Enthalpy Change of a Reaction — Standard Enthalpy and Hess's Law

The Enthalpy Change of a Reaction

Every substance carries a certain enthalpy. When a reaction runs, one set of substances disappears and another appears, and the enthalpy of the contents of the flask changes. That change is what a chemist measures as heat at constant pressure.

Key Point (Definition): The reaction enthalpy $\Delta_r H$ is the enthalpy of the products minus the enthalpy of the reactants, each substance counted with its stoichiometric coefficient:

$$\Delta_r H = \sum_i a_i H_{\mathrm{products}} - \sum_i b_i H_{\mathrm{reactants}}$$

Here $a_i$ and $b_i$ are the coefficients of the products and the reactants in the balanced equation, and $H$ is the molar enthalpy of each species. The subscript $r$ marks it as a reaction quantity.

For the combustion of methane,

$$\mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)}$$

$$\Delta_r H = \left[ H_m(\mathrm{CO_2},g) + 2H_m(\mathrm{H_2O},l) \right] - \left[ H_m(\mathrm{CH_4},g) + 2H_m(\mathrm{O_2},g) \right]$$

where $H_m$ is the molar enthalpy of each species. The coefficient $2$ appears twice because two moles of oxygen are consumed and two moles of water are made.

The sign carries the physics

If the products sit lower in enthalpy than the reactants, $\Delta_r H$ is negative and the flask gives out heat: the reaction is exothermic. If the products sit higher, $\Delta_r H$ is positive and heat must be supplied: the reaction is endothermic.

Methane burning releases $890.3\,\mathrm{kJ}$ for every mole of methane, so $\Delta_r H = -890.3\,\mathrm{kJ\,mol^{-1}}$. Calcium carbonate decomposing absorbs $178.3\,\mathrm{kJ}$ per mole, so for that change $\Delta_r H = +178.3\,\mathrm{kJ\,mol^{-1}}$.

What "per mole" means here

The unit $\mathrm{kJ\,mol^{-1}}$ on a reaction enthalpy is not per mole of any one substance. It is per mole of reaction as written — one full turn of the balanced equation, consuming $a$ moles of this and $b$ moles of that.

Balance the same chemistry a different way and the number changes:

$$\mathrm{Fe_2O_3(s)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{Fe(s)} + 3\,\mathrm{H_2O(l)}; \qquad \Delta_r H = -33.2\,\mathrm{kJ\,mol^{-1}}$$

$$\tfrac{1}{2}\mathrm{Fe_2O_3(s)} + \tfrac{3}{2}\mathrm{H_2(g)} \rightarrow \mathrm{Fe(s)} + \tfrac{3}{2}\mathrm{H_2O(l)}; \qquad \Delta_r H = -16.6\,\mathrm{kJ\,mol^{-1}}$$

Both statements describe the same reduction. The second processes half as much material and releases half as much heat. Enthalpy change is an extensive quantity, and the equation you write fixes how much "one mole of reaction" is.

Key Point: A value of $\Delta_r H$ means nothing until the balanced equation it belongs to is written down beside it.

Thermochemical Equations and the Rules for Handling Them

Key Point (Definition): A thermochemical equation is a balanced chemical equation written together with the value of its $\Delta_r H$ and the physical state of every substance in it.

$$\mathrm{C_2H_5OH(l)} + 3\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{CO_2(g)} + 3\,\mathrm{H_2O(l)}; \qquad \Delta_r H = -1367\,\mathrm{kJ\,mol^{-1}}$$

That single line says: burning one mole of liquid ethanol in oxygen gas, at constant temperature and pressure, to give carbon dioxide gas and liquid water, releases $1367\,\mathrm{kJ}$.

Three rules govern what you may do to such an equation. Each one follows from enthalpy being a state function and an extensive property.

Rule 1: physical states must always be shown

Enthalpy depends on the state of aggregation, so $\Delta_r H$ does too. Compare the two ways of writing the combustion of hydrogen:

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}; \qquad \Delta_r H = -285.8\,\mathrm{kJ\,mol^{-1}}$$

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(g)}; \qquad \Delta_r H = -241.8\,\mathrm{kJ\,mol^{-1}}$$

The chemistry is identical; only the state of the product differs. The gap of $44.0\,\mathrm{kJ\,mol^{-1}}$ is exactly the enthalpy needed to vaporise one mole of water at $298\,\mathrm{K}$. Leaving off the $(l)$ or $(g)$ turns a precise statement into a vague one, and where allotropes exist the allotrope must be named too: $\mathrm{C(graphite)}$ and $\mathrm{C(diamond)}$ give different numbers.

Rule 2: reversing the equation reverses the sign

Running a reaction backwards takes the system from the old products to the old reactants. The enthalpy difference is the same size with the opposite sense.

$$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}; \qquad \Delta_r H = -92.4\,\mathrm{kJ\,mol^{-1}}$$

$$2\,\mathrm{NH_3(g)} \rightarrow \mathrm{N_2(g)} + 3\,\mathrm{H_2(g)}; \qquad \Delta_r H = +92.4\,\mathrm{kJ\,mol^{-1}}$$

What is released on the way out must be supplied on the way back, otherwise energy could be manufactured by cycling round and back.

