Enthalpy Changes and Reaction Pathway Diagrams

AS · 13 min

Almost every chemical reaction either gives out heat or takes it in, and chemists measure that energy as an enthalpy change, $\Delta H$ . This note sets up the language of energetics: exothermic and endothermic reactions, reaction pathway diagrams with activation energy, standard conditions, and the four standard enthalpy changes whose definitions you must be able to write word for word. Definitions in this topic are some of the most reliable marks on Paper 2, and the equations that go with them underpin every Hess's law calculation.

Enthalpy and enthalpy change

Chemists divide the world into the system (the reacting chemicals) and the surroundings (everything else: the solvent, the container, the air, the thermometer).

Enthalpy, $H$ , is the total energy content of a system that is available as heat at constant pressure. You can never measure $H$ itself, only the change in it.

Definition

The enthalpy change, $\Delta H$ , of a reaction is the heat energy change measured at constant pressure.

\Delta H = H_{\text{products}} - H_{\text{reactants}}

Enthalpy changes are given in $\text{kJ mol}^{-1}$ : "per mole" means per mole of reaction as written in the equation, or per mole of the substance named in a definition.

Exothermic and endothermic reactions

Key result

	exothermic	endothermic
heat energy	released to the surroundings	absorbed from the surroundings
sign of $\Delta H$	negative	positive
enthalpy of products	lower than reactants	higher than reactants
temperature of surroundings	rises	falls
examples	combustion, neutralisation, respiration, most oxidation reactions, displacement reactions	thermal decomposition (for example $\ce{CaCO3 -> CaO + CO2}$ ), photosynthesis, dissolving ammonium nitrate, electrolysis

The sign convention is from the system's point of view: in an exothermic reaction the system loses energy, so $\Delta H$ is negative even though the thermometer reading goes up. Always write the sign, including "+" for endothermic values: $\Delta H = +178\ \text{kJ mol}^{-1}$ , not $178\ \text{kJ mol}^{-1}$ .

Where the energy comes from

Chemical reactions involve breaking bonds in the reactants and making new bonds in the products.

Breaking bonds needs energy: it is endothermic.
Making bonds releases energy: it is exothermic.

If more energy is released making the new bonds than is used breaking the old ones, the reaction is exothermic overall. If more is used than released, it is endothermic. You will use this to calculate $\Delta H$ from bond energies in a later note.

Reaction pathway diagrams

A reaction pathway diagram (also called an energy profile) plots enthalpy against the progress of the reaction. Reactants and products are drawn as levels; between them, the curve rises to a peak.

The peak exists because bonds must start to break before new ones can form. The particles must collide with enough energy to get over this barrier. The height of the barrier above the reactants is the activation energy.

Definition

Activation energy, $E_A$ , is the minimum energy required for a collision to be effective (that is, for a reaction to occur).

Exothermic reaction. Products are below reactants. The arrow for $\Delta H$ points down from the reactant level to the product level; the arrow for $E_A$ points up from the reactant level to the peak.

y = 6 - 4*(1 + tanh(1.2*(x - 5)))/2 + 5*exp(-(x - 5)^2/2) (4.59, 6) -> (4.59, 9.51) (8.5, 6) -> (8.5, 2)

(Vertical axis: enthalpy; horizontal axis: progress of reaction. The reactants are at the left level, $6$ , and the products at the right level, $2$ . The upward arrow is $E_A$ ; the downward arrow on the right is $\Delta H$ , negative.)

Endothermic reaction. Products are above reactants, so the $\Delta H$ arrow points up. The activation energy is still measured from the reactants to the peak, so it is always at least as large as $\Delta H$ .

y = 2 + 4*(1 + tanh(1.2*(x - 5)))/2 + 5*exp(-(x - 5)^2/2) (5.41, 2) -> (5.41, 9.51) (8.5, 2) -> (8.5, 6)

(Reactants at level $2$ , products at level $6$ . The long upward arrow is $E_A$ ; the shorter upward arrow on the right is $\Delta H$ , positive.)

