannot measure the heat of turning white copper(II) sulfate into blue crystals in a beaker. Hess's law gets round this: because energy is conserved, the enthalpy change for a reaction is the same whatever route you take, so you can calculate it from reactions that can be measured. This note shows how to construct energy cycles with enthalpy changes of formation and of combustion, how to combine equations algebraically, and how to design an experiment that uses Hess's law. Hess cycle calculations are worth several marks in almost every Paper 2.

Hess's law

Definition

Hess's law: the total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place, provided the initial and final conditions are the same.

Think of enthalpy like height on a hill. Whether you walk straight up the path or zig-zag, the change in height between the bottom and the top is the same. In the same way, the enthalpy change between reactants and products is fixed by their enthalpies, not by the steps in between.

Hess's law follows from the law of conservation of energy. If two routes between the same reactants and products had different enthalpy changes, you could go one way and come back the other and create energy from nothing.

Building an energy cycle

An energy cycle (or Hess cycle) shows the reaction you want as one side of a triangle, and an indirect route through some common intermediate as the other two sides.

Method

Solving any Hess cycle

Write the balanced equation for the reaction whose $\Delta H$ you want, with state symbols, across the top.
Choose the third corner:
- elements in their standard states, if you are given $\Delta H_f$ values;
- combustion products ( $\ce{CO2}$ and $\ce{H2O}$ ), if you are given $\Delta H_c$ values.
Draw arrows in the direction of the defined process: formation arrows go from the elements; combustion arrows go to the combustion products.
Write the enthalpy change on each arrow, multiplied by the number of moles in the equation.
Apply Hess's law: going round the two routes from the same start to the same finish gives the same total. When you travel against an arrow, change its sign.
Calculate, and include the sign and units.

Cycles using enthalpy changes of formation

A Hess cycle using enthalpy changes of formation. Both formation arrows point up from the elements. Route 1 (elements → reactants → products) equals route 2 (elements → products), so ΔH_r = ΣΔH_f(products) − ΣΔH_f(reactants).

The direct route from elements to products is $\sum \Delta H_f(\text{products})$ . The indirect route goes from elements to reactants and then to products: $\sum \Delta H_f(\text{reactants}) + \Delta H_r$ . Setting them equal:

Key result

\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})

Remember: $\Delta H_f$ of any element in its standard state is zero.

Cycles using enthalpy changes of combustion

A Hess cycle using enthalpy changes of combustion. Both combustion arrows point down to the same combustion products. Route 1 (reactants → combustion products) equals route 2 (reactants → products → combustion products), so ΔH_r = ΣΔH_c(reactants) − ΣΔH_c(products).

Here both reactants and products burn to the same $\ce{CO2}$ and $\ce{H2O}$ . The direct route from reactants to combustion products is $\sum \Delta H_c(\text{reactants})$ ; the indirect route is $\Delta H_r + \sum \Delta H_c(\text{products})$ . So:

Key result

\Delta H_r = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})

Note the order is reversed compared with the formation equation, because the combustion arrows point away from the reaction.

This is the standard way to find enthalpy changes of formation of organic compounds, which can almost never be measured directly but whose combustion is easy to measure.

Tip

Do not memorise the two formulas blindly: draw the cycle every time. Examiners give a mark for a correct cycle (or correct use of the formula), and a cycle protects you from sign errors. The formulas are a check.

Combining equations algebraically

Instead of drawing a triangle, you can add and subtract equations like algebra. Whatever you do to an equation, do the same to its $\Delta H$ :

reverse an equation: change the sign of $\Delta H$ ;
multiply an equation by a number: multiply $\Delta H$ by the same number;
add equations: add their $\Delta H$ values. Species that appear on both sides cancel.