Rule 3: multiplying the equation multiplies $\Delta_r H$

Multiply every coefficient by $n$ and the enthalpy change is multiplied by $n$ as well, fractions included.

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}; \qquad \Delta_r H = -285.8\,\mathrm{kJ\,mol^{-1}}$$

$$2\,\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{H_2O(l)}; \qquad \Delta_r H = -571.6\,\mathrm{kJ\,mol^{-1}}$$

The coefficients in a thermochemical equation stand for moles, never molecules, which is why writing $\tfrac{1}{2}\mathrm{O_2}$ is perfectly legal — half a mole of oxygen is an ordinary quantity.

Key Point: Reverse it, flip the sign. Scale it, scale the number. Add two of them, add their $\Delta_r H$ values. These three moves are the whole toolkit for the calculations in this section.

[Board] A thermochemical equation loses marks for a missing state symbol even when the arithmetic is right. Write $(s)$, $(l)$, $(g)$ or $(aq)$ on every species, every time.

Standard States and the Standard Reaction Enthalpy

Enthalpy changes shift with pressure and temperature, so tabulated values have to be quoted under an agreed set of conditions. That agreed set is the standard state.

Key Point (Definition): The standard state of a substance at a specified temperature is its pure form at a pressure of $1\,\mathrm{bar}$. Data are usually tabulated at $298\,\mathrm{K}$.

The standard state of ethanol at $298\,\mathrm{K}$ is pure liquid ethanol at $1\,\mathrm{bar}$. The standard state of iron at $500\,\mathrm{K}$ is pure solid iron at $1\,\mathrm{bar}$. Temperature is not part of the definition — it must be stated separately — but pressure is fixed at $1\,\mathrm{bar}$ and the substance must be pure.

Standard conditions are marked with a superscript circle on the symbol.

Key Point (Definition): The standard reaction enthalpy $\Delta_r H^{\circ}$ is the enthalpy change of a reaction in which all reactants and all products are in their standard states.

The reference state of an element

Formation data need one more agreement, because there is no absolute zero of enthalpy to measure from.

Key Point (Definition): The reference state of an element is its most stable state of aggregation at $298\,\mathrm{K}$ and $1\,\mathrm{bar}$. By convention, an element in its reference state is assigned $\Delta_f H^{\circ} = 0$.

The reference state of hydrogen is $\mathrm{H_2(g)}$, of oxygen $\mathrm{O_2(g)}$, of carbon $\mathrm{C(graphite)}$, of sulphur $\mathrm{S(rhombic)}$, of bromine $\mathrm{Br_2(l)}$, of mercury $\mathrm{Hg(l)}$. Diamond, ozone and $\mathrm{S(monoclinic)}$ are elements but not reference states, and their formation enthalpies are not zero — $\Delta_f H^{\circ}$ for $\mathrm{C(diamond)}$ is $+1.90\,\mathrm{kJ\,mol^{-1}}$.

Standard enthalpy of formation

Key Point (Definition): The standard enthalpy of formation $\Delta_f H^{\circ}$ is the enthalpy change when one mole of a compound is formed in its standard state from its elements in their reference states.

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}; \qquad \Delta_f H^{\circ} = -285.8\,\mathrm{kJ\,mol^{-1}}$$

$$\mathrm{C(graphite,s)} + 2\,\mathrm{H_2(g)} \rightarrow \mathrm{CH_4(g)}; \qquad \Delta_f H^{\circ} = -74.8\,\mathrm{kJ\,mol^{-1}}$$

$$2\,\mathrm{C(graphite,s)} + 3\,\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{C_2H_5OH(l)}; \qquad \Delta_f H^{\circ} = -277.7\,\mathrm{kJ\,mol^{-1}}$$

Two conditions must hold together. Exactly one mole of the compound is produced, and the starting materials are elements in their reference states. Both of these fail the test:

$$\mathrm{CaO(s)} + \mathrm{CO_2(g)} \rightarrow \mathrm{CaCO_3(s)}; \qquad \Delta_r H^{\circ} = -178.3\,\mathrm{kJ\,mol^{-1}}$$

$$\mathrm{H_2(g)} + \mathrm{Br_2(l)} \rightarrow 2\,\mathrm{HBr(g)}; \qquad \Delta_r H^{\circ} = -72.8\,\mathrm{kJ\,mol^{-1}}$$

The first builds calcium carbonate from two compounds, not from calcium, carbon and oxygen. The second starts from elements correctly but makes two moles of product; dividing every coefficient by two repairs it:

$$\tfrac{1}{2}\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{Br_2(l)} \rightarrow \mathrm{HBr(g)}; \qquad \Delta_f H^{\circ} = -36.4\,\mathrm{kJ\,mol^{-1}}$$

$\Delta_f H^{\circ}$ is a special case of $\Delta_r H^{\circ}$ — the case where the reaction happens to be a formation reaction.

