Enthalpy and heat capacity

Almost no chemistry and almost no engineering happens in a sealed rigid box, which is awkward, because the First Law is written for internal energy and internal energy is what a sealed rigid box measures.

Heat capacity is not one number

Ask how much heat it takes to warm something by a degree and the answer is a heat capacity, $C = δ Q / d T$ . The notation already contains a warning. Heat is written $δ Q$ rather than $d Q$ because it is not the change in any property of the system: it depends on the path taken, as the previous lesson established. So a ratio built out of it cannot be a property either, not until the path is pinned down.

Pin it down two ways and you get the two heat capacities that matter. Hold the volume fixed and no $p d V$ work is possible, so the First Law $d U = δ Q - p d V$ collapses to $δ Q = d U$ , and

C_{V} = {(\frac{\partial U}{\partial T})}_{V}

This is a genuine partial derivative of a state function, so it is a property of the substance. Hold the pressure fixed instead and the system may expand while it warms, doing work on its surroundings, so some of the heat supplied never becomes internal energy at all. The constant-pressure heat capacity $C_{p}$ is therefore a different number, and it must be the larger of the two.

Both are extensive: double the sample and you double the capacity. Divide by mass to get specific heat capacity (liquid water is $4.18$ J g⁻¹ K⁻¹, which is remarkably large and is why oceans moderate climate) or by amount of substance to get molar heat capacity, in J mol⁻¹ K⁻¹. Molar values are the ones to compare across substances, because they count the same number of particles each time.

Enthalpy, and why the combination is worth a name

Now do the constant-pressure case properly rather than by hand-waving. Take a system at fixed external pressure $p$ , doing only expansion work. Over a finite change the First Law gives $Δ U = Q_{p} - p Δ V$ , since $p$ is constant and comes out of the integral. Rearranged,

Q_{p} = Δ U + p Δ V = (U_{2} + p V_{2}) - (U_{1} + p V_{1})

The heat is the change in the quantity $U + p V$ , evaluated at the two end states and nowhere else. That combination is worth naming, so define the enthalpy

H = U + p V

Every symbol on the right is a state function, so $H$ is one too, and the result just derived reads $Δ H = Q_{p}$ : at constant pressure, with only expansion work, the heat absorbed is the change in a state function. That is the whole point. Heat is normally path-dependent, but constrain the path to constant pressure and the ambiguity disappears, because the $p Δ V$ term that the surroundings absorb has been folded into the bookkeeping in advance.

This is why enthalpy dominates chemistry. A reaction in an open flask is at constant pressure, held there by the atmosphere, and the heat it gives out is therefore a property of the reaction rather than of the flask. Enthalpy is not a new form of energy and nothing is stored in the $p V$ term in any physical sense: it is the correct accounting for the commonest constraint. The corresponding heat capacity is

C_{p} = {(\frac{\partial H}{\partial T})}_{p}

which stands in exactly the same relation to $H$ as $C_{V}$ does to $U$ .

Why $C_{p}$ exceeds $C_{V}$

For an ideal gas the difference can be derived exactly, with no appeal to intuition. Start from the definition and substitute the equation of state: $H = U + p V = U + n R T$ . Differentiate with respect to temperature at constant pressure. The internal energy of an ideal gas depends on temperature alone, a fact Joule's expansion experiment established, so $d U / d T$ is $C_{V}$ whichever variable is held fixed, and the second term differentiates to $n R$ . Hence

C_{p} - C_{V} = n R

or per mole, $C_{p, m} - C_{V, m} = R = 8.314$ J mol⁻¹ K⁻¹. The physical reading is the one anticipated above. Warm a gas by one kelvin at constant volume and all the heat raises $U$ . Warm it by one kelvin at constant pressure and it must expand by $Δ V = n R Δ T / p$ to keep the pressure constant, doing work $p Δ V = n R Δ T$ on the surroundings. The extra $R$ per mole per kelvin is exactly that work.

Example. $2.00$ mol of nitrogen ( $C_{V, m} = 20.81$ J mol⁻¹ K⁻¹, $C_{p, m} = 29.12$ J mol⁻¹ K⁻¹) is warmed from $300$ K to $350$ K, once at constant volume and once at constant pressure. How much heat does each path take, and where does the difference go?