For example, to find $\Delta H$ for $\ce{C(s) + 1/2O2(g) -> CO(g)}$ (which cannot be measured, since some $\ce{CO2}$ always forms):

\begin{aligned} &\ce{C(s) + O2(g) -> CO2(g)} && \Delta H = -393.5\ \text{kJ mol}^{-1} \\ &\ce{CO2(g) -> CO(g) + 1/2O2(g)} && \Delta H = +283.0\ \text{kJ mol}^{-1}\ \text{(reversed combustion of CO)} \end{aligned}

Adding: $\ce{C(s) + 1/2O2(g) -> CO(g)}$ , $\Delta H = -393.5 + 283.0 = -110.5\ \text{kJ mol}^{-1}$ .

Worked examples

Routine: a cycle with enthalpy changes of formation

Calculate the enthalpy change for the reduction of iron(III) oxide by carbon monoxide:

\ce{Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)}

$\Delta H_f$ / $\text{kJ mol}^{-1}$ : $\ce{Fe2O3(s)}$ $-824.2$ ; $\ce{CO(g)}$ $-110.5$ ; $\ce{CO2(g)}$ $-393.5$ .

Solution

$\Delta H_f(\ce{Fe(s)}) = 0$ (element).

\sum \Delta H_f(\text{products}) = 2(0) + 3(-393.5) = -1180.5

\sum \Delta H_f(\text{reactants}) = -824.2 + 3(-110.5) = -1155.7

\Delta H_r = -1180.5 - (-1155.7) = -24.8\ \text{kJ mol}^{-1}

Enthalpy change of formation from combustion data

Calculate the standard enthalpy change of formation of ethanol, $\ce{C2H5OH(l)}$ .

$\Delta H_c$ / $\text{kJ mol}^{-1}$ : $\ce{C(s)}$ $-393.5$ ; $\ce{H2(g)}$ $-285.8$ ; $\ce{C2H5OH(l)}$ $-1367.3$ .

Solution

Target equation: $\ce{2C(s) + 3H2(g) + 1/2O2(g) -> C2H5OH(l)}$ .

Cycle: the reactants ( $\ce{2C + 3H2 + 1/2O2}$ ) burn to $\ce{2CO2 + 3H2O}$ ; ethanol also burns to $\ce{2CO2 + 3H2O}$ . Oxygen has no enthalpy change of combustion.

\sum \Delta H_c(\text{reactants}) = 2(-393.5) + 3(-285.8) = -787.0 - 857.4 = -1644.4

\Delta H_f = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products}) = -1644.4 - (-1367.3) = -277.1\ \text{kJ mol}^{-1}

An enthalpy change that cannot be measured directly

Anhydrous copper(II) sulfate is white; the hydrated crystals, $\ce{CuSO4.5H2O}$ , are blue. The enthalpy change for

\ce{CuSO4(s) + 5H2O(l) -> CuSO4.5H2O(s)}

cannot be measured directly, because it is impossible to control the reaction so that exactly five moles of water are taken up and the product is a pure solid. Instead, both solids are dissolved separately in water:

$\ce{CuSO4(s) + aq -> CuSO4(aq)}$ , $\Delta H_1 = -66.5\ \text{kJ mol}^{-1}$
$\ce{CuSO4.5H2O(s) + aq -> CuSO4(aq)}$ , $\Delta H_2 = +11.7\ \text{kJ mol}^{-1}$

Calculate the enthalpy change of hydration of anhydrous copper(II) sulfate.

Solution

Cycle: from $\ce{CuSO4(s)}$ , either go directly to $\ce{CuSO4(aq)}$ ( $\Delta H_1$ ), or first form $\ce{CuSO4.5H2O(s)}$ ( $\Delta H$ , unknown) and then dissolve it ( $\Delta H_2$ ). Both routes end at the same solution.

\Delta H_1 = \Delta H + \Delta H_2

\Delta H = \Delta H_1 - \Delta H_2 = -66.5 - (+11.7) = -78.2\ \text{kJ mol}^{-1}

Using two cycles in one question

$\Delta H_c$ / $\text{kJ mol}^{-1}$ : $\ce{C(s)}$ $-393.5$ ; $\ce{H2(g)}$ $-285.8$ ; $\ce{C2H4(g)}$ $-1411.0$ ; $\ce{C2H6(g)}$ $-1559.7$ .

