Thermodynamics

In this chapter: system and surroundings, state functions, internal energy, heat and work, the first law, pressure-volume work, reversible and free expansion, enthalpy and the relation between ΔH and ΔU, extensive and intensive properties, heat capacity and calorimetry, standard enthalpies (phase change, formation, combustion, atomisation, bond, lattice, solution), Hess's law, entropy, Gibbs energy and its link with equilibrium.

Basic terms

• System is the part of the universe we study; everything else is the surroundings. An open system exchanges both energy and matter with the surroundings (reactants in an open beaker), a closed system exchanges energy but not matter (a stoppered flask), and an isolated system exchanges neither (an ideal thermos flask).
• The state of a system is described by state variables such as p, V, T and n. A state function depends only on the initial and final states, not on the path: U, H, S, G, p, V and T are state functions. Heat (q) and work (w) depend on the path.

Internal energy and the first law

Internal energy U is the total energy of the system (chemical, electrical, mechanical). It changes when heat passes into or out of the system, or when work is done on or by it. In an adiabatic process there is no heat exchange, so ΔU = w_ad.

This is the first law of thermodynamics: the energy of an isolated system is constant. Energy can change form but can be neither created nor destroyed.

Pressure-volume work

• In free expansion (into vacuum, p_ex = 0) no work is done, whether the process is reversible or not.
• For an ideal gas at constant temperature ΔU = 0, so q = −w. For isothermal free expansion of an ideal gas, q = 0, w = 0 and ΔU = 0.
• For an adiabatic change, q = 0 and ΔU = w_ad.

Worked example: 1 mol of an ideal gas expands isothermally and reversibly at 300 K from 10 L to 20 L. Find w, q and ΔU. (R = 8.314 J K⁻¹ mol⁻¹)
Solution: w = −2.303 × 1 × 8.314 × 300 × log 2 = −2.303 × 2494.2 × 0.3010 = −1729 J ≈ −1.73 kJ. The temperature is constant, so ΔU = 0 and q = −w = +1.73 kJ. The gas absorbs heat and uses all of it to do work on the surroundings.

Enthalpy

Most reactions are carried out at constant (atmospheric) pressure, so we define enthalpy H = U + pV. At constant volume the heat change equals ΔU (q_V = ΔU); at constant pressure it equals ΔH (q_p = ΔH). For reactions involving gases,

Worked example: For N₂(g) + 3H₂(g) → 2NH₃(g), Δ_rH = −92.4 kJ mol⁻¹ at 298 K. Find Δ_rU.
Solution: Δn_g = 2 − 4 = −2. Δ_rU = ΔH − Δn_gRT = −92.4 − (−2)(8.314 × 10⁻³)(298) = −92.4 + 4.96 = −87.4 kJ mol⁻¹. Notice R is taken in kJ so the units match.

Extensive properties depend on the amount of matter: mass, volume, U, H, heat capacity. Intensive properties do not: temperature, pressure, density, molar heat capacity.

Heat capacity and calorimetry

Heat q = C ΔT, where C is the heat capacity. Per mole it is the molar heat capacity C_m; per gram it is the specific heat c, so q = mcΔT. For an ideal gas, C_p − C_V = R. A bomb calorimeter works at constant volume and measures ΔU; a calorimeter open to the atmosphere (constant pressure) measures ΔH.

Enthalpy change of a reaction

The standard state of a substance is its pure form at 1 bar and a specified temperature, usually 298 K; standard values carry the symbol °. A balanced equation with its Δ_rH and the physical states of all species is a thermochemical equation. If the equation is reversed, the sign of ΔH changes; if it is multiplied by a number, ΔH is multiplied too.

Hess's law of constant heat summation

If a reaction takes place in several steps, its standard enthalpy change is the sum of the enthalpy changes of the steps, at the same temperature. This follows from H being a state function. It lets us find enthalpies that cannot be measured directly, such as the enthalpy of formation of CO (carbon always burns partly to CO₂).

Spontaneity

A spontaneous process takes place on its own without outside help once it starts; it need not be fast. A negative ΔH favours spontaneity but does not decide it: ice melts above 0 °C and NH₄Cl dissolves in water although both are endothermic.

Entropy

Entropy S is a measure of the degree of randomness of a system. It increases from solid to liquid to gas, and when a gas is heated or a solid dissolves. For a reversible change at temperature T, ΔS = q_rev/T (unit J K⁻¹). For a spontaneous process in an isolated system, ΔS > 0; in general ΔS_total = ΔS_system + ΔS_surroundings > 0 for a spontaneous change (the second law). The entropy of a perfectly crystalline substance approaches zero as the temperature approaches absolute zero.

Gibbs energy

ΔG < 0: spontaneous. ΔG > 0: non-spontaneous (the reverse reaction is spontaneous). ΔG = 0: the system is at equilibrium. The signs of ΔH and ΔS decide how temperature matters.

Gibbs energy and equilibrium

A large negative Δ_rG° means K is much greater than 1 and the reaction goes almost to completion; a large positive value means very little product forms.

Worked example: For a reaction, Δ_rH° = +30 kJ mol⁻¹ and Δ_rS° = +100 J K⁻¹ mol⁻¹. Above what temperature does it become spontaneous (assume ΔH and ΔS do not change with T)?
Solution: At the changeover, ΔG = 0, so T = ΔH/ΔS = 30,000 J mol⁻¹ ÷ 100 J K⁻¹ mol⁻¹ = 300 K. Above 300 K, the TΔS term exceeds ΔH and ΔG becomes negative.

Common mistakes: (1) Using the older physics convention (w positive when the system does work) inside the chemistry formula ΔU = q + w. (2) Counting liquid water in Δn_g; only gases count. (3) Mixing J and kJ: ΔS is usually in J K⁻¹ while ΔH is in kJ. (4) Subtracting in the wrong order in the bond enthalpy formula (reactants minus products, the reverse of the formation formula). (5) Thinking every exothermic reaction is spontaneous.

JEE and NEET focus

• Sign of q, w and ΔU in expansion, compression, isothermal and adiabatic processes.
• Work in reversible isothermal and irreversible (constant external pressure) expansion of an ideal gas.
• ΔH and ΔU conversion using Δn_gRT.
• Δ_rH° from enthalpies of formation, from enthalpies of combustion (Hess's law) and from bond enthalpies.
• Sign of ΔS for common processes, and ΔG = ΔH − TΔS with the temperature at which a reaction becomes spontaneous.
• Relation between Δ_rG° and the equilibrium constant.

Practice questions