At constant volume, $Q_{V} = n C_{V, m} Δ T = 2.00 \times 20.81 \times 50 = 2081$ J. At constant pressure, $Q_{p} = n C_{p, m} Δ T = 2.00 \times 29.12 \times 50 = 2912$ J. The difference, $2912 - 2081 = 831$ J, is exactly $n R Δ T = 2.00 \times 8.314 \times 50 = 831$ J: the work the expanding gas does on the surroundings to hold its pressure steady.

Now you. $3.00$ mol of argon ( $C_{V, m} = 12.47$ J mol⁻¹ K⁻¹, $C_{p, m} = 20.79$ J mol⁻¹ K⁻¹) is warmed by $50$ K. Find $Q_{V}$ , $Q_{p}$ , and check that their difference is $n R Δ T$ .

Answer

$Q_{V} = 3.00 \times 12.47 \times 50 = 1871$ J and $Q_{p} = 3.00 \times 20.79 \times 50 = 3119$ J. The difference is $3119 - 1871 = 1248$ J, matching $n R Δ T = 3.00 \times 8.314 \times 50 = 1247$ J to within rounding.

The ratio $γ = C_{p} / C_{V}$ appears throughout gas dynamics and in the adiabatic relation $p V^{γ} = constant$ , so it is worth tracking alongside the capacities themselves.

For a solid or a liquid the same difference exists but is tiny, because the thermal expansion is tiny: copper's $C_{p}$ and $C_{V}$ differ by under two per cent at room temperature. For gases the difference is never negligible, and using the wrong one is a standard way to be wrong by forty per cent.

Equipartition and the size of $C_{V}$

Kinetic theory gives the values, not merely the difference. The equipartition theorem of classical statistical mechanics says that each independent quadratic term in a molecule's energy carries an average of $\frac{1}{2} k_{B} T$ , and therefore contributes $\frac{1}{2} R$ per mole to $C_{V, m}$ .

A monatomic gas has three such terms, the kinetic energies along $x$ , $y$ and $z$ . So $C_{V, m} = \frac{3}{2} R = 12.47$ J mol⁻¹ K⁻¹, $C_{p, m} = \frac{5}{2} R = 20.79$ , and $γ = 5 / 3 \approx 1.67$ . Argon at $298$ K measures $C_{V, m} = 12.5$ and $γ = 1.67$ . The agreement is essentially exact, and it holds for helium, neon and krypton too.

A diatomic molecule adds rotation about the two axes perpendicular to the bond (rotation about the bond itself involves negligible moment of inertia) for two more terms, predicting $C_{V, m} = \frac{5}{2} R = 20.79$ and $γ = 7 / 5 = 1.40$ . Nitrogen at $298$ K measures $C_{p, m} = 29.12$ J mol⁻¹ K⁻¹, hence $C_{V, m} = 29.12 - 8.31 = 20.81$ , and $γ = 1.40$ . Again the prediction lands.

Carbon dioxide breaks the pattern. It is linear, so on the same counting it should also give $\frac{5}{2} R$ . Its measured $C_{p, m}$ at $298$ K is $37.1$ J mol⁻¹ K⁻¹, giving $C_{V, m} = 28.8$ and $γ = 1.29$ . Something is absorbing energy that the count of translations and rotations does not include.

Where the classical picture fails

The missing contribution is vibration, and the interesting question is not why carbon dioxide has it but why nitrogen does not. A vibrating bond has two quadratic terms, kinetic and potential, so classical equipartition says every diatomic should show $C_{V, m} = \frac{7}{2} R = 29.1$ at all temperatures. Nitrogen shows $\frac{5}{2} R$ . The classical theory is not slightly off here; it predicts a contribution that is simply absent.

Cooling makes it worse. Hydrogen's molar $C_{V}$ is $\frac{5}{2} R$ near room temperature, falls as it is cooled below about $100$ K, and reaches $\frac{3}{2} R$ near $60$ K, the value for a monatomic gas. The rotational contribution vanishes. Nothing in classical mechanics permits a degree of freedom to switch off: a mode either exists or it does not.

