Thermodynamics & Thermochemistry
The bookkeeping of energy in chemical change — heat, work, enthalpy, entropy, and the free energy that decides what happens
- How to define a system, distinguish state from path functions, and apply the first law \(\Delta U=q+w\).
- The meaning of enthalpy and the link \(\Delta H=\Delta U+\Delta n_g RT\).
- The two heat capacities and the relation \(C_P-C_V=R\).
- How to combine reactions with Hess's law and use formation and bond enthalpies.
- The idea of entropy and the second law of thermodynamics.
- How Gibbs free energy \(\Delta G=\Delta H-T\Delta S\) decides spontaneity.
System, Surroundings & State Functions
Thermodynamics begins by drawing a boundary. The system is the part of the universe under study; everything else is the surroundings. A system is open (exchanges matter and energy), closed (energy only) or isolated (neither).
A state function depends only on the present state, not the route taken — internal energy, enthalpy, entropy, pressure, temperature. A path function depends on how the change was carried out — heat and work. Properties are extensive if they scale with amount (mass, volume) or intensive if they do not (temperature, density).
The First Law of Thermodynamics
Energy is conserved. The internal energy \(U\) of a system changes only when heat is exchanged or work is done. Taking heat absorbed by the system and work done on the system as positive (the IUPAC convention):
Here \(q>0\) when heat flows in and \(w>0\) when work is done on the system (compression). For an expansion against an external pressure, \(w=-P_{\text{ext}}\Delta V\) is negative — the system spends energy pushing back the surroundings.
Older texts write \(\Delta U=q-w\), taking \(w\) as work done by the system. Both are correct if used consistently — but mixing them is the single most common error in this chapter. This book uses \(\Delta U=q+w\) with \(w\) done on the system.
Work of Expansion
When a gas expands, it does pressure–volume work. For an irreversible expansion against a constant external pressure, \(w=-P_{\text{ext}}\Delta V\). For a reversible isothermal expansion of an ideal gas — done in infinitesimal steps, always at equilibrium — the work is larger in magnitude:
Enthalpy
Most reactions happen at constant (atmospheric) pressure, not constant volume. The heat exchanged at constant pressure is captured by a new state function, the enthalpy \(H=U+PV\). At constant pressure \(q_P=\Delta H\), while at constant volume \(q_V=\Delta U\).
where \(\Delta n_g\) is the change in moles of gas. For reactions with no gas-mole change, \(\Delta H=\Delta U\). A reaction is exothermic if \(\Delta H<0\) (heat released) and endothermic if \(\Delta H>0\) (heat absorbed).
Heat Capacities
The heat capacity is the heat needed to raise a system's temperature by one kelvin. It differs at constant volume \((C_V)\) and constant pressure \((C_P)\), because at constant pressure some heat also does expansion work.
Thermochemical Equations
A thermochemical equation gives the enthalpy change for a balanced reaction with states specified. Several standard enthalpies recur, each defined per mole at \(298\ \text{K}\) and \(1\ \text{bar}\) (the standard state, marked \(^{\circ}\)).
| Enthalpy of… | Defined as the \(\Delta H\) when… |
|---|---|
| Formation \((\Delta H_f^{\circ})\) | 1 mol of a compound forms from its elements in their standard states |
| Combustion \((\Delta H_c^{\circ})\) | 1 mol of a substance burns completely in oxygen |
| Neutralisation | 1 mol of water forms from acid + base in dilute solution |
| Atomisation | 1 mol of gaseous atoms forms from the element |
| Solution | 1 mol of solute dissolves in a large excess of solvent |
By convention the standard enthalpy of formation of any element in its reference form (e.g. \(\ce{O2}(g)\), graphite) is zero.
Hess's Law of Constant Heat Summation
Because enthalpy is a state function, the total enthalpy change of a reaction is the same whether it happens in one step or many. This is Hess's law — and it lets us find enthalpies that are impossible to measure directly by adding known steps.
Bond Enthalpies
Breaking bonds costs energy; forming them releases it. For gas-phase reactions, the reaction enthalpy is the energy spent breaking reactant bonds minus the energy recovered forming product bonds.
Entropy & the Second Law
The first law tells us energy is conserved, but not which way a change will go. The second law supplies direction through entropy \(S\) — a measure of disorder, or of the number of ways energy and matter can be arranged.
For a reversible exchange of heat at temperature \(T\), \(\Delta S=\dfrac{q_{\text{rev}}}{T}\). Entropy rises when solids melt, liquids vaporise, gases are produced, or substances mix. The third law sets the entropy of a perfect crystal at \(0\ \text{K}\) to zero, giving entropies an absolute scale.
Gibbs Free Energy & Spontaneity
Tracking the entropy of the whole universe is awkward. Gibbs combined the system's own enthalpy and entropy into one criterion that needs only system quantities — the free energy \(G=H-TS\).
A process is spontaneous when \(\Delta G<0\), at equilibrium when \(\Delta G=0\), and non-spontaneous when \(\Delta G>0\). At equilibrium \(\Delta G^{\circ}=-RT\ln K\), tying free energy directly to the equilibrium constant.
| \(\Delta H\) | \(\Delta S\) | Spontaneity |
|---|---|---|
| \(-\) | \(+\) | Spontaneous at all temperatures |
| \(+\) | \(-\) | Non-spontaneous at all temperatures |
| \(-\) | \(-\) | Spontaneous at low temperature |
| \(+\) | \(+\) | Spontaneous at high temperature |
Putting It to Work
Problem. A system absorbs \(150\ \text{J}\) of heat and has \(50\ \text{J}\) of work done on it. Find \(\Delta U\).
