Thermodynamics

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 5.

Why does ice melt in your hand?

Melting ice needs heat, yet it happens on its own in a warm room. Burning petrol releases a lot of heat, yet a tank of petrol sits safely until a spark arrives. Snap a cold pack and it chills, click a hand warmer and it heats. Thermodynamics is the chemistry of these questions. For any change it tells you how much energy moves and whether the change can happen at all. It never tells you how fast. That one limitation decides half the exam traps in this chapter.

The language: system, surroundings and boundary

The system is the part of the universe you are studying, the surroundings are everything else that can interact with it, and the boundary (real or imagined) separates them. Systems are named by what can cross that boundary.

Processes are named by what is held constant.

State functions and path functions

Think of climbing a hill. Your height gained depends only on where you start and end. The distance you walked depends on the route. A state function is like height: it depends only on the present state of the system (p, V, T, n, U, H, S, G). A path function is like distance walked: it depends on how you got there (heat q and work w). That is why ΔU, ΔH, ΔS and ΔG can be added up around any route, while q and w cannot.

Properties also divide by size.

A ratio of two extensive properties is intensive: density = mass/volume, molar volume = V/n. Students lose marks by calling heat capacity intensive. It is extensive, while specific heat and molar heat capacity are intensive.

The first law: energy is conserved

The internal energy U is the total energy stored in the system: kinetic, potential, vibrational, rotational, electronic and chemical. You cannot measure it, but you can change it in two ways, by heat flow or by work. Joule's paddle-wheel experiments showed that the same amount of adiabatic work, whether done mechanically, electrically or by friction, always gives the same rise in temperature, which proves U is a state function.

$$\Delta U = q + w$$

IUPAC sign convention: whatever enters the system is positive.

For an isolated system q = 0 and w = 0, so ΔU = 0: the energy of the universe is constant.

Pause and predict: a gas expands into an evacuated flask at constant temperature (free expansion). What are q, w and ΔU?

Check your answer: all three are zero. There is nothing to push against (pext = 0) so w = 0. An ideal gas at the same temperature has the same U, so ΔU = 0. Then q = ΔU − w = 0. The volume changed, but no work was done because no force opposed it.

Work of expansion: why reversible is the maximum

When a gas expands against a constant external pressure, $w = -p_{\text{ext}}\,\Delta V$. The minus sign is there because the gas loses energy to the surroundings. If pext is much smaller than the gas pressure, the gas rushes out and delivers little work. If pext is kept almost equal to the gas pressure at every instant, the gas expands extremely slowly through a series of equilibrium states, and the work is as large as it can ever be. That ideal path is a reversible process. For an ideal gas at constant temperature,

$$w_{\text{rev}} = -2.303\,nRT\log\dfrac{V_2}{V_1}$$

Because $p_1V_1 = p_2V_2$ at constant T, the ratio $V_2/V_1$ equals $p_1/p_2$. On a pV graph, the area under the curve is the work. A one-step expansion is a single rectangle, a many-step expansion is a staircase that fills more area, and the reversible curve fills the most.

Worked example: irreversible versus reversible work. One mole of an ideal gas at 300 K expands isothermally from 1 L to 10 L. Find the work (a) against a constant external pressure of 1 bar and (b) reversibly. (R = 8.314 J K⁻¹ mol⁻¹; 1 bar L = 100 J.)

The common wrong step: using the pressure of the gas instead of the external pressure in w = −pext ΔV, or dropping the minus sign for an expansion.

Beyond the textbook. For the exam: reversible work is the maximum work obtainable from a process, and irreversible work is always smaller. In reality: no real process is perfectly reversible, because a truly reversible expansion would take infinite time. It is a limit that real systems only approach. Why it matters: the gap between reversible and real work is the reason no engine or machine converts all its heat input into work.

Enthalpy: heat measured in an open beaker

Most laboratory reactions happen in open vessels at constant pressure. If a gas is produced it pushes the atmosphere back, so part of the energy released is spent on expansion work. The bookkeeping quantity that automatically includes that work is enthalpy.

$$H = U + pV \qquad\text{and, at constant pressure,}\qquad \Delta H = q_p$$

Starting from ΔU = q + w with $w = -p\,\Delta V$, we get $q_p = \Delta U + p\,\Delta V = \Delta H$. Since H, U, p and V are state functions, H is one too. For a reaction involving gases, $p\,\Delta V = \Delta n_g RT$, so

$$\Delta H = \Delta U + \Delta n_g RT$$

where $\Delta n_g$ = moles of gaseous products − moles of gaseous reactants

For solids and liquids $p\,\Delta V$ is negligible, so ΔH ≈ ΔU. ΔH and ΔU are equal only when Δng = 0.