Quantum mechanics supplies what is missing. The energy of each mode is quantised, and a mode contributes fully only when $k_{B} T$ is comfortably larger than its level spacing, and freezes out when it is not. The characteristic temperature of a mode is the spacing divided by $k_{B}$ . Hydrogen's rotational spacing corresponds to about $85$ K, which is why its rotation dies at the temperature it does. Nitrogen's bond vibrates at $2359$ cm⁻¹, a characteristic temperature near $3390$ K, so at $298$ K the mode is almost entirely in its ground state and contributes nothing. Carbon dioxide's bending vibration sits at only $667$ cm⁻¹, around $960$ K, low enough to be partly excited at room temperature, which is precisely the excess seen in its $C_{p}$ . Heat capacity is one of the places where the quantum nature of matter is visible in a bench measurement.

Formation enthalpies and Hess's law

Enthalpy has no absolute zero, so what gets tabulated is always a difference. The convention fixes one: the standard enthalpy of formation $Δ_{f} H^{ominus}$ is the enthalpy change forming one mole of a substance from its elements in their standard states at $10^{5}$ Pa, and an element in its standard state is assigned zero by definition. Graphite is zero and diamond is $+ 1.9$ kJ mol⁻¹, because graphite is the stable form.

Because $H$ is a state function, $Δ H$ around any closed path is zero, and a reaction enthalpy is the same whether the reaction runs in one step or twenty. That is Hess's law, stated by Germain Hess in $1840$ , before the First Law itself was settled. It is not an extra postulate. It is what "state function" means, applied to chemistry, and it lets a reaction enthalpy be computed for a reaction nobody can run cleanly, by routing through elements:

Δ_{r} H^{ominus} = \sum Δ_{f} H^{ominus} (products) - \sum Δ_{f} H^{ominus} (reactants)

Take the combustion of methane, $C(g) + 2 (g) \to C(g) + 2 O(l)$ . The tabulated values at $298$ K are $Δ_{f} H^{ominus} = - 74.6$ kJ mol⁻¹ for methane, $- 393.5$ for carbon dioxide, $- 285.8$ for liquid water, and zero for oxygen, an element in its standard state. The products sum to $- 393.5 + 2 (- 285.8) = - 965.1$ kJ mol⁻¹ and the reactants to $- 74.6$ , so $Δ_{r} H^{ominus} = - 965.1 - (- 74.6) = - 890.5$ kJ mol⁻¹, against a directly measured value of $- 890.8$ . Note that the water must be liquid: taking it as vapour changes the answer by $2 \times 44 = 88$ kJ, the difference between the higher and lower heating values of natural gas, and a real source of confusion in engineering data.

Example. Find the standard enthalpy of combustion of ethanol, $OH(l) + 3 (g) \to 2 C(g) + 3 O(l)$ , given $Δ_{f} H^{ominus} = - 277.7$ kJ mol⁻¹ for liquid ethanol.

The products sum to $2 (- 393.5) + 3 (- 285.8) = - 1644.4$ kJ mol⁻¹ and the reactants to $- 277.7$ , oxygen being zero. So $Δ_{r} H^{ominus} = - 1644.4 - (- 277.7) = - 1366.7$ kJ mol⁻¹, against a measured $- 1366.8$ .

Now you. Do the same for propane, $(g) + 5 (g) \to 3 C(g) + 4 O(l)$ , given $Δ_{f} H^{ominus} = - 104.7$ kJ mol⁻¹ for propane.

Answer

Products: $3 (- 393.5) + 4 (- 285.8) = - 2323.7$ kJ mol⁻¹. Reactants: $- 104.7$ . So $Δ_{r} H^{ominus} = - 2323.7 - (- 104.7) = - 2219.0$ kJ mol⁻¹.

The Born-Haber cycle for sodium chloride is the same argument applied to an ionic lattice, where it extracts a lattice enthalpy that cannot be measured directly at all. It is worked through in lesson six of the Atoms and Elements course, and there is no point repeating it here.

Calorimetry: two vessels, two quantities

Measuring these numbers means choosing which constraint to impose. A bomb calorimeter seals the sample in a rigid steel vessel under excess oxygen and immerses it in a stirred water bath. The volume is fixed, so no work is done and the measured heat is $Δ U$ , not $Δ H$ . A coffee-cup calorimeter, an insulated vessel open to the atmosphere, holds the pressure fixed instead and measures $Δ H$ directly, which suits dissolutions and neutralisations.