Building a Reaction Enthalpy from Formation Enthalpies

Absolute enthalpies cannot be measured, but the definition of $\Delta_f H^{\circ}$ supplies a fixed floor: the elements in their reference states, all at zero. Every compound is then quoted by how far it sits above or below that floor.

Take any reaction from reactants to products by a detour: pull the reactants apart into their elements, then assemble the products from those elements. Pulling the reactants apart costs the negative of their formation enthalpies; building the products releases theirs. The detour and the direct route must give the same total.

Key Point: $\Delta_r H^{\circ} = \sum_i a_i \Delta_f H^{\circ}(\text{products}) - \sum_i b_i \Delta_f H^{\circ}(\text{reactants})$, with $a_i$ and $b_i$ the stoichiometric coefficients in the balanced equation.

Worked through: decomposing limestone

$$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$$

Every coefficient is $1$, so

$$\Delta_r H^{\circ} = \left[ \Delta_f H^{\circ}(\mathrm{CaO},s) + \Delta_f H^{\circ}(\mathrm{CO_2},g) \right] - \Delta_f H^{\circ}(\mathrm{CaCO_3},s)$$

$$= \left[ (-635.1) + (-393.5) \right] - (-1206.9) = -1028.6 + 1206.9 = +178.3\,\mathrm{kJ\,mol^{-1}}$$

The positive sign says the kiln must be heated, which matches how lime is actually made.

Worked through: reducing iron(III) oxide

$$\mathrm{Fe_2O_3(s)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{Fe(s)} + 3\,\mathrm{H_2O(l)}$$

Iron and hydrogen are elements in their reference states, so both contribute zero:

$$\Delta_r H^{\circ} = 3(-285.8) - (-824.2) = -857.4 + 824.2 = -33.2\,\mathrm{kJ\,mol^{-1}}$$

Some values worth carrying

Substance	$\Delta_f H^{\circ}$ / $\mathrm{kJ\,mol^{-1}}$	Substance	$\Delta_f H^{\circ}$ / $\mathrm{kJ\,mol^{-1}}$
$\mathrm{CO_2(g)}$	$-393.5$	$\mathrm{CaCO_3(s)}$	$-1206.9$
$\mathrm{CO(g)}$	$-110.5$	$\mathrm{CaO(s)}$	$-635.1$
$\mathrm{H_2O(l)}$	$-285.8$	$\mathrm{Fe_2O_3(s)}$	$-824.2$
$\mathrm{H_2O(g)}$	$-241.8$	$\mathrm{Al_2O_3(s)}$	$-1675.7$
$\mathrm{CH_4(g)}$	$-74.8$	$\mathrm{SO_2(g)}$	$-296.8$
$\mathrm{C_2H_6(g)}$	$-84.7$	$\mathrm{NO(g)}$	$+90.3$
$\mathrm{C_2H_2(g)}$	$+226.8$	$\mathrm{HCl(g)}$	$-92.3$
$\mathrm{C_2H_5OH(l)}$	$-277.7$	$\mathrm{C(graphite)},\ \mathrm{O_2(g)},\ \mathrm{H_2(g)}$	$0$

A negative $\Delta_f H^{\circ}$ marks a compound that lies below its elements in enthalpy, which is one reason such compounds are common and stable. A positive value, as for ethyne, marks a compound that stores energy relative to its elements.

[NEET] The commonest slip in this calculation is dropping the coefficient. In $3\,\mathrm{H_2O(l)}$ the formation enthalpy is used three times. The second commonest is forgetting that the elements are zero and hunting for a value that is not in the table.

Hess's Law of Constant Heat Summation

Some reaction enthalpies cannot be measured directly. Burning graphite in a limited supply of oxygen is meant to give carbon monoxide, but some carbon dioxide always forms alongside it, so the heat measured belongs to a mixture and not to the clean reaction

$$\mathrm{C(graphite,s)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CO(g)}; \qquad \Delta_r H^{\circ} = ?$$

Enthalpy is a state function, and its change depends only on the initial and final states. The route taken between them is irrelevant. That single fact rescues the calculation.

Key Point (Hess's Law): If a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate steps into which the overall reaction may be divided, at the same temperature.

For a change from $\mathrm{A}$ to $\mathrm{B}$ carried out directly with enthalpy change $\Delta_r H$, or through intermediates with steps $\Delta_r H_1$, $\Delta_r H_2$, $\Delta_r H_3$:

$$\Delta_r H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3 + \cdots$$

The carbon monoxide problem solved

Two reactions that can be measured cleanly:

$$\mathrm{C(graphite,s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}; \qquad \Delta_r H^{\circ} = -393.5\,\mathrm{kJ\,mol^{-1}} \qquad (i)$$

$$\mathrm{CO(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}; \qquad \Delta_r H^{\circ} = -283.0\,\mathrm{kJ\,mol^{-1}} \qquad (ii)$$

The target has $\mathrm{CO}$ on the right, while equation $(ii)$ has it on the left. Reversing $(ii)$ moves it across and flips the sign:

$$\mathrm{CO_2(g)} \rightarrow \mathrm{CO(g)} + \tfrac{1}{2}\mathrm{O_2(g)}; \qquad \Delta_r H^{\circ} = +283.0\,\mathrm{kJ\,mol^{-1}} \qquad (iii)$$

Adding $(i)$ and $(iii)$, the $\mathrm{CO_2}$ appearing on the right of one and the left of the other cancels, and one $\mathrm{O_2}$ on the left is partly cancelled by the $\tfrac{1}{2}\mathrm{O_2}$ on the right:

$$\mathrm{C(graphite,s)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CO(g)}$$

$$\Delta_r H^{\circ} = (-393.5) + (+283.0) = -110.5\,\mathrm{kJ\,mol^{-1}}$$

That number appears in every table as $\Delta_f H^{\circ}(\mathrm{CO},g)$, and it was never measured directly.