Solution. Both heat in and work on the system are positive:
Problem. For \(\ce{N2(g) + 3H2(g) -> 2NH3(g)}\) at \(298\ \text{K}\), \(\Delta U=-92.2\ \text{kJ}\). Find \(\Delta H\).
Solution. Gas moles change from \(4\to2\), so \(\Delta n_g=-2\):
Problem. Given \(\ce{C(s) + O2 -> CO2}\), \(\Delta H=-393\), and \(\ce{CO + \tfrac12 O2 -> CO2}\), \(\Delta H=-283\ \text{kJ}\). Find \(\Delta H\) for \(\ce{C(s) + \tfrac12 O2 -> CO}\).
Solution. Subtract the second equation from the first:
Problem. Find \(\Delta H_{\text{rxn}}^{\circ}\) for \(\ce{CH4 + 2O2 -> CO2 + 2H2O}\) given \(\Delta H_f^{\circ}\) values \(\ce{CH4}=-74.8,\ \ce{CO2}=-393.5,\ \ce{H2O}=-285.8\ \text{kJ mol}^{-1}\).
Solution. Products minus reactants \((\Delta H_f^{\circ}\text{ of }\ce{O2}=0)\):
Problem. Estimate \(\Delta H\) for \(\ce{H2 + Cl2 -> 2HCl}\) given bond enthalpies \(\ce{H-H}=436,\ \ce{Cl-Cl}=242,\ \ce{H-Cl}=431\ \text{kJ mol}^{-1}\).
Solution. Bonds broken minus bonds formed:
Problem. A reaction has \(\Delta H=+30\ \text{kJ}\) and \(\Delta S=+100\ \text{J K}^{-1}\). Above what temperature does it become spontaneous?
Solution. Spontaneity needs \(\Delta G=\Delta H-T\Delta S<0\), i.e. \(T>\tfrac{\Delta H}{\Delta S}\):
Above \(300\ \text{K}\) the entropy term wins and the reaction proceeds.
Chapter Summary
Open, closed, isolated; state functions ignore path, heat and work do not.
\(\Delta U=q+w\); expansion work \(w=-P_{\text{ext}}\Delta V\).
\(q_P=\Delta H\); \(\Delta H=\Delta U+\Delta n_g RT\); \(C_P-C_V=R\).
Hess's law adds steps; formation and bond enthalpies give \(\Delta H_{\text{rxn}}\).
\(\Delta S_{\text{univ}}>0\) for spontaneity; \(\Delta S=q_{\text{rev}}/T\).
\(\Delta G=\Delta H-T\Delta S\); \(\Delta G<0\) spontaneous; \(\Delta G^{\circ}=-RT\ln K\).
Problems
State your sign convention, mark each quantity as a state or path function, and keep energy units consistent (use \(R=8.314\ \text{J mol}^{-1}\text{K}^{-1}\)). Difficulty rises down the list.
- A gas does \(40\ \text{J}\) of work on its surroundings while absorbing \(100\ \text{J}\) of heat. Find \(\Delta U\).
- Calculate the work done when \(1\ \text{mol}\) of an ideal gas expands isothermally and reversibly from \(1\ \text{L}\) to \(10\ \text{L}\) at \(300\ \text{K}\).
- For \(\ce{2CO(g) + O2(g) -> 2CO2(g)}\), \(\Delta H=-566\ \text{kJ}\). Find \(\Delta U\) at \(298\ \text{K}\).
- How much heat is needed to raise the temperature of \(2\ \text{mol}\) of a monatomic ideal gas by \(50\ \text{K}\) at constant volume?
- Using Hess's law, find \(\Delta H\) for \(\ce{C(graphite) -> C(diamond)}\) from the combustion enthalpies \(-393.5\) and \(-395.4\ \text{kJ mol}^{-1}\).
- Estimate \(\Delta H\) for the hydrogenation \(\ce{C2H4 + H2 -> C2H6}\) from bond enthalpies.
- The standard enthalpy of combustion of ethanol is \(-1367\ \text{kJ mol}^{-1}\). Find the heat released by burning \(23\ \text{g}\) of ethanol.
- Predict the sign of \(\Delta S\) for \(\ce{CaCO3(s) -> CaO(s) + CO2(g)}\) and explain.
- For a reaction \(\Delta H=-40\ \text{kJ}\) and \(\Delta S=-120\ \text{J K}^{-1}\). Below what temperature is it spontaneous?
- At \(298\ \text{K}\) a reaction has \(\Delta G^{\circ}=-11.4\ \text{kJ}\). Find its equilibrium constant \(K\).
- One mole of ice melts at \(273\ \text{K}\) with \(\Delta H_{\text{fus}}=6.0\ \text{kJ mol}^{-1}\). Find \(\Delta S\) for melting.
- A reaction at \(500\ \text{K}\) has \(\Delta H=+50\ \text{kJ}\) and \(K=4.0\). Find \(\Delta S\).