Worked example: from ΔH to ΔU. For N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K, ΔH = −92.4 kJ. Find ΔU.

The common wrong step: using R = 8.314 with ΔH in kJ without converting. Always change R to 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ or ΔH to joules.

Heat capacity and calorimetry

The heat needed to raise a system's temperature by one kelvin is its heat capacity: $q = C\,\Delta T$. Per mole it is the molar heat capacity Cm, and per gram it is the specific heat c. At constant volume $q = C_v\Delta T = \Delta U$, and at constant pressure $q = C_p\Delta T = \Delta H$. Heating a gas at constant pressure takes more heat for the same ΔT, because the gas must also push the atmosphere back as it expands. For an ideal gas that extra heat per mole is exactly R:

$$C_p - C_v = R$$

For a monatomic ideal gas $C_v = \tfrac{3}{2}R$ and $C_p = \tfrac{5}{2}R$. Two instruments turn these ideas into numbers.

The calorimeter's own heat capacity (its constant) must be found first by a calibration run with a known heat.

Reaction enthalpy and its many names

The enthalpy change for a reaction is the difference between the enthalpies of products and reactants:

$$\Delta_r H = \sum H_{\text{products}} - \sum H_{\text{reactants}}$$

It is quoted for the equation exactly as written, so the equation must show physical states. A superscript ° means standard conditions: each substance in its pure form at 1 bar (usually at 298 K). A thermochemical equation obeys three rules: coefficients mean moles, reversing the equation reverses the sign of ΔH, and multiplying the equation by a factor multiplies ΔH by the same factor.

The reference state of an element is its most stable form at 1 bar and the stated temperature. That is graphite for carbon, O₂(g) for oxygen and liquid for bromine. Therefore ΔfH° is zero for C(graphite), O₂(g), Br₂(l) and Hg(l), but not zero for diamond, ozone or Br₂(g).

Beyond the textbook. For the exam: the C–H bond enthalpy in CH₄ is 416 kJ mol⁻¹ (one quarter of the 1665 kJ needed to atomise methane). In reality: if you break the four C–H bonds of methane one at a time, each step needs a slightly different amount of energy. 416 is only the average. Why it matters: bond enthalpy calculations give approximate answers, usually within a few percent of calorimetric values, and they work only for gas phase species.

Hess's law: enthalpy takes any road

Because H is a state function, the total ΔrH for a reaction is the same whether it happens in one step or in several. Add the steps, add the enthalpies. This lets you calculate reactions that cannot be measured directly. Burning carbon in a limited supply of oxygen always gives some CO₂ along with CO, so ΔH for C + ½O₂ → CO cannot be measured in a calorimeter. Hess's law gets around it.

Worked example: the unmeasurable CO. Given (i) C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹ and (ii) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ mol⁻¹, find ΔH for C(graphite) + ½O₂(g) → CO(g).

The common wrong step: reversing the equation but forgetting to flip the sign of ΔH.

Using formation enthalpies. For any reaction,

$$\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$$

Worked example: methane burning. Find ΔcH° of CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) using ΔfH°: CH₄ −74.8, CO₂ −393.5, H₂O(l) −285.8 kJ mol⁻¹.

Using bond enthalpies. Breaking bonds costs energy and forming bonds releases it, so for a gas phase reaction

$$\Delta_r H^\circ = \sum (\text{bond enthalpies of reactants}) - \sum (\text{bond enthalpies of products})$$

Notice the order is the reverse of the formation enthalpy formula. Reactants minus products here, products minus reactants there.

Worked example: chlorination of methane. Using bond enthalpies (kJ mol⁻¹) C–H 414, Cl–Cl 243, C–Cl 339, H–Cl 431, find ΔH for CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g).

The common wrong step: subtracting the other way round and getting +113. Also, if any species is a solid or liquid, add its phase change enthalpy first, otherwise the method does not apply.

Lattice enthalpy and the Born–Haber cycle

You cannot measure the lattice enthalpy of NaCl directly, but you can build it from steps that you can measure, using Hess's law around a closed cycle. Forming NaCl(s) from Na(s) and ½Cl₂(g) is broken into five steps.

Worked example: lattice enthalpy of NaCl. Steps 1 to 4 add up to 108.4 + 496 + 121 − 348.6 = +376.8. The total must be −411.2, so step 5 = −411.2 − 376.8 = −788.0. The lattice enthalpy, defined for breaking the solid into ions, is +788 kJ mol⁻¹.

Whether a salt dissolves depends on $$\Delta_{\text{sol}}H^\circ = \Delta_{\text{lattice}}H^\circ + \Delta_{\text{hyd}}H^\circ$$. For NaCl the hydration enthalpy is −784 kJ mol⁻¹, so ΔsolH° = 788 − 784 = +4 kJ mol⁻¹. Hydration of the ions nearly pays the bill for pulling the crystal apart. Most ionic solids dissolve because hydration enthalpy is large and negative.