Method

Drawing a reaction pathway diagram

Draw and label the axes: enthalpy (or energy) vertically, progress of reaction (or reaction pathway) horizontally.
Draw a horizontal line for the reactants, labelled with their formulas.
Draw the product line lower (exothermic) or higher (endothermic), labelled.
Join them with a smooth curve that rises to a single peak.
Draw $E_A$ as an arrow from the reactant line up to the peak.
Draw $\Delta H$ as an arrow from the reactant line to the product line (pointing down for exothermic, up for endothermic), labelled with its value and sign.

The activation energy of the reverse reaction is measured from the products to the peak. For an exothermic forward reaction:

E_A(\text{reverse}) = E_A(\text{forward}) + |\Delta H|

You will add a second, lower curve for a catalysed reaction in the kinetics notes.

Standard conditions

The value of $\Delta H$ depends on temperature, pressure and the physical states of the substances. To compare values fairly, chemists quote standard enthalpy changes, shown by the symbol $\ominus$ : $\Delta H^{\ominus}$ .

Key result

Standard conditions (as used in this syllabus):

temperature $298\ \text{K}$ ( $25\ ^\circ\text{C}$ );
pressure $101\ \text{kPa}$ ;
every substance in its standard state: its normal physical state under these conditions (for example $\ce{H2O(l)}$ , $\ce{O2(g)}$ , $\ce{C(s)}$ as graphite, $\ce{Br2(l)}$ , $\ce{I2(s)}$ );
solutions at a concentration of $1.00\ \text{mol dm}^{-3}$ .

The four standard enthalpy changes

Definition

Standard enthalpy change of reaction, $\Delta H_r^{\ominus}$ : the enthalpy change when the amounts of reactants shown in the equation react to give products under standard conditions.

Standard enthalpy change of formation, $\Delta H_f^{\ominus}$ : 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, $\Delta H_c^{\ominus}$ : the enthalpy change when one mole of a substance is burnt in an excess of oxygen under standard conditions.

Standard enthalpy change of neutralisation, $\Delta H_{\text{neut}}^{\ominus}$ : the enthalpy change when one mole of water is formed by the reaction of an acid with an alkali under standard conditions.

Each definition fixes what there is one mole of. That decides how the equation must be written.

Enthalpy change of reaction

$\Delta H_r$ depends on the equation as written. If the equation is doubled, $\Delta H_r$ doubles; if it is reversed, the sign changes.

\ce{N2(g) + 3H2(g) -> 2NH3(g)} \qquad \Delta H_r^{\ominus} = -92\ \text{kJ mol}^{-1}

\ce{2NH3(g) -> N2(g) + 3H2(g)} \qquad \Delta H_r^{\ominus} = +92\ \text{kJ mol}^{-1}

Enthalpy change of formation

One mole of the compound is the only product; the reactants are its elements in their standard states. Fractions are often needed.

\ce{2C(s) + 3H2(g) + 1/2O2(g) -> C2H5OH(l)} \qquad \Delta H_f^{\ominus} = -277\ \text{kJ mol}^{-1}

\ce{Na(s) + 1/2Cl2(g) -> NaCl(s)} \qquad \Delta H_f^{\ominus} = -411\ \text{kJ mol}^{-1}

The enthalpy change of formation of an element in its standard state is zero by definition: forming $\ce{O2(g)}$ from $\ce{O2(g)}$ involves no change. Many $\Delta H_f$ values cannot be measured directly (carbon and hydrogen do not react to give ethanol), so they are found using Hess's law.

Enthalpy change of combustion

One mole of the substance burnt reacts with excess oxygen; combustion is complete, so carbon goes to $\ce{CO2}$ and hydrogen to $\ce{H2O(l)}$ . Combustion is always exothermic.

\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l)} \qquad \Delta H_c^{\ominus} = -1367\ \text{kJ mol}^{-1}

\ce{H2(g) + 1/2O2(g) -> H2O(l)} \qquad \Delta H_c^{\ominus} = -286\ \text{kJ mol}^{-1}

Notice the second equation is also the formation of water: $\Delta H_c(\ce{H2}) = \Delta H_f(\ce{H2O})$ . Likewise $\Delta H_c(\ce{C, graphite}) = \Delta H_f(\ce{CO2}) = -394\ \text{kJ mol}^{-1}$ .