Why the law cannot fail

Suppose the two routes from graphite to carbon dioxide gave different totals. Go up by the larger route and come back down by the smaller, and the cycle would return the system to exactly its starting state while leaving a surplus of energy behind. Energy would have been created out of nothing. Hess's law is the first law of thermodynamics wearing a chemical uniform.

Key Point: Hess's law holds because $H$ is a state function. Any quantity that is a state function obeys the same summation rule, which is why the identical arithmetic works later for entropy and Gibbs energy.

Assembling a Target Equation from Given Equations

Most Hess's law questions hand you two or three thermochemical equations and one target. The work is bookkeeping, and it goes faster with a fixed routine.

Step 1. Write the target equation and mark which species must end up on the left and which on the right.

Step 2. Take each given equation in turn. Find a species in it that appears in the target and only there. Reverse the equation if that species is on the wrong side; multiply it if the coefficient is wrong.

Step 3. Add the adjusted equations. Everything not in the target must cancel between the two sides.

Step 4. Add the adjusted $\Delta_r H$ values, reversing signs where you reversed an equation and scaling where you scaled one.

A three-equation build

Find $\Delta_r H^{\circ}$ for $\ 2\,\mathrm{C(graphite,s)} + \mathrm{H_2(g)} \rightarrow \mathrm{C_2H_2(g)}$, given

$$\mathrm{C_2H_2(g)} + \tfrac{5}{2}\mathrm{O_2(g)} \rightarrow 2\,\mathrm{CO_2(g)} + \mathrm{H_2O(l)}; \qquad \Delta_r H^{\circ} = -1299.6\,\mathrm{kJ\,mol^{-1}} \qquad (i)$$

$$\mathrm{C(graphite,s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}; \qquad \Delta_r H^{\circ} = -393.5\,\mathrm{kJ\,mol^{-1}} \qquad (ii)$$

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}; \qquad \Delta_r H^{\circ} = -285.8\,\mathrm{kJ\,mol^{-1}} \qquad (iii)$$

Ethyne is a product in the target but a reactant in $(i)$, so $(i)$ is reversed. Graphite is needed as $2$ moles on the left, so $(ii)$ is doubled. Hydrogen is needed as $1$ mole on the left, so $(iii)$ is used unchanged.

$$2 \times (ii): \quad 2\,\mathrm{C(graphite,s)} + 2\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{CO_2(g)}; \qquad \Delta H = 2(-393.5) = -787.0$$

$$(iii): \quad \mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}; \qquad \Delta H = -285.8$$

$$\text{reverse } (i): \quad 2\,\mathrm{CO_2(g)} + \mathrm{H_2O(l)} \rightarrow \mathrm{C_2H_2(g)} + \tfrac{5}{2}\mathrm{O_2(g)}; \qquad \Delta H = +1299.6$$

Adding the three: the $2\,\mathrm{CO_2}$ cancels, the $\mathrm{H_2O}$ cancels, and the oxygen tallies as $2 + \tfrac{1}{2} - \tfrac{5}{2} = 0$. What survives is the target.

$$\Delta_r H^{\circ} = -787.0 - 285.8 + 1299.6 = +226.8\,\mathrm{kJ\,mol^{-1}}$$

Ethyne lies well above its elements in enthalpy, which is why it burns so fiercely in an oxyacetylene torch.

The same job as a single formula

That build is exactly the formation formula in disguise. Combustion equations are formation equations for $\mathrm{CO_2}$ and $\mathrm{H_2O}$ turned round, so whenever every given equation is a combustion, the shortcut is

$$\Delta_r H^{\circ} = \sum b_i \Delta_c H^{\circ}(\text{reactants}) - \sum a_i \Delta_c H^{\circ}(\text{products})$$

with the reactants and products the other way round compared with the formation version, because combustion enthalpies point down from a substance while formation enthalpies point up to it.

Key Point: Two checks catch nearly every error. Confirm that the adjusted equations really add to the target with everything else cancelling, and confirm that the sign of the answer matches the chemistry you expect.

[JEE Main] When the given data are combustion enthalpies, the two routines conflict if you mix them. Use the cancellation method or the combustion formula, not half of each.

Solved Examples

Question 1: Combustion enthalpy of methane from formation data

Calculate $\Delta_r H^{\circ}$ for $\ \mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)}$. Take $\Delta_f H^{\circ}$ values of $-74.8$, $-393.5$ and $-285.8\,\mathrm{kJ\,mol^{-1}}$ for $\mathrm{CH_4(g)}$, $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$.