Spontaneity: why enthalpy alone is not enough

A spontaneous process is one that can proceed on its own without outside help. Water runs downhill, heat flows from hot to cold, a gas fills any container. The first guess is that systems always go to lower energy, so every exothermic reaction should be spontaneous. This is wrong in both directions. Ice melts above 273 K though ΔH is positive, and ½N₂ + O₂ → NO₂ has ΔH = +33.2 kJ mol⁻¹ but is feasible. Two ideal gases mix spontaneously with ΔH ≈ 0. Something other than energy must drive these changes. That something is entropy.

Entropy: counting the ways

Entropy S measures how many ways the energy and particles of a system can be arranged, loosely called disorder or randomness. It is a state function, and for a reversible change $\Delta S = q_{\text{rev}}/T$ (unit J K⁻¹ mol⁻¹). A gas has far more arrangements than a liquid, and a liquid more than a crystal.

The second law: for a spontaneous process the total entropy of the system plus surroundings increases. $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$, and ΔStotal = 0 at equilibrium. Heat released by the system raises the entropy of the surroundings, and at constant T and p

$$\Delta S_{\text{surr}} = -\dfrac{\Delta H_{\text{sys}}}{T}$$

This is why exothermic reactions tend to be spontaneous: a large −ΔH makes ΔSsurr large and positive. A reaction that decreases the system's own entropy can still occur if it dumps enough heat into the surroundings.

Pause and predict: water freezes spontaneously at −10 °C, although its own entropy decreases. How is the second law still satisfied?

Check your answer: freezing is exothermic. The heat released raises the entropy of the colder surroundings by more than the system's entropy falls, so ΔStotal is positive. At exactly 0 °C the two effects balance and ΔStotal = 0, the equilibrium point.

Gibbs energy: one number that decides

Tracking both ΔSsys and ΔSsurr is clumsy. Substituting ΔSsurr = −ΔH/T into ΔStotal and multiplying by −T gives a single quantity for the system alone, the Gibbs energy, $G = H - TS$:

$$\Delta G = \Delta H - T\Delta S \qquad\text{and}\qquad \Delta G = -T\,\Delta S_{\text{total}}$$

The sign of ΔG is the sign of the tug of war between ΔH (energy) and TΔS (disorder). Temperature decides who wins.

At the switching point ΔG = 0, so $T = \Delta H / \Delta S$.

Worked example: when does ice melt? For H₂O(s) → H₂O(l), ΔfusH = 6.00 kJ mol⁻¹ at 273 K. Show that ice melts at 298 K but not at 250 K.

The common wrong step: putting ΔH in kJ and ΔS in J K⁻¹ into the same equation. Convert ΔS to kJ (0.022 kJ K⁻¹ mol⁻¹) or ΔH to joules before you multiply.

Beyond the textbook. For the exam: ΔG < 0 means a reaction is spontaneous, but it says nothing about rate. In reality: diamond turning into graphite has a negative ΔG at room temperature, so diamond is thermodynamically unstable. Yet diamonds last for ever because the carbon atoms would have to rearrange through a huge energy barrier. Why it matters: thermodynamics decides if a reaction can go, kinetics decides if you will ever see it, and catalysts only help with the second.

The third law and absolute entropy

As a pure crystal is cooled towards 0 K its particles settle into a single perfect arrangement, so the entropy of a pure crystalline substance is zero at 0 K. This sets a zero point, which lets us calculate absolute entropies by adding up qrev/T as the substance is warmed. That is why tables list S° values for substances at 298 K, while for H or G only changes are meaningful.

Gibbs energy and the equilibrium constant

At equilibrium the reaction has no further tendency to go either way, so ΔrG = 0. The general expression is $\Delta_r G = \Delta_r G^\circ + RT\ln Q$, and when Q = K it gives

$$\Delta_r G^\circ = -RT\ln K = -2.303\,RT\log K$$

Worked example: from ΔG° to K. For a reaction at 298 K, ΔrG° = −17.1 kJ mol⁻¹. Find K. (R = 8.314 J K⁻¹ mol⁻¹.)

The common wrong step: leaving ΔrG° in kJ while R is in J, which gives log K = 0.003 and K ≈ 1.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Before this: Some Basic Concepts of Chemistry gives you the mole and stoichiometry that every ΔH per mole calculation depends on.
• Next: Equilibrium uses ΔG° = −RT ln K to explain why K has the size it does, and shows how temperature shifts it.
• Forward link: Chemical Kinetics is the other half of the story, since it handles how fast a spontaneous reaction goes, and Electrochemistry turns ΔG into cell voltage.