CBSE Class 11 Chemistry Thermodynamics Notes Set 08

These Class 11 CBSE Chemistry notes for Chapter 5 Thermodynamics explain system and surroundings, state functions, heat and work with the IUPAC sign convention, the first law, enthalpy and its measurement, Hess's law and the different enthalpies of reaction. They then move to entropy, the second and third laws and Gibbs energy as the criterion of spontaneity. Eight solved examples show the numerical working step by step.

Important Notes and Formulas

Basic Terms of Thermodynamics

1. System is the part of the universe chosen for study; everything else that can exchange energy with it is the surroundings. The real or imaginary surface that separates them is the boundary.
2. Open system: exchanges both matter and energy with the surroundings (a beaker of boiling water without a lid). Closed system: exchanges energy but not matter (a sealed flask of reactants). Isolated system: exchanges neither matter nor energy (a tightly closed thermos flask is a good approximation).
3. The state of a system is fixed by its measurable properties such as pressure p, volume V, temperature T and amount n. A change in any one of them changes the state.
4. Extensive properties depend on the amount of matter (mass, volume, internal energy, enthalpy, entropy, heat capacity). Intensive properties do not (temperature, pressure, density, molar volume, refractive index). The ratio of two extensive properties is intensive.
5. A state function depends only on the initial and final states and not on the path (U, H, S, G, p, V, T). Heat q and work w are path functions because their values depend on how the change is carried out.
6. A system is in thermodynamic equilibrium when there is no change of macroscopic properties with time: mechanical, thermal and chemical equilibrium all hold.

Work, Heat and Internal Energy

The internal energy U of a system is the sum of all kinetic, potential, electronic and nuclear energies of its particles. Its absolute value cannot be measured; only the change ΔU = U_final − U_initial can be. U is a state function and is extensive. For an ideal gas U depends only on temperature.

Internal energy can be changed in two ways, by heat (q), which flows because of a temperature difference, and by work (w). A change brought about only by work in an insulated system is called adiabatic (q = 0).

Pressure–Volume Work

For a gas changing in volume from V_i to V_f against a constant external pressure p_ex, w = −p_ex(V_f − V_i) = −p_exΔV. On expansion ΔV > 0 so w is negative; on compression ΔV < 0 so w is positive. The magnitude of the work is the shaded area under the p–V curve. The unit is J (1 L bar = 100 J; 1 L atm = 101.3 J).

1. Free expansion is expansion into a vacuum (p_ex = 0). No work is done, so w = 0. For an ideal gas in an insulated container, q = 0 also, hence ΔU = 0.
2. In a reversible process the system is always in equilibrium with the surroundings, and the driving force is infinitesimal. For the isothermal reversible expansion or compression of n mol of an ideal gas: w_rev = −2.303 nRT log (V_f/V_i). A reversible expansion gives the maximum work obtainable.
3. For an isothermal process in an ideal gas ΔU = 0, so q = −w.
4. In a cyclic process the system returns to its initial state, so ΔU = 0 and q = −w. The net work equals the area enclosed by the cycle on a p–V diagram.

First Law of Thermodynamics

The energy of an isolated system is constant. Energy can neither be created nor destroyed; it can only change from one form to another. For a closed system, ΔU = q + w. At constant volume w = 0, so ΔU = q_V. For an adiabatic change ΔU = w_ad.

Enthalpy (H)

H = U + pV. It is a state function and an extensive property. At constant pressure, q_p = ΔU + pΔV = ΔH, so the heat exchanged at constant pressure is the enthalpy change. The two heats are related by ΔH = ΔU + Δn_gRT, where Δn_g is the moles of gaseous products minus the moles of gaseous reactants. If Δn_g = 0, or if no gases take part, ΔH ≈ ΔU.

Heat Capacity and Calorimetry

1. Heat capacity C = q/ΔT (J K⁻¹) is the heat needed to raise the temperature of the substance by 1 K. It is extensive. Specific heat capacity c is per gram, and molar heat capacity C_m is per mole.
2. q = m c ΔT = C ΔT. At constant volume q_V = C_VΔT = ΔU, and at constant pressure q_p = C_pΔT = ΔH.
3. For one mole of an ideal gas C_p − C_V = R, and C_p is always larger because extra heat goes into expansion work.
4. Calorimetry: ΔU is measured in a bomb calorimeter (a sealed steel vessel at constant volume, q_V = ΔU) and ΔH in an open, constant-pressure calorimeter. The heat released is q = −C_calΔT, where C_cal includes the water and the vessel.