Answer:

I use products minus reactants, each with its coefficient. Oxygen is an element in its reference state, so it contributes zero.

$$\Delta_r H^{\circ} = \left[ (-393.5) + 2(-285.8) \right] - \left[ (-74.8) + 2(0) \right]$$

$$= (-393.5 - 571.6) + 74.8 = -965.1 + 74.8 = -890.3\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta_r H^{\circ} = -890.3\,\mathrm{kJ\,mol^{-1}}$

Watch out: The $2$ in front of $\mathrm{H_2O}$ multiplies its formation enthalpy. Using $-285.8$ once gives $-604.5$, which is a long way out.

Question 2: The same reaction with water as vapour

Repeat the previous calculation with the water produced as $\mathrm{H_2O(g)}$, for which $\Delta_f H^{\circ} = -241.8\,\mathrm{kJ\,mol^{-1}}$, and account for the difference.

Answer:

$$\Delta_r H^{\circ} = \left[ (-393.5) + 2(-241.8) \right] - (-74.8) = -877.1 + 74.8 = -802.3\,\mathrm{kJ\,mol^{-1}}$$

The value is less negative than before by $890.3 - 802.3 = 88.0\,\mathrm{kJ\,mol^{-1}}$.

Two moles of water are left as vapour instead of liquid. Vaporising a mole of water at $298\,\mathrm{K}$ costs $44.0\,\mathrm{kJ}$, so two moles cost $88.0\,\mathrm{kJ}$ — energy that stays locked in the vapour instead of being released.

Ans: $\Delta_r H^{\circ} = -802.3\,\mathrm{kJ\,mol^{-1}}$; the $88.0\,\mathrm{kJ\,mol^{-1}}$ gap is $2 \times \Delta_{vap}H$ of water

Watch out: This is the reason state symbols are compulsory. A combustion value quoted without them could be either number.

Question 3: Reversing and scaling a thermochemical equation

For $\ \mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$, $\Delta_r H^{\circ} = -92.4\,\mathrm{kJ\,mol^{-1}}$. Write $\Delta_r H^{\circ}$ for (a) the decomposition of $2$ moles of ammonia into its elements, and (b) the formation of one mole of ammonia.

Answer:

(a) Decomposition is the given equation run backwards, so the magnitude is unchanged and the sign flips.

$$2\,\mathrm{NH_3(g)} \rightarrow \mathrm{N_2(g)} + 3\,\mathrm{H_2(g)}; \qquad \Delta_r H^{\circ} = +92.4\,\mathrm{kJ\,mol^{-1}}$$

(b) One mole of ammonia means halving every coefficient, so I halve the enthalpy change as well.

$$\tfrac{1}{2}\mathrm{N_2(g)} + \tfrac{3}{2}\mathrm{H_2(g)} \rightarrow \mathrm{NH_3(g)}; \qquad \Delta_r H^{\circ} = \tfrac{1}{2}(-92.4) = -46.2\,\mathrm{kJ\,mol^{-1}}$$

Since this makes exactly one mole of the compound from elements in their reference states, it is also $\Delta_f H^{\circ}(\mathrm{NH_3},g)$.

Ans: (a) $+92.4\,\mathrm{kJ\,mol^{-1}}$; (b) $-46.2\,\mathrm{kJ\,mol^{-1}}$

Question 4: Which equation defines a formation enthalpy

Three equations are given below. Identify the one whose $\Delta_r H^{\circ}$ is a standard enthalpy of formation, and repair the others where possible.

(a) $\ \mathrm{CaO(s)} + \mathrm{CO_2(g)} \rightarrow \mathrm{CaCO_3(s)}$

(b) $\ \mathrm{H_2(g)} + \mathrm{Br_2(l)} \rightarrow 2\,\mathrm{HBr(g)};\ \Delta_r H^{\circ} = -72.8\,\mathrm{kJ\,mol^{-1}}$

(c) $\ \mathrm{C(graphite,s)} + 2\,\mathrm{H_2(g)} \rightarrow \mathrm{CH_4(g)}$

Answer:

A formation equation needs two things at once: exactly one mole of the compound as the only product, and elements in their reference states as the only reactants.

(a) fails on the reactants. Calcium carbonate is being built from two compounds, so this is an ordinary $\Delta_r H^{\circ}$. It cannot be repaired by rescaling; the reactants would have to be $\mathrm{Ca(s)}$, $\mathrm{C(graphite)}$ and $\mathrm{O_2(g)}$.

(b) has the right reactants but makes two moles of product. Dividing every coefficient by $2$ fixes it, and the enthalpy change halves:

$$\tfrac{1}{2}\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{Br_2(l)} \rightarrow \mathrm{HBr(g)}; \qquad \Delta_f H^{\circ} = -36.4\,\mathrm{kJ\,mol^{-1}}$$

(c) passes both tests. Graphite and $\mathrm{H_2(g)}$ are reference states and one mole of methane is formed.