Enthalpy change of neutralisation

One mole of water is formed. For any strong acid with any strong alkali, the reaction is the same:

\ce{H+(aq) + OH-(aq) -> H2O(l)} \qquad \Delta H_{\text{neut}}^{\ominus} \approx -57\ \text{kJ mol}^{-1}

Sulfuric acid releases two $\ce{H+}$ per formula unit, so the equation for one mole of water uses half a mole of acid:

\ce{1/2H2SO4(aq) + NaOH(aq) -> 1/2Na2SO4(aq) + H2O(l)}

With a weak acid such as ethanoic acid, the value is slightly less exothermic (about $-55\ \text{kJ mol}^{-1}$ ), because some energy is used to complete the dissociation of the weak acid as its $\ce{H+}$ ions are removed.

Watch out

$\Delta H_f$ needs one mole of compound from elements: $\ce{2Na(s) + Cl2(g) -> 2NaCl(s)}$ is not a formation equation (two moles formed), and $\ce{CO(g) + 1/2O2(g) -> CO2(g)}$ is not one either (CO is not an element).
$\Delta H_c$ needs one mole of fuel: $\ce{2H2 + O2 -> 2H2O}$ gives $2 \times \Delta H_c$ .
$\Delta H_{\text{neut}}$ is per mole of water, not per mole of acid. For sulfuric acid, one mole of acid gives two moles of water.
State symbols are part of the equation. $\ce{H2O(g)}$ instead of $\ce{H2O(l)}$ changes the value by the enthalpy of vaporisation.

Worked examples

Routine: signs and temperature changes

When solid ammonium chloride dissolves in water, the temperature of the water falls. When magnesium is added to hydrochloric acid, the temperature rises. State the sign of $\Delta H$ for each process and explain.

Solution

Ammonium chloride dissolving: the system takes in heat from the surroundings (the water), so the water cools. The process is endothermic and $\Delta H$ is positive.

Magnesium with acid: the system gives out heat to the surroundings, so the solution warms. The reaction is exothermic and $\Delta H$ is negative.

Writing equations for definitions

Write equations, with state symbols, for: (a) the standard enthalpy change of formation of ethanoic acid, $\ce{CH3COOH(l)}$ ; (b) the standard enthalpy change of combustion of propane, $\ce{C3H8(g)}$ ; (c) the standard enthalpy change of formation of calcium carbonate.

Solution

(a) $\ce{2C(s) + 2H2(g) + O2(g) -> CH3COOH(l)}$

(b) $\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}$

In each, check: one mole of the named substance; elements in their standard states for formation; water as a liquid for combustion.

Choosing the right equation

Which of these equations has an enthalpy change equal to a standard enthalpy change of formation?

A. $\ce{2H2(g) + O2(g) -> 2H2O(l)}$

B. $\ce{C(s) + O2(g) -> CO2(g)}$

C. $\ce{H2(g) + Cl2(g) -> 2HCl(g)}$

D. $\ce{Mg(g) + 1/2O2(g) -> MgO(s)}$

Solution

B. One mole of $\ce{CO2}$ is formed from carbon and oxygen, both in their standard states.

A forms two moles of water. C forms two moles of HCl. D uses $\ce{Mg(g)}$ , but magnesium's standard state is $\ce{Mg(s)}$ .

Reading a reaction pathway diagram

For the reaction $\ce{2SO2(g) + O2(g) -> 2SO3(g)}$ , $\Delta H = -197\ \text{kJ mol}^{-1}$ and the activation energy of the uncatalysed forward reaction is $+260\ \text{kJ mol}^{-1}$ (an illustrative value). Sketch the reaction pathway diagram and calculate the activation energy of the reverse reaction.

Solution

Sketch: enthalpy on the vertical axis, progress of reaction on the horizontal. Reactants $\ce{2SO2 + O2}$ on a higher level; products $\ce{2SO3}$ on a level $197\ \text{kJ mol}^{-1}$ lower; a curve rising to a peak $260\ \text{kJ mol}^{-1}$ above the reactants. Arrow $E_A$ up from reactants to the peak; arrow $\Delta H$ down from reactants to products.