Enthalpies of Reaction

The standard state of a substance is its pure form at 1 bar pressure (and usually 298 K); it is shown by the superscript °. The standard enthalpy of any element in its reference (most stable) state is zero, e.g. C(graphite), O₂(g), Br₂(l), S(rhombic).

1. A thermochemical equation is a balanced equation that gives the physical states and the value of ΔH for the amounts shown.
2. If the equation is multiplied by a factor, ΔH is multiplied by the same factor. If the equation is reversed, the sign of ΔH changes.
3. Δ_rH° = Σ a_i Δ_fH°(products) − Σ b_i Δ_fH°(reactants), with a_i and b_i the stoichiometric coefficients.

Sublimation: Δ_subH° = Δ_fusH° + Δ_vapH°. The mean bond enthalpy is used to estimate reaction enthalpies in the gas phase: Δ_rH° = Σ bond enthalpies of reactants − Σ bond enthalpies of products.

Hess's Law of Constant Heat Summation

The enthalpy change of a reaction is the same whether it takes place in one step or in several steps, because H is a state function. If A → B directly has ΔH, and A → C → B has ΔH₁ and ΔH₂, then ΔH = ΔH₁ + ΔH₂. It is used to find enthalpies that cannot be measured directly, such as C(graphite) → CO(g).

Spontaneity and Entropy

1. A spontaneous process proceeds on its own once started and does not need continuous outside help (a gas expanding into a vacuum, heat flowing from hot to cold, rusting of iron). A non-spontaneous process occurs only when energy is supplied continuously.
2. Exothermicity alone is not the criterion: some endothermic changes (dissolving NH₄Cl, melting of ice above 0 °C) are spontaneous. Spontaneity also depends on the disorder of the system.
3. Entropy (S) is a state function that measures the randomness or disorder of a system. S(gas) > S(liquid) > S(solid). Dissolving a solid, melting, vaporising, and a reaction that increases the moles of gas all raise S.
4. For a reversible change at temperature T: ΔS = q_rev/T (J K⁻¹ mol⁻¹). For a phase change at equilibrium: Δ_fusS = Δ_fusH/T_f and Δ_vapS = Δ_vapH/T_b.
5. Second law: the total entropy of the system and its surroundings increases in every spontaneous process, ΔS_total = ΔS_sys + ΔS_surr > 0. At equilibrium ΔS_total = 0.
6. Third law: the entropy of a perfectly crystalline substance is zero at absolute zero (0 K). Because of this, absolute (standard) entropies S° can be tabulated, and Δ_rS° = ΣS°(products) − ΣS°(reactants).
7. The entropy of the surroundings changes by ΔS_surr = −ΔH_sys/T at constant pressure.

Gibbs Energy and Spontaneity

Gibbs energy is defined as G = H − TS. At constant temperature and pressure ΔG = ΔH − TΔS (the Gibbs–Helmholtz relation), and ΔG = −TΔS_total. So the process is spontaneous when ΔG < 0, at equilibrium when ΔG = 0, and non-spontaneous when ΔG > 0. The decrease in Gibbs energy also gives the maximum non-expansion (useful) work.

For a reaction with ΔH and ΔS of the same sign, the changeover temperature is T = ΔH/ΔS (where ΔG = 0). Standard Gibbs energy is related to the equilibrium constant by Δ_rG° = −RT ln K = −2.303 RT log K. If Δ_rG° < 0, then K > 1 and products are favoured; if Δ_rG° > 0, then K < 1. Also Δ_rG° = ΣΔ_fG°(products) − ΣΔ_fG°(reactants). A negative ΔG tells only whether a reaction is feasible, not how fast it is: many feasible reactions need activation energy to start.

Solved Examples

Example 1. Calculate the change in internal energy in each case. (i) A system absorbs 15 kJ of heat and does 5 kJ of work. (ii) 5 kJ of work is done on the system and 15 kJ of heat is given out by the system.
Answer:
(i) q = +15 kJ and the work is done by the system, so w = −5 kJ. ΔU = q + w = 15 + (−5) = +10 kJ. The internal energy increases.
(ii) w = +5 kJ (work on the system) and q = −15 kJ. ΔU = −15 + 5 = −10 kJ. The internal energy decreases.