Ans: (c) is a formation equation; (b) becomes one on halving, giving $\Delta_f H^{\circ}(\mathrm{HBr},g) = -36.4\,\mathrm{kJ\,mol^{-1}}$; (a) cannot be rewritten as one

Watch out: $\mathrm{Br_2}$ must be written as a liquid. Bromine vapour is not the reference state, and $\Delta_f H^{\circ}$ measured from $\mathrm{Br_2(g)}$ would be smaller by the enthalpy of vaporisation.

Question 5: The thermite reaction

Calculate $\Delta_r H^{\circ}$ for $\ 2\,\mathrm{Al(s)} + \mathrm{Fe_2O_3(s)} \rightarrow \mathrm{Al_2O_3(s)} + 2\,\mathrm{Fe(s)}$, given $\Delta_f H^{\circ}(\mathrm{Al_2O_3},s) = -1675.7$ and $\Delta_f H^{\circ}(\mathrm{Fe_2O_3},s) = -824.2\,\mathrm{kJ\,mol^{-1}}$.

Answer:

Aluminium and iron are elements in their reference states, so only the two oxides carry a value.

$$\Delta_r H^{\circ} = \left[ (-1675.7) + 2(0) \right] - \left[ 2(0) + (-824.2) \right]$$

$$= -1675.7 + 824.2 = -851.5\,\mathrm{kJ\,mol^{-1}}$$

The strongly negative value is why the thermite mixture melts the iron it produces.

Ans: $\Delta_r H^{\circ} = -851.5\,\mathrm{kJ\,mol^{-1}}$

Question 6: Formation enthalpy of benzene from its combustion

One mole of benzene is burned completely at $298\,\mathrm{K}$ to $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$, liberating $3267.6\,\mathrm{kJ}$. Find $\Delta_f H^{\circ}$ of benzene, given $\Delta_f H^{\circ}$ of $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$ as $-393.5$ and $-285.8\,\mathrm{kJ\,mol^{-1}}$.

Answer:

The combustion equation, balanced for one mole of benzene, is

$$\mathrm{C_6H_6(l)} + \tfrac{15}{2}\mathrm{O_2(g)} \rightarrow 6\,\mathrm{CO_2(g)} + 3\,\mathrm{H_2O(l)}; \qquad \Delta_c H^{\circ} = -3267.6\,\mathrm{kJ\,mol^{-1}}$$

Applying the formation formula to this equation, with $x$ standing for $\Delta_f H^{\circ}(\mathrm{C_6H_6},l)$:

$$-3267.6 = \left[ 6(-393.5) + 3(-285.8) \right] - x$$

$$-3267.6 = (-2361.0 - 857.4) - x = -3218.4 - x$$

$$x = -3218.4 + 3267.6 = +49.2\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta_f H^{\circ}(\mathrm{C_6H_6},l) = +49.2\,\mathrm{kJ\,mol^{-1}}$, against a tabulated $+49.0\,\mathrm{kJ\,mol^{-1}}$ — the small gap is rounding in the formation data used

Watch out: The unknown is a reactant, so it enters with a minus sign. Writing $-3267.6 = -3218.4 + x$ delivers $-49.2$, the right size with the wrong sign.

Question 7: Carbon monoxide by Hess's law

Find $\Delta_r H^{\circ}$ for $\ \mathrm{C(graphite,s)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CO(g)}$ from

$(i)\ \mathrm{C(graphite,s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)};\ \Delta_r H^{\circ} = -393.5\,\mathrm{kJ\,mol^{-1}}$

$(ii)\ \mathrm{CO(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)};\ \Delta_r H^{\circ} = -283.0\,\mathrm{kJ\,mol^{-1}}$

Answer:

The target needs $\mathrm{CO}$ on the right, and equation $(ii)$ has it on the left. I reverse $(ii)$ and flip its sign.

$$\mathrm{CO_2(g)} \rightarrow \mathrm{CO(g)} + \tfrac{1}{2}\mathrm{O_2(g)}; \qquad \Delta_r H^{\circ} = +283.0\,\mathrm{kJ\,mol^{-1}}$$

Adding this to $(i)$: $\mathrm{CO_2}$ cancels completely, and on the oxygen side $1$ mole on the left against $\tfrac{1}{2}$ mole on the right leaves $\tfrac{1}{2}$ mole on the left. The sum is the target equation.

$$\Delta_r H^{\circ} = (-393.5) + (+283.0) = -110.5\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta_r H^{\circ} = -110.5\,\mathrm{kJ\,mol^{-1}}$, which is $\Delta_f H^{\circ}(\mathrm{CO},g)$

Watch out: Subtracting the two given values as $-393.5 - (-283.0)$ happens to give the same $-110.5$ here only because the reversal and the subtraction are the same operation. Doing it by inspection stops working the moment scaling is involved.

Question 8: Methane formation from combustion data

Calculate $\Delta_f H^{\circ}$ of methane from

$(i)\ \mathrm{C(graphite,s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)};\ \Delta_r H^{\circ} = -393.5$

$(ii)\ \mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)};\ \Delta_r H^{\circ} = -285.8$

$(iii)\ \mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)};\ \Delta_r H^{\circ} = -890.3$

All values in $\mathrm{kJ\,mol^{-1}}$.