The reverse reaction starts from the products, which are $197\ \text{kJ mol}^{-1}$ below the reactants, so its barrier is:

E_A(\text{reverse}) = 260 + 197 = +457\ \text{kJ mol}^{-1}

Exam-hard: interpreting neutralisation data

The enthalpy change of neutralisation for hydrochloric acid with sodium hydroxide is $-57.1\ \text{kJ mol}^{-1}$ , and for nitric acid with potassium hydroxide it is $-57.3\ \text{kJ mol}^{-1}$ . For ethanoic acid with sodium hydroxide it is $-55.2\ \text{kJ mol}^{-1}$ .

(a) Explain why the first two values are almost identical. (b) Explain why the value for ethanoic acid is less exothermic. (c) Predict the enthalpy change when $1.00\ \text{mol}$ of sulfuric acid is completely neutralised by sodium hydroxide.

Solution

(a) Strong acids and strong alkalis are fully dissociated in solution. In both reactions the only change is $\ce{H+(aq) + OH-(aq) -> H2O(l)}$ ; the other ions are spectators. The same reaction gives the same enthalpy change per mole of water.

(b) Ethanoic acid is a weak acid, only partially dissociated. As $\ce{H+}$ ions are neutralised, more $\ce{CH3COOH}$ molecules dissociate, and this dissociation is endothermic (energy is needed to break the O–H bond). Some of the energy released by forming water is used up, so the overall value is less negative.

(c) $\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}$ : one mole of sulfuric acid forms two moles of water, so $\Delta H \approx 2 \times (-57.1) \approx -114\ \text{kJ}$ (for the reaction as written).

Exam tip

Learn the four definitions word for word. Mark schemes typically split each into two marks: (1) the quantity ("one mole of compound", "one mole of substance", "one mole of water"); (2) the conditions ("from its elements in their standard states", "in excess oxygen", "under standard conditions").
Always include the sign and units of $\Delta H$ . A missing "+" on an endothermic value often costs the mark.
On reaction pathway diagrams, the $E_A$ arrow must start at the reactants line and the $\Delta H$ arrow must go from reactants to products. Label axes and both lines.
Standard conditions are $298\ \text{K}$ and $101\ \text{kPa}$ ; "room temperature and pressure" is not accepted.

Summary

$\Delta H$ : heat energy change at constant pressure. Exothermic: $\Delta H$ negative, surroundings warm up. Endothermic: $\Delta H$ positive, surroundings cool.
Bond breaking is endothermic; bond making is exothermic.
Reaction pathway diagram: enthalpy against progress; $E_A$ from reactants to peak; $\Delta H$ from reactants to products.
Activation energy: minimum energy required for a collision to be effective.
Standard conditions: $298\ \text{K}$ , $101\ \text{kPa}$ , standard states, $1.00\ \text{mol dm}^{-3}$ ; symbol $\ominus$ .
$\Delta H_r$ (amounts in the equation), $\Delta H_f$ (one mole of compound from elements in standard states), $\Delta H_c$ (one mole burnt in excess oxygen), $\Delta H_{\text{neut}}$ (one mole of water from acid and alkali).
$\Delta H_f$ of an element in its standard state is zero.

Practice

Question

Define (a) standard enthalpy change of formation, (b) standard enthalpy change of combustion.
State the standard conditions used for enthalpy changes.
Write an equation for the standard enthalpy change of formation of (a) $\ce{NH3(g)}$ , (b) $\ce{C6H12O6(s)}$ , (c) $\ce{Al2O3(s)}$ .
Write an equation for the standard enthalpy change of combustion of (a) methanol, $\ce{CH3OH(l)}$ , (b) butane.
Explain why the standard enthalpy change of formation of oxygen gas is zero.
Sketch and label a reaction pathway diagram for the endothermic decomposition $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ , $\Delta H = +178\ \text{kJ mol}^{-1}$ .
Explain, in terms of bonds, why combustion reactions are exothermic.
The enthalpy change of the reaction $\ce{C(s) + 1/2O2(g) -> CO(g)}$ is $-111\ \text{kJ mol}^{-1}$ . State, with a reason, which standard enthalpy change(s) this represents.
For a reaction with $\Delta H = +45\ \text{kJ mol}^{-1}$ , the activation energy of the reverse reaction is $+80\ \text{kJ mol}^{-1}$ . Calculate the activation energy of the forward reaction and sketch the diagram.
$50.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ hydrochloric acid is mixed with $50.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ barium hydroxide, $\ce{Ba(OH)2}$ . Using $\Delta H_{\text{neut}} = -57.1\ \text{kJ mol}^{-1}$ , calculate the heat energy released, and write the equation for the reaction whose enthalpy change equals $\Delta H_{\text{neut}}$ for this acid and alkali.

