Thermodynamics of Chemical Reactions and Equations of State

Chapter 1 built the equations of state that describe how much volume a gas occupies at given pressure and temperature. This chapter asks a different question: how much energy moves between a chemical system and its surroundings as that state changes, and how is that energy accounted for exactly? Classical thermodynamics answers this with two state functions — internal energy $U$ and enthalpy $H$ — and two conservation/direction laws that govern them. The equations of state from Chapter 1 are not left behind: they reappear here as the engine that lets the abstract thermodynamic machinery produce real numbers, from the work done by an expanding van der Waals gas to the cooling of a real gas forced through a throttle.

You should have to hand the van der Waals equation, the critical constants, and the Boyle temperature from Chapter 1. This chapter develops the first law, enthalpy and heat capacities, the internal-pressure concept that measures real-gas non-ideality energetically, the Joule–Thomson effect, a compact but rigorous introduction to the second law and Gibbs energy, and Kirchhoff's law for the temperature dependence of reaction enthalpies. The next chapter uses the Gibbs energy introduced here to define the chemical potential and build the theory of solutions.

1. The First Law of Thermodynamics

The internal energy $U$ of a system is the total energy contained in it — translational, rotational, vibrational and potential energy of its constituent particles — and it is a state function: its value depends only on the current state of the system, not on the path by which that state was reached. The first law of thermodynamics asserts that $U$ changes only through heat and work crossing the system boundary:

using the IUPAC convention in which $q$ and $w$ are both positive when energy flows into the system. Unlike $U$, neither $q$ nor $w$ is a state function individually — both depend on the path taken between two states — but their sum, $dU$, does not. This is the content of the first law: energy is conserved, and internal energy bookkeeping does not care how the energy arrived.

Key distinction. $U$ is a state function (exact differential $dU$); $q$ and $w$ are path functions (inexact differentials, often written $\delta q, \delta w$). Only sums and differences of state functions between two fixed states are path-independent.

2. Reversible and Irreversible Work of Expansion

For a system doing only expansion work against an external pressure $p_{\text{ext}}$, the work done on the system as its volume changes by $dV$ is

the minus sign reflecting that the system loses energy as it expands ($dV>0$) against an opposing external pressure. Two limiting cases matter:

• Free expansion ($p_{\text{ext}}=0$, e.g. expansion into a vacuum): $w=0$ regardless of $\Delta V$.
• Reversible expansion: $p_{\text{ext}}$ is adjusted at every instant to equal the system's own pressure $p$ (quasi-statically, through a continuous sequence of near-equilibrium states). This extracts the maximum possible work from an expansion, since $p_{\text{ext}}=p$ is the largest external pressure the system can push against without the process becoming impossible ($p_{\text{ext}}>p$ would compress it instead).

For a reversible isothermal expansion of a perfect gas from $V_1$ to $V_2$ at temperature $T$, substitute $p=nRT/V$ and integrate:

For a van der Waals gas, substitute instead $p = \dfrac{nRT}{V-nb} - \dfrac{an^2}{V^2}$ (Chapter 1) and integrate term by term:

Worked example. One mole of CO$_2$ ($a=3.610~\text{dm}^6\,\text{atm}\,\text{mol}^{-2}$, $b=0.0429~\text{dm}^3\,\text{mol}^{-1}$) expands reversibly and isothermally at $300~\text{K}$ from $V_1=0.500~\text{dm}^3$ to $V_2=5.00~\text{dm}^3$. The perfect-gas prediction is

The van der Waals result uses $V_1-nb=0.4571$, $V_2-nb=4.9571$:

The van der Waals gas does less work expanding ($5.29~\text{kJ}$ against $5.74~\text{kJ}$) than the perfect-gas model predicts — a direct energetic consequence of the lower pressure that intermolecular attraction produces at every volume along the path, exactly the effect Chapter 1 quantified.

3. Enthalpy and Heat Capacities

Enthalpy is defined as $H \equiv U + pV$, a state function (being a combination of state functions) chosen so that, at constant pressure, $dH = dq_p$: enthalpy change equals heat absorbed at constant pressure, the condition under which most chemistry is actually done (open vessels, atmospheric pressure).