Answer:

The target is $\ \mathrm{C(graphite,s)} + 2\,\mathrm{H_2(g)} \rightarrow \mathrm{CH_4(g)}$.

Graphite is on the left in $(i)$ and in the target, so $(i)$ is used as it is. Hydrogen is needed as $2$ moles, so $(ii)$ is doubled. Methane must end on the right, so $(iii)$ is reversed.

$$(i): \quad \Delta H = -393.5$$

$$2 \times (ii): \quad 2\,\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{H_2O(l)}; \qquad \Delta H = -571.6$$

$$\text{reverse } (iii): \quad \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)} \rightarrow \mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)}; \qquad \Delta H = +890.3$$

Adding: $\mathrm{CO_2}$ cancels, $2\,\mathrm{H_2O}$ cancels, and oxygen tallies as $1 + 1 - 2 = 0$. The target survives.

$$\Delta_f H^{\circ} = -393.5 - 571.6 + 890.3 = -74.8\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta_f H^{\circ}(\mathrm{CH_4},g) = -74.8\,\mathrm{kJ\,mol^{-1}}$

Watch out: Doubling $(ii)$ must double its enthalpy too. Carrying $-285.8$ forward while writing $2\,\mathrm{H_2O}$ in the equation gives $+211.0$, an answer with the wrong sign entirely.

Question 9: A cycle with three steps

For a change $\mathrm{A} \rightarrow \mathrm{B}$ carried out through the intermediates $\mathrm{C}$ and $\mathrm{D}$, the measured step enthalpies are $\mathrm{A} \rightarrow \mathrm{C}$, $+56\,\mathrm{kJ}$; $\mathrm{C} \rightarrow \mathrm{D}$, $-118\,\mathrm{kJ}$; $\mathrm{D} \rightarrow \mathrm{B}$, $+21\,\mathrm{kJ}$. Find $\Delta H$ for the direct change $\mathrm{A} \rightarrow \mathrm{B}$ and for $\mathrm{B} \rightarrow \mathrm{A}$.

Answer:

The three steps chain head to tail from $\mathrm{A}$ to $\mathrm{B}$, so by Hess's law I add them.

$$\Delta H(\mathrm{A} \rightarrow \mathrm{B}) = 56 - 118 + 21 = -41\,\mathrm{kJ}$$

The reverse change has the same magnitude and the opposite sign.

$$\Delta H(\mathrm{B} \rightarrow \mathrm{A}) = +41\,\mathrm{kJ}$$

Ans: $-41\,\mathrm{kJ}$ forward, $+41\,\mathrm{kJ}$ reverse

Question 10: Ethyne from its elements

Find $\Delta_r H^{\circ}$ for $\ 2\,\mathrm{C(graphite,s)} + \mathrm{H_2(g)} \rightarrow \mathrm{C_2H_2(g)}$ given the combustion enthalpies $\Delta_c H^{\circ}(\mathrm{C_2H_2},g) = -1299.6$, $\Delta_c H^{\circ}(\mathrm{C,graphite}) = -393.5$ and $\Delta_c H^{\circ}(\mathrm{H_2},g) = -285.8\,\mathrm{kJ\,mol^{-1}}$, with water produced as a liquid.

Answer:

Burning graphite gives $\mathrm{CO_2}$ and burning hydrogen gives $\mathrm{H_2O(l)}$, so those two combustion enthalpies are also formation enthalpies. The ethyne equation has to be reversed to put ethyne on the right.

$$2 \times \text{(graphite burning)}: \quad \Delta H = 2(-393.5) = -787.0$$

$$1 \times \text{(hydrogen burning)}: \quad \Delta H = -285.8$$

$$\text{reverse (ethyne burning)}: \quad 2\,\mathrm{CO_2(g)} + \mathrm{H_2O(l)} \rightarrow \mathrm{C_2H_2(g)} + \tfrac{5}{2}\mathrm{O_2(g)}; \qquad \Delta H = +1299.6$$

Adding the three, $2\,\mathrm{CO_2}$ and $\mathrm{H_2O}$ cancel and the oxygen balances as $2 + \tfrac{1}{2} - \tfrac{5}{2} = 0$.

$$\Delta_r H^{\circ} = -787.0 - 285.8 + 1299.6 = +226.8\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta_r H^{\circ} = +226.8\,\mathrm{kJ\,mol^{-1}}$

Watch out: A positive answer here is correct, not a slip. Ethyne is an endothermic compound sitting above graphite and hydrogen in enthalpy.

Question 11: Formation enthalpy of ethane

The standard enthalpy of combustion of ethane is $-1559.7\,\mathrm{kJ\,mol^{-1}}$, with water formed as a liquid. Using $\Delta_f H^{\circ}(\mathrm{CO_2},g) = -393.5$ and $\Delta_f H^{\circ}(\mathrm{H_2O},l) = -285.8\,\mathrm{kJ\,mol^{-1}}$, find $\Delta_f H^{\circ}$ of ethane.