Answers

(a) The enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. (b) The enthalpy change when one mole of a substance is burnt in an excess of oxygen under standard conditions.
$298\ \text{K}$ and $101\ \text{kPa}$ , substances in their standard states, solutions at $1.00\ \text{mol dm}^{-3}$ .
(a) $\ce{1/2N2(g) + 3/2H2(g) -> NH3(g)}$ . (b) $\ce{6C(s) + 6H2(g) + 3O2(g) -> C6H12O6(s)}$ . (c) $\ce{2Al(s) + 3/2O2(g) -> Al2O3(s)}$ .
(a) $\ce{CH3OH(l) + 3/2O2(g) -> CO2(g) + 2H2O(l)}$ . (b) $\ce{C4H10(g) + 13/2O2(g) -> 4CO2(g) + 5H2O(l)}$ .
Oxygen gas is an element in its standard state. Forming it "from its elements in their standard states" means forming $\ce{O2(g)}$ from $\ce{O2(g)}$ : there is no change, so no enthalpy change.
Axes: enthalpy (vertical), progress of reaction (horizontal). $\ce{CaCO3(s)}$ on a lower line; $\ce{CaO(s) + CO2(g)}$ on a line $178\ \text{kJ mol}^{-1}$ higher. A curve rising from reactants to a peak above the products level, then falling to the products line. $E_A$ arrow from reactants up to the peak; $\Delta H = +178\ \text{kJ mol}^{-1}$ arrow pointing up from reactants to products.
Energy is needed to break the bonds in the fuel and oxygen, but more energy is released when the very strong bonds in $\ce{CO2}$ (C=O) and $\ce{H2O}$ (O–H) are formed. Energy released by bond making exceeds energy absorbed by bond breaking, so overall energy is released.
It is the standard enthalpy change of formation of carbon monoxide: one mole of CO formed from its elements in their standard states. It is not the enthalpy change of combustion of carbon, because combustion is complete (carbon is burnt in excess oxygen to $\ce{CO2}$ ).
For an endothermic reaction the products are $45\ \text{kJ mol}^{-1}$ above the reactants. $E_A(\text{reverse}) = E_A(\text{forward}) - \Delta H$ , so $E_A(\text{forward}) = 80 + 45 = +125\ \text{kJ mol}^{-1}$ . Sketch: products line higher than reactants by 45; peak 125 above reactants and 80 above products.
$n(\ce{HCl}) = 0.0500 \times 2.00 = 0.100\ \text{mol}$ of $\ce{H+}$ ; $n(\ce{Ba(OH)2}) = 0.0500 \times 1.00 = 0.0500\ \text{mol}$ , giving $0.100\ \text{mol}$ of $\ce{OH-}$ . So $0.100\ \text{mol}$ of water forms. Heat released $= 0.100 \times 57.1 = 5.71\ \text{kJ}$ . Equation for one mole of water: $\ce{HCl(aq) + 1/2Ba(OH)2(aq) -> 1/2BaCl2(aq) + H2O(l)}$ .

Enthalpy and enthalpy change

Exothermic and endothermic reactions

Where the energy comes from

Reaction pathway diagrams

Standard conditions

The four standard enthalpy changes

Enthalpy change of reaction

Enthalpy change of formation

Enthalpy change of combustion

Enthalpy change of neutralisation

Worked examples

Practice

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