The heat capacity at constant volume, $C_V = \left(\dfrac{\partial U}{\partial T}\right)_V$, and the heat capacity at constant pressure, $C_p = \left(\dfrac{\partial H}{\partial T}\right)_p$, measure how much heat is needed per unit temperature rise under each constraint. For a perfect gas, $H = U + nRT$, so

Joule's law (justified in the next section) states that for a perfect gas $U$ depends on $T$ alone, so $(\partial U/\partial T)_p = (\partial U/\partial T)_V = C_V$, giving Mayer's relation:

4. Internal Pressure and the Thermodynamic Equation of State

The internal pressure, $\pi_T \equiv \left(\dfrac{\partial U}{\partial V}\right)_T$, measures how internal energy changes with volume at constant temperature — physically, the energy cost of pulling molecules apart against their mutual attraction (or pushing them together against repulsion), separate from any change in their kinetic energy. Joule attempted to measure this by letting a gas expand freely (adiabatically, into a vacuum, so $q=0$ and $w=0$, hence $\Delta U=0$) and looking for a temperature change; within the precision available to him, none was found for a near-ideal gas, establishing Joule's law: $\pi_T = 0$ for a perfect gas, i.e. $U = U(T)$ only.

A general thermodynamic identity — the thermodynamic equation of state, derivable by combining the first and second laws through a Maxwell relation (the result is quoted here without proof) — expresses the internal pressure in terms of the equation of state alone:

For a perfect gas, $p=nRT/V$ gives $\left(\dfrac{\partial p}{\partial T}\right)_V = nR/V$, so $T(\partial p/\partial T)_V = nRT/V = p$, and $\pi_T = p-p=0$ — recovering Joule's law exactly.

For a van der Waals gas, $p = \dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}$ gives $\left(\dfrac{\partial p}{\partial T}\right)_V = \dfrac{nR}{V-nb}$, so

(the last step uses the van der Waals equation itself to replace $nRT/(V-nb)$). Therefore

Internal energy does depend on volume for a van der Waals gas, growing as the gas expands — exactly as expected, since expanding against attractive forces (without any compensating change in kinetic energy, $T$ fixed) stores energy as increased separation of attracting molecules.

5. The General Relation Between $C_p$ and $C_V$

For any substance, not just a perfect gas, a parallel thermodynamic argument (again using the first and second laws and a Maxwell relation, quoted here without proof) gives the exact relation

For a perfect gas, $\pi_T=0$ and $p = nRT/V$ combine with $(\partial V/\partial T)_p = nR/p$ to reproduce $C_p-C_V = p\cdot nR/p = nR$, consistent with Section 3. For a real gas, $\pi_T \ne 0$ and $C_p - C_V$ departs from $nR$ — the correction is largest exactly where the gas is least ideal, linking this chapter's thermodynamics directly back to Chapter 1's equation-of-state corrections.

6. The Joule–Thomson Effect

A more sensitive experiment than Joule's free expansion is the Joule–Thomson (throttling) experiment: a gas is forced continuously through a porous plug or valve from high pressure to low pressure, with the whole apparatus thermally insulated. This process occurs at constant enthalpy ($\Delta H = 0$ for the throttled gas — a standard result of the steady-flow energy balance), and the temperature change observed defines the Joule–Thomson coefficient:

A perfect gas has $\mu_{\text{JT}}=0$ identically. A real gas does not, because $H=H(T,p)$ depends on $p$ through intermolecular forces; using $(\partial H/\partial p)_T = V_m - T(\partial V_m/\partial T)_p$ (the enthalpy analogue of the internal-pressure identity of Section 4) together with the low-pressure expansion of the van der Waals equation, $V_m \approx RT/p + B(T)$ with $B(T) = b - a/RT$ (Chapter 1), gives after differentiation and simplification:

This changes sign at the Joule–Thomson inversion temperature, $T_{\text{inv}} = \dfrac{2a}{Rb}$ — exactly twice the Boyle temperature $T_B = a/Rb$ found in Chapter 1. Below $T_{\text{inv}}$, $\mu_{\text{JT}}>0$: throttling cools the gas (attraction dominates — expanding against attraction absorbs energy, which must come from the gas's own kinetic energy, lowering $T$). Above $T_{\text{inv}}$, $\mu_{\text{JT}}<0$: throttling heats the gas (the excluded-volume effect dominates).