(a) Calculate $\Delta H_f$ of ethene. (b) Calculate $\Delta H$ for the hydrogenation of ethene, $\ce{C2H4(g) + H2(g) -> C2H6(g)}$ .

Solution

(a) $\ce{2C(s) + 2H2(g) -> C2H4(g)}$ :

\Delta H_f = 2(-393.5) + 2(-285.8) - (-1411.0) = -787.0 - 571.6 + 1411.0 = +52.4\ \text{kJ mol}^{-1}

Ethene has a positive enthalpy of formation: it is less stable than its elements.

(b) Reactants $\ce{C2H4 + H2}$ and product $\ce{C2H6}$ all burn to $\ce{2CO2 + 3H2O}$ :

\Delta H = [-1411.0 + (-285.8)] - (-1559.7) = -1696.8 + 1559.7 = -137.1\ \text{kJ mol}^{-1}

Exam-hard: a cycle with several compounds

The first step in making nitric acid is the oxidation of ammonia:

\ce{4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)}

$\Delta H_f$ / $\text{kJ mol}^{-1}$ : $\ce{NH3(g)}$ $-46.1$ ; $\ce{NO(g)}$ $+90.3$ ; $\ce{H2O(g)}$ $-241.8$ .

(a) Calculate $\Delta H_r$ . (b) The enthalpy change of vaporisation of water is $+44.0\ \text{kJ mol}^{-1}$ . Calculate $\Delta H_r$ if the water were formed as a liquid. (c) Explain why the answer to (a) is not the standard enthalpy change of reaction.

Solution

(a)

\sum \Delta H_f(\text{products}) = 4(+90.3) + 6(-241.8) = 361.2 - 1450.8 = -1089.6

\sum \Delta H_f(\text{reactants}) = 4(-46.1) + 5(0) = -184.4

\Delta H_r = -1089.6 - (-184.4) = -905.2\ \text{kJ mol}^{-1}

(b) Condensing water releases $44.0\ \text{kJ mol}^{-1}$ : $\ce{H2O(g) -> H2O(l)}$ , $\Delta H = -44.0$ . For six moles: $6 \times (-44.0) = -264.0$ .

\Delta H_r = -905.2 - 264.0 = -1169.2\ \text{kJ mol}^{-1}

(c) Under standard conditions ( $298\ \text{K}$ ), water's standard state is liquid. The value in (a) has water as a gas, so it is not the standard enthalpy change; (b) is the value with all substances in their standard states.

Using Hess's law experimentally: decomposition of calcium carbonate

The decomposition $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ needs a very high temperature, so its enthalpy change cannot be measured directly in a calorimeter. Instead:

React a weighed sample of calcium carbonate (about $2.5\ \text{g}$ ) with excess hydrochloric acid ( $50\ \text{cm}^3$ of $2\ \text{mol dm}^{-3}$ ) in a polystyrene cup, measuring $\Delta T$ . This gives $\Delta H_1$ for $\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)}$ .
Repeat with a weighed sample of calcium oxide (about $1.4\ \text{g}$ ) to find $\Delta H_2$ for $\ce{CaO(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l)}$ .
The two reactions end with the same solution, so $\Delta H_{\text{decomposition}} = \Delta H_1 - \Delta H_2$ .

Points examiners ask about:

Acid must be in excess so all the solid reacts; the moles come from the mass of solid.
In step 1, some of the energy change is lost as $\ce{CO2}$ escapes, and spray may be lost; add the solid in small portions and use a loosely fitting lid.
$\Delta T$ in step 1 is small, so its percentage uncertainty is large; a thermometer reading to $0.1\ ^\circ\text{C}$ improves this.
Calcium oxide absorbs water and $\ce{CO2}$ from the air, so use freshly heated oxide and weigh it quickly.