Answer:

The balanced combustion of one mole of ethane is

$$\mathrm{C_2H_6(g)} + \tfrac{7}{2}\mathrm{O_2(g)} \rightarrow 2\,\mathrm{CO_2(g)} + 3\,\mathrm{H_2O(l)}$$

Applying the formation formula with $x = \Delta_f H^{\circ}(\mathrm{C_2H_6},g)$:

$$-1559.7 = \left[ 2(-393.5) + 3(-285.8) \right] - x = (-787.0 - 857.4) - x = -1644.4 - x$$

$$x = -1644.4 + 1559.7 = -84.7\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta_f H^{\circ}(\mathrm{C_2H_6},g) = -84.7\,\mathrm{kJ\,mol^{-1}}$

Watch out: The $\tfrac{7}{2}$ in front of oxygen tempts people to clear the fraction by doubling the equation. Doubling it also doubles the combustion enthalpy, and forgetting that gives half the right answer for $x$.

Question 12: How much heat from a given mass

How much heat is released when $11.5\,\mathrm{g}$ of ethanol is burned completely at $298\,\mathrm{K}$ and $1\,\mathrm{bar}$? For $\ \mathrm{C_2H_5OH(l)} + 3\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{CO_2(g)} + 3\,\mathrm{H_2O(l)}$, $\Delta_r H^{\circ} = -1367\,\mathrm{kJ\,mol^{-1}}$. Molar mass of ethanol $= 46.0\,\mathrm{g\,mol^{-1}}$.

Answer:

The equation is written for one mole of ethanol, so the tabulated value applies to one mole.

$$n = \frac{11.5}{46.0} = 0.250\,\mathrm{mol}$$

$$q_p = n \times \Delta_r H^{\circ} = 0.250 \times (-1367) = -341.8\,\mathrm{kJ}$$

The negative sign means the system loses this energy, so $341.8\,\mathrm{kJ}$ of heat is given out to the surroundings.

Ans: $341.8\,\mathrm{kJ}$ released, and $\Delta H = -341.8\,\mathrm{kJ}$ for the system

Question 13: From $\Delta_r H^{\circ}$ to $\Delta_r U^{\circ}$

For $\ \mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)}$, $\Delta_r H^{\circ} = -890.3\,\mathrm{kJ\,mol^{-1}}$ at $298\,\mathrm{K}$. Find $\Delta_r U^{\circ}$. Take $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$.

Answer:

I count only gaseous moles. On the right there is $1$ mole of gas, since the water is liquid. On the left there are $1 + 2 = 3$ moles.

$$\Delta n_g = 1 - 3 = -2$$

From $\Delta H = \Delta U + \Delta n_g RT$:

$$\Delta_r U^{\circ} = \Delta_r H^{\circ} - \Delta n_g RT = -890.3 - (-2)(8.314 \times 10^{-3})(298)$$

$$= -890.3 + 4.955 = -885.3\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta_r U^{\circ} = -885.3\,\mathrm{kJ\,mol^{-1}}$

Watch out: The two moles of liquid water are not counted in $\Delta n_g$. Counting them would give $\Delta n_g = 0$ and hide the correction altogether.

Question 14: Building sulphur trioxide in two steps

Calculate $\Delta_f H^{\circ}$ of $\mathrm{SO_3(g)}$ from

$(i)\ \mathrm{S(rhombic,s)} + \mathrm{O_2(g)} \rightarrow \mathrm{SO_2(g)};\ \Delta_r H^{\circ} = -296.8\,\mathrm{kJ\,mol^{-1}}$

$(ii)\ 2\,\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{SO_3(g)};\ \Delta_r H^{\circ} = -198.4\,\mathrm{kJ\,mol^{-1}}$

Answer:

The target is $\ \mathrm{S(rhombic,s)} + \tfrac{3}{2}\mathrm{O_2(g)} \rightarrow \mathrm{SO_3(g)}$, making exactly one mole of $\mathrm{SO_3}$.

Equation $(i)$ already gives one mole of $\mathrm{SO_2}$ from sulphur, so I keep it. Equation $(ii)$ makes two moles of $\mathrm{SO_3}$, so I halve it.

$$\tfrac{1}{2} \times (ii): \quad \mathrm{SO_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{SO_3(g)}; \qquad \Delta H = \tfrac{1}{2}(-198.4) = -99.2$$

Adding this to $(i)$, the $\mathrm{SO_2}$ cancels and the oxygen adds to $\tfrac{3}{2}$ moles on the left.

$$\Delta_f H^{\circ} = -296.8 + (-99.2) = -396.0\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta_f H^{\circ}(\mathrm{SO_3},g) = -396.0\,\mathrm{kJ\,mol^{-1}}$

Watch out: Adding $-296.8$ and $-198.4$ without halving gives $-495.2$, which belongs to $\mathrm{S(rhombic,s)} + \mathrm{SO_2(g)} + 2\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{SO_3(g)}$ — two moles of $\mathrm{SO_3}$, and a mole of $\mathrm{SO_2}$ consumed that the target never mentions.