Worked example. For N$_2$ ($a=1.352~\text{dm}^6\,\text{atm}\,\text{mol}^{-2}$, $b=0.0387~\text{dm}^3\,\text{mol}^{-1}$, $C_{p,m}\approx29.1~\text{J}\,\text{mol}^{-1}\text{K}^{-1} = 0.2872~\text{dm}^3\,\text{atm}\,\text{mol}^{-1}\text{K}^{-1}$) at $T=300~\text{K}$:

a positive coefficient, meaning N$_2$ cools on throttling at room temperature — consistent with its measured value and with $T=300~\text{K}$ being far below its inversion temperature, $T_{\text{inv}} = 2a/Rb = 2(425.7~\text{K}) = 851.4~\text{K}$ (twice the Boyle temperature computed in Chapter 1).

7. The Second Law, Entropy and Gibbs Energy

The first law forbids energy from being created or destroyed but says nothing about which processes actually occur spontaneously — a gas never spontaneously compresses itself into one corner of a room, even though doing so would violate no energy conservation law. The second law of thermodynamics supplies the missing direction, through a new state function, entropy $S$, defined via $dS = \dfrac{dq_{\text{rev}}}{T}$ for a reversible path between two states (entropy, like $U$ and $H$, is path-independent even though $q$ is not, because the reversible-path definition is shown to be unique to the two end states). The Clausius inequality, in its original system-level form, states that for any real (possibly irreversible) process exchanging heat $dq$ with its surroundings at temperature $T$,

with equality only in the reversible limit. Adding the corresponding entropy change of the surroundings, $dS_{\text{surroundings}} = -dq/T$ for the actual heat transferred, gives the equivalent universe-level statement used throughout this chapter,

again with equality only for a reversible process — spontaneous processes always increase the entropy of the universe. For a reversible isothermal expansion of a perfect gas, $dq_{\text{rev}} = -dw = nRT\,dV/V$, so $\Delta S = nR\ln(V_2/V_1)$, positive for an expansion, consistent with the intuitive picture of entropy as a measure of the number of microstates accessible to a system (more volume, more accessible positions, higher entropy).

The second-law criterion $dS_{\text{universe}}\ge0$ is a statement about the universe (system plus surroundings), which is often inconvenient for predicting the behaviour of a chemical system alone at constant temperature and pressure. The Gibbs energy, $G \equiv H - TS$, is constructed precisely to repackage the second law as a statement about the system alone: at constant $T$ and $p$, it can be shown that $dS_{\text{universe}} \ge 0$ is equivalent to

A process at constant temperature and pressure is spontaneous if and only if it decreases the Gibbs energy of the system; equilibrium is reached when $G$ is at a minimum ($dG=0$). This single criterion — requiring no bookkeeping of the surroundings at all — is the practical workhorse of chemical thermodynamics, and the next chapter builds the chemical potential, $\mu_i = (\partial G/\partial n_i)_{T,p,n_{j\ne i}}$, directly on top of it to describe mixtures and solutions.

8. Standard Reaction Enthalpies and Kirchhoff's Law

The standard reaction enthalpy, $\Delta_rH^{\ominus}$, is the enthalpy change when reactants in their standard states (pure, at $1~\text{bar}$, in their stated physical state) are converted completely to products in their standard states, per mole of reaction as written. Because $H$ is a state function, $\Delta_rH^{\ominus}$ can be built from tabulated standard enthalpies of formation, $\Delta_rH^{\ominus} = \sum \nu_i \Delta_fH^{\ominus}(\text{products}) - \sum \nu_i \Delta_fH^{\ominus}(\text{reactants})$ (Hess's law), independent of the reaction path.

Reaction enthalpies are tabulated at a reference temperature (conventionally $298~\text{K}$) but chemical processes often occur elsewhere. Since $H$ is a state function, differentiating $\Delta_rH = \sum\nu_iH_i(\text{products}) - \sum\nu_iH_i(\text{reactants})$ with respect to $T$ at constant $p$, and using $(\partial H_i/\partial T)_p = C_{p,i}$ for each species, gives Kirchhoff's law:

If $\Delta_rC_p$ is approximately constant over the temperature range of interest, this integrates to $\Delta_rH(T_2) = \Delta_rH(T_1) + \Delta_rC_p(T_2-T_1)$.