Watch out

Forgetting the multipliers. If the equation has $\ce{3CO2}$ , use $3 \times \Delta H_f(\ce{CO2})$ .
Products minus reactants for combustion data. With $\Delta H_c$ values it is reactants minus products. Draw the cycle to avoid this.
Giving elements a non-zero ΔHf, or giving oxygen a $\Delta H_c$ . Both are zero (oxygen does not burn; water does not burn either).
Ignoring state symbols. $\Delta H_f$ of $\ce{H2O(l)}$ ( $-285.8$ ) differs from $\ce{H2O(g)}$ ( $-241.8$ ).

Exam tip

Questions often say "construct an energy cycle" or "use a Hess cycle". Draw a labelled triangle with the target equation along the top and arrows in the right directions; this can earn a mark even if the arithmetic goes wrong.
Show each sum separately (products, reactants), then the subtraction. Keep the sign in front of every number.
Give the final answer with sign and units ( $\text{kJ mol}^{-1}$ ), to the precision of the data.
When asked why an enthalpy change "cannot be measured directly", name a specific reason: the reaction does not happen (carbon and hydrogen do not form methane), other products form ( $\ce{CO2}$ with $\ce{CO}$ ), or it is too slow or needs very high temperature.

Summary

Hess's law: the total enthalpy change is independent of the route, provided the initial and final conditions are the same (conservation of energy).
With $\Delta H_f$ : $\Delta H_r = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$ ; elements have $\Delta H_f = 0$ .
With $\Delta H_c$ : $\Delta H_r = \sum \Delta H_c(\text{reactants}) - \sum \Delta H_c(\text{products})$ .
Reversing an equation changes the sign of $\Delta H$ ; multiplying multiplies $\Delta H$ .
Used to find enthalpy changes that cannot be measured: formation of organic compounds, hydration of salts, thermal decomposition.
Always draw the cycle and use the multipliers from the equation.

Practice

Question

State Hess's law.
Explain why the enthalpy change of formation of methane cannot be measured directly, and how it can be found.
Calculate $\Delta H_r$ for $\ce{2SO2(g) + O2(g) -> 2SO3(g)}$ . $\Delta H_f$ : $\ce{SO2(g)}$ $-296.8$ , $\ce{SO3(g)}$ $-395.7\ \text{kJ mol}^{-1}$ .
Calculate $\Delta H_r$ for $\ce{NH3(g) + HCl(g) -> NH4Cl(s)}$ . $\Delta H_f$ : $\ce{NH3(g)}$ $-46.1$ , $\ce{HCl(g)}$ $-92.3$ , $\ce{NH4Cl(s)}$ $-314.4\ \text{kJ mol}^{-1}$ .
Calculate $\Delta H_f$ of propane, $\ce{C3H8(g)}$ . $\Delta H_c$ : $\ce{C(s)}$ $-393.5$ , $\ce{H2(g)}$ $-285.8$ , $\ce{C3H8(g)}$ $-2219.2\ \text{kJ mol}^{-1}$ .
Calculate $\Delta H_f$ of glucose, $\ce{C6H12O6(s)}$ , given $\Delta H_c(\ce{C6H12O6}) = -2803\ \text{kJ mol}^{-1}$ and the values in question 5.
In the calcium carbonate experiment, a student found $\Delta H_1 = -17\ \text{kJ mol}^{-1}$ (for $\ce{CaCO3}$ with acid) and $\Delta H_2 = -195\ \text{kJ mol}^{-1}$ (for $\ce{CaO}$ with acid). Calculate the enthalpy change of decomposition of calcium carbonate and state whether it is exothermic or endothermic.
Use the equations below to calculate $\Delta H$ for $\ce{C(s) + 1/2O2(g) -> CO(g)}$ : $\ce{C(s) + O2(g) -> CO2(g)}$ , $\Delta H = -393.5\ \text{kJ mol}^{-1}$ ; $\ce{CO(g) + 1/2O2(g) -> CO2(g)}$ , $\Delta H = -283.0\ \text{kJ mol}^{-1}$ .
The thermite reaction is $\ce{2Al(s) + Fe2O3(s) -> Al2O3(s) + 2Fe(s)}$ . $\Delta H_f$ : $\ce{Al2O3(s)}$ $-1675.7$ , $\ce{Fe2O3(s)}$ $-824.2\ \text{kJ mol}^{-1}$ . Calculate $\Delta H_r$ and the energy released per gram of reaction mixture ( $A_r$ : Al 27.0, Fe 55.8, O 16.0).
Ethanol is made industrially by $\ce{C2H4(g) + H2O(g) -> C2H5OH(l)}$ . Use $\Delta H_c(\ce{C2H4}) = -1411.0$ and $\Delta H_c(\ce{C2H5OH}) = -1367.3\ \text{kJ mol}^{-1}$ , together with $\Delta H_{\text{vap}}(\ce{H2O}) = +44.0\ \text{kJ mol}^{-1}$ , to calculate $\Delta H$ for this reaction. Explain why water does not appear with a combustion value in your cycle.