Example 2. Calculate w, q and ΔU when 0.75 mol of an ideal gas expands isothermally and reversibly at 27 °C from 15 L to 25 L. (R = 8.314 J K⁻¹ mol⁻¹)
Answer:
T = 27 + 273 = 300 K. w = −2.303 nRT log (V_f/V_i) = −2.303 × 0.75 × 8.314 × 300 × log (25/15)
= −2.303 × 0.75 × 8.314 × 300 × 0.2218 = −955.5 J (negative: the system does work).
For an isothermal change of an ideal gas ΔU = 0, so q = −w = +955.5 J.

Example 3. A 5 L cylinder contains 10 mol of oxygen at 27 °C. All the gas leaks suddenly into the atmosphere (1.0 atm) and the cylinder becomes empty. Calculate the work done by the gas. (R = 0.0821 L atm K⁻¹ mol⁻¹; 1 L atm = 101.3 J)
Answer:
The gas finally occupies V_f = nRT/p = (10 × 0.0821 × 300)/1.0 = 246.3 L, so ΔV = 246.3 − 5 = 241.3 L.
w = −p_exΔV = −1.0 atm × 241.3 L = −241.3 L atm = −241.3 × 101.3 J = −24 444 J (about −24.4 kJ).
The negative sign shows that the work is done by the gas on the surroundings.

Example 4. The heat of combustion of benzene in a bomb calorimeter (constant volume) is −3263.9 kJ mol⁻¹ at 25 °C. Calculate the heat of combustion at constant pressure.
Answer:
C₆H₆(l) + 7½O₂(g) → 6CO₂(g) + 3H₂O(l). Only O₂ and CO₂ are gases, so Δn_g = 6 − 7.5 = −1.5 mol.
ΔH = ΔU + Δn_gRT = −3263.9 kJ + (−1.5 mol)(8.314 × 10⁻³ kJ K⁻¹ mol⁻¹)(298 K) = −3263.9 − 3.7
ΔH = −3267.6 kJ mol⁻¹.

Example 5. The enthalpies of combustion of graphite and diamond are −393.5 and −395.4 kJ mol⁻¹. Calculate the enthalpy change for C(graphite) → C(diamond).
Answer:
(i) C(graphite) + O₂(g) → CO₂(g); ΔH = −393.5 kJ mol⁻¹
(ii) C(diamond) + O₂(g) → CO₂(g); ΔH = −395.4 kJ mol⁻¹
Subtract (ii) from (i): C(graphite) → C(diamond); ΔH = −393.5 − (−395.4) = +1.9 kJ mol⁻¹.

Example 6. Calculate the enthalpy change for H₂(g) + Br₂(g) → 2HBr(g). The bond enthalpies of H–H, Br–Br and H–Br are 435, 192 and 364 kJ mol⁻¹ respectively.
Answer:
Δ_rH = Σ (bond enthalpies of reactants) − Σ (bond enthalpies of products) = (435 + 192) − (2 × 364)
= 627 − 728 = −101 kJ. The reaction is exothermic.

Example 7. Calculate the entropy change when 1 mol of ice melts at 273 K. The enthalpy of fusion of ice is 6025 J mol⁻¹.
Answer:
Melting at the melting point is a reversible process, so Δ_fusS = Δ_fusH/T_f = 6025 J mol⁻¹ / 273 K
= 22.1 J K⁻¹ mol⁻¹. The entropy increases because the liquid is more disordered than the solid.

Example 8. For the reaction A + B → C + D, ΔH = −10 000 J mol⁻¹ and ΔS = −33.3 J K⁻¹ mol⁻¹. (i) At what temperatures is the forward reaction spontaneous? (ii) When does it reverse?
Answer:
At equilibrium ΔG = 0, so T = ΔH/ΔS = (−10 000)/(−33.3) = 300.3 K.
(i) ΔG = ΔH − TΔS = −10 000 + 33.3 T. This is negative only when T < 300.3 K, so the forward reaction is spontaneous below 300.3 K.
(ii) Above 300.3 K, ΔG is positive, so the reverse reaction becomes spontaneous.