A bomb is calibrated rather than computed, by burning a substance whose energy of combustion is known to high precision. Benzoic acid is the international standard for this, at $- 26.43$ kJ g⁻¹. Burning $1.000$ g of it and observing the bath rise by $2.641$ K gives a calorimeter constant $C_{cal} = 26.43 / 2.641 = 10.01$ kJ K⁻¹. Now burn $1.000$ g of glucose in the same apparatus and suppose the rise is $1.555$ K. The heat released is $10.01 \times 1.555 = 15.57$ kJ, and glucose has molar mass $180.16$ g mol⁻¹, so $Δ_{c} U = - 15.57 \times 180.16 = - 2805$ kJ mol⁻¹.

To convert, note that $Δ H = Δ U + Δ (p V)$ , and for a reaction where the condensed phases contribute negligible volume, only the gases matter: $Δ (p V) = Δ n_{gas} R T$ if they are ideal. So

Δ H = Δ U + Δ n_{gas} R T

For glucose, $(s) + 6 (g) \to 6 C(g) + 6 O(l)$ , six moles of gas are consumed and six produced, so $Δ n_{gas} = 0$ and $Δ_{c} H = Δ_{c} U = - 2805$ kJ mol⁻¹. The value from formation enthalpies is $6 (- 393.5) + 6 (- 285.8) - (- 1273.3) = - 2802.7$ kJ mol⁻¹, which agrees to within the precision of the temperature reading.

Methane shows the correction biting. There $Δ n_{gas} = 1 - 3 = - 2$ , and $R T = 8.314 \times 298.15 = 2479$ J mol⁻¹, so $Δ_{c} U = Δ_{c} H - Δ n_{gas} R T = - 890.5 + 4.96 = - 885.5$ kJ mol⁻¹. Half a per cent, which is far larger than a good calorimeter's error, so the conversion is never optional.

Example. Ethanol burns as $OH(l) + 3 (g) \to 2 C(g) + 3 O(l)$ with $Δ_{c} H = - 1366.7$ kJ mol⁻¹ at $298.15$ K. What would a bomb calorimeter measure?

The gas count is $2 - 3$ , so $Δ n_{gas} = - 1$ , and $R T = 2.479$ kJ mol⁻¹. The bomb measures $Δ_{c} U = Δ_{c} H - Δ n_{gas} R T = - 1366.7 - (- 1) (2.479) = - 1364.2$ kJ mol⁻¹.

Now you. Propane burns as $(g) + 5 (g) \to 3 C(g) + 4 O(l)$ with $Δ_{c} H = - 2219.0$ kJ mol⁻¹ at $298.15$ K. Find $Δ_{c} U$ .

Answer

Gas moles go from $1 + 5 = 6$ to $3$ , so $Δ n_{gas} = - 3$ and $Δ_{c} U = - 2219.0 - (- 3) (2.479) = - 2211.6$ kJ mol⁻¹.

Exothermic is not the same as spontaneous

A reaction with $Δ H < 0$ releases heat and is exothermic; one with $Δ H > 0$ absorbs it and is endothermic. It is very tempting to go one step further and say that reactions happen because they release energy, and for most of the nineteenth century respectable chemists did say exactly that. Marcellin Berthelot's principle of maximum work, stated in $1867$ , held that every spontaneous change is the one that releases the most heat.

It is false, and the counterexamples are on any bench. Dissolve ammonium nitrate in water and it dissolves eagerly while cooling the beaker sharply: $Δ_{sol} H^{ominus} = + 25.7$ kJ mol⁻¹, which is how instant cold packs work. Put ice in a room at $10$ degrees Celsius and it melts, absorbing $6.01$ kJ mol⁻¹, entirely spontaneously. Both processes go uphill in enthalpy and both go anyway.

So enthalpy cannot be the criterion for change. Something else is being maximised alongside it, something that the dissolved ions and the liquid water have more of than the crystal did, and it is not energy. Naming that quantity, measuring it, and combining it with $Δ H$ into a single criterion is the work of the second half of this course. The Second Law arrives first, and entropy after it.