Answers

The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.
Carbon and hydrogen do not react together to form methane under normal conditions (and any reaction would form a mixture of products), so the heat change cannot be measured. Instead, measure the enthalpy changes of combustion of carbon, hydrogen and methane, and use a Hess cycle: $\Delta H_f(\ce{CH4}) = \Delta H_c(\ce{C}) + 2\Delta H_c(\ce{H2}) - \Delta H_c(\ce{CH4})$ .
$\Delta H_r = 2(-395.7) - [2(-296.8) + 0] = -791.4 + 593.6 = -197.8\ \text{kJ mol}^{-1}$ .
$\Delta H_r = -314.4 - [(-46.1) + (-92.3)] = -314.4 + 138.4 = -176.0\ \text{kJ mol}^{-1}$ .
$\ce{3C(s) + 4H2(g) -> C3H8(g)}$ . $\Delta H_f = 3(-393.5) + 4(-285.8) - (-2219.2) = -1180.5 - 1143.2 + 2219.2 = -104.5\ \text{kJ mol}^{-1}$ .
$\ce{6C(s) + 6H2(g) + 3O2(g) -> C6H12O6(s)}$ . $\Delta H_f = 6(-393.5) + 6(-285.8) - (-2803) = -2361.0 - 1714.8 + 2803 = -1272.8 \approx -1273\ \text{kJ mol}^{-1}$ .
$\Delta H = \Delta H_1 - \Delta H_2 = -17 - (-195) = +178\ \text{kJ mol}^{-1}$ : endothermic.
Reverse the second equation ( $+283.0$ ) and add it to the first: $\Delta H = -393.5 + 283.0 = -110.5\ \text{kJ mol}^{-1}$ .
$\Delta H_r = -1675.7 - (-824.2) = -851.5\ \text{kJ mol}^{-1}$ (Al and Fe are elements). Mass of mixture $= 2(27.0) + [2(55.8) + 3(16.0)] = 54.0 + 159.6 = 213.6\ \text{g}$ . Energy per gram $= \dfrac{851.5}{213.6} = 3.99\ \text{kJ g}^{-1}$ .
First find $\Delta H$ for $\ce{C2H4(g) + H2O(l) -> C2H5OH(l)}$ using combustion data. Water is already fully oxidised: it is itself a combustion product, so it does not burn and contributes nothing. $\Delta H = \Delta H_c(\ce{C2H4}) - \Delta H_c(\ce{C2H5OH}) = -1411.0 - (-1367.3) = -43.7\ \text{kJ mol}^{-1}$ . With steam instead: add $\ce{H2O(g) -> H2O(l)}$ , $\Delta H = -44.0$ . Total $= -43.7 + (-44.0) = -87.7\ \text{kJ mol}^{-1}$ .

Hess's law

Building an energy cycle

Cycles using enthalpy changes of formation

Cycles using enthalpy changes of combustion

Combining equations algebraically

Worked examples

Practice

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