Worked example. For the Haber process, $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$, $\Delta_rH^{\ominus}(298~\text{K}) = -92.2~\text{kJ}\,\text{mol}^{-1}$. Using $C_{p,m}(\text{N}_2)=29.1$, $C_{p,m}(\text{H}_2)=28.8$, $C_{p,m}(\text{NH}_3)=35.1~\text{J}\,\text{mol}^{-1}\text{K}^{-1}$:

The reaction becomes noticeably more exothermic at higher temperature in this (approximate, constant-$\Delta C_p$) treatment, because the products have a lower combined heat capacity than the reactants.

9. Common Exam Pitfalls

• Dropping the minus sign in $dw=-p_{\text{ext}}dV$, or confusing the IUPAC sign convention ($w>0$ when work is done on the system) with the older "engineering" convention. State which convention is in use and stay consistent throughout a calculation.
• Using $p$ (the system's own pressure) instead of $p_{\text{ext}}$ in an irreversible expansion. They coincide only in the reversible limit; an irreversible expansion against a constant external pressure uses $w=-p_{\text{ext}}\Delta V$ directly, no integration needed.
• Assuming $C_p - C_V = nR$ for a real gas. Mayer's relation is a perfect-gas result; the general relation of Section 5 involves the internal pressure $\pi_T$ and reduces to $nR$ only when $\pi_T=0$.
• Forgetting that $\mu_{\text{JT}}$ can be positive or negative. Below the inversion temperature throttling cools a gas (used industrially to liquefy gases); above it, throttling heats the gas — the sign is not fixed, and depends on $T$ relative to $T_{\text{inv}}=2a/Rb$.
• Using $dS_{\text{system}}\ge0$ alone as the spontaneity criterion. The Clausius inequality requires $dS_{\text{universe}}=dS_{\text{system}}+dS_{\text{surroundings}}\ge0$; the Gibbs energy criterion $dG\le0$ is what packages this into a system-only statement, but only at constant $T$ and $p$.
• Applying Kirchhoff's law with reactant/product $C_p$ values swapped or with the wrong stoichiometric coefficients. $\Delta_rC_p$ follows the same "products minus reactants, each weighted by its stoichiometric coefficient" pattern as $\Delta_rH$ itself.

Summary

• The first law, $dU=dq+dw$, makes $U$ a state function; $dw=-p_{\text{ext}}dV$ for expansion work, maximised in magnitude for a reversible process where $p_{\text{ext}}=p$ at every instant.
• Reversible isothermal work: $w=-nRT\ln(V_2/V_1)$ for a perfect gas; $w=-nRT\ln\!\left[\frac{V_2-nb}{V_1-nb}\right]+an^2\!\left(\frac1{V_1}-\frac1{V_2}\right)$ for a van der Waals gas.
• Enthalpy $H=U+pV$ gives $dH=dq_p$; Mayer's relation $C_p-C_V=nR$ holds for a perfect gas only.
• Internal pressure $\pi_T=(\partial U/\partial V)_T = T(\partial p/\partial T)_V - p$ is zero for a perfect gas (Joule's law) and equals $a/V_m^2$ for a van der Waals gas.
• The general relation $C_p-C_V=(\pi_T+p)(\partial V/\partial T)_p$ reduces to Mayer's relation only when $\pi_T=0$.
• The Joule–Thomson coefficient $\mu_{\text{JT}}=(\partial T/\partial p)_H \approx (2a/RT-b)/C_{p,m}$ changes sign at the inversion temperature $T_{\text{inv}}=2a/Rb=2T_B$.
• The second law, $dS_{\text{universe}}\ge0$, and the Gibbs energy, $G=H-TS$ with spontaneity criterion $dG\le0$ at constant $T,p$, set up the chemical potential of the next chapter.
• Kirchhoff's law, $(\partial\Delta_rH/\partial T)_p=\Delta_rC_p$, gives the temperature dependence of reaction enthalpies from heat-capacity data.