Midterm 1 : 2nd Law of Thermodynamics (fixed)

Understand he statistical or probability approach to entropy or microscopic definition in terms of microstates. Analogies with coin tossing examples

Core definitions

• A microstate is one specific, fully detailed arrangement of every particle consistent with the system’s macroscopic constraints.

• A macrostate is a bulk description (e.g., “2 particles on each side”) that many microstates can produce.

• W is the number of microstates belonging to a macrostate. Larger W means higher probability.

• The fundamental postulate of statistical mechanics: every accessible microstate is equally probable. A macrostate is likely only because it contains many microstates.

Coin-tossing model

• Each coin has 2 outcomes, and independent coins multiply, so N coins give 2^N total microstates.

• The number of microstates in the macrostate “k heads” is the binomial coefficient (the rows of Pascal’s triangle):

$$W = \frac{N!}{k!\,(N-k)!}$$

• 2 coins: 1, 2, 1 (total 4)

• 3 coins: 1, 3, 3, 1 (total 8)

• 4 coins: 1, 4, 6, 4, 1 (total 16)

• Mixed outcomes such as HHTT are the most probable because there are more ways to realize them.

Mapping to particles (partition removed)

• Each of the 4 particles is independently in the left or right half, like a coin.

• The (4,0) distribution has W = 1, so P = 1/16 (least likely).

• The (2,2) distribution has W = 6, so P = 6/16 = 37.5% (most likely).

• Scaling up: the chance that all N particles are on one side is 2^(−N). For N = 100 that is about 10^(−30). For a mole of gas it is effectively zero. The even split dominates, and the distribution narrows as 1/√N.

Linking to the second law and spontaneity

1. A spontaneous process moves toward the macrostate with the most microstates.

2. High entropy corresponds to a large number of microstates (energy and matter dispersed).

3. Therefore spontaneous change corresponds to increasing entropy.

• The second law is statistical: a decrease is not forbidden, only overwhelmingly improbable for macroscopic N.

Why crystals still exist

• Solid → liquid → gas increases disorder, so entropy alone would prevent crystals.

• Crystals form because of the competing tendency, minimization of energy/enthalpy through interatomic and intermolecular forces.

• The balance of the two tendencies is captured by:

$$\Delta G = \Delta H - T\Delta S$$

Common pitfalls

• “Disorder” is a heuristic. The rigorous idea is the count of accessible microstates.

• A specific sequence (HHHH) is exactly as likely as any other specific sequence (HTHT). Only macrostates differ in probability.

• Real gas particles are indistinguishable, which adds an N! correction to W. It does not affect entropy changes for volume changes.

Worked example: probability of exactly 5 heads vs. 10 heads in 10 tosses

• Total microstates = 2^10 = 1024

$W(5\text{H}) = \frac{10!}{5!\,5!} = 252 \quad\Rightarrow\quad P = \frac{252}{1024} = 24.6\%$$W(10\text{H}) = 1 \quad\Rightarrow\quad P = \frac{1}{1024} = 0.098\%$

Explain Boltzmann’s expression for entropy and how it relates to the macroscopic definition given earlier

Boltzmann’s expression

$$S = k_B \ln W$$

• S = entropy (J·K⁻¹)

• kB = 1.38 × 10⁻²³ J·K⁻¹ (Boltzmann constant)

• W = number of microstates (dimensionless)

• The molar gas constant is R = NA·kB, so kB is “per particle” and R is “per mole.”

Why a logarithm? (derivation from two requirements)

• Entropy is extensive: for two independent systems, S = S₁ + S₂.

• Microstates multiply: W = w₁w₂ (3 coins: 2 × 2 × 2 = 8, not 2 + 2 + 2 = 6).

• The only function that turns a product into a sum is the logarithm:

$$\ln W = \ln(w_1 w_2) = \ln w_1 + \ln w_2$$

• So S = kB ln W satisfies both requirements.

Connection to the macroscopic definition

Microscopic route (isothermal doubling of volume):

• Doubling V gives each particle twice as many positions, so for one particle w₂ = 2w₁:

$$\Delta S = k_B \ln\frac{w_2}{w_1} = k_B \ln 2$$

• For N particles, w₂/w₁ = 2^N:

$$\Delta S = k_B \ln 2^N = N k_B \ln 2$$

• For one mole:

$$\Delta S = N_A k_B \ln 2 = R\ln 2$$

Macroscopic route:

• For an isothermal ideal gas, ΔU = 0, so q_rev = −w_rev:

$$q_{rev} = nRT\ln\frac{V_2}{V_1}$$

• For n = 1 and V₂ = 2V₁, q_rev = RT ln 2, so:

$$\Delta S = \frac{q_{rev}}{T} = R\ln 2$$

• The two results are identical, so the statistical and thermodynamic definitions are equivalent.

General volume result: W is proportional to V^N, so:

$$\Delta S = N k_B \ln\frac{V_2}{V_1} = nR\ln\frac{V_2}{V_1}$$

Assumptions and pitfalls

• This counts positional microstates only. At constant T the kinetic-energy distribution is unchanged and cancels.

• Ideal gas (non-interacting particles) is assumed.

• S is a state function. ΔS is the same for reversible and irreversible paths, but only q_rev may be used in q/T. For irreversible paths, ΔS > q_irr/T.

• In a free expansion into vacuum, q = 0, yet ΔS(sys) = nR ln(V₂/V₁) > 0.

Worked example: entropy gain per molecule on doubling the volume

$$\Delta S = k_B\ln 2 = (1.38\times10^{-23}\ \text{J·K}^{-1})(0.693) = 9.57\times10^{-24}\ \text{J·K}^{-1}$$

Per mole:

$\Delta S = R\ln 2 = (8.314\ \text{J·K}^{-1}\text{mol}^{-1})(0.693) = 5.76\ \text{J·K}^{-1}\text{mol}^{-1}$

Determine the entropy change for the isothermal expansion or compression of an ideal gas

Key result (constant T, ideal gas)

$$\Delta S = nR\ln\frac{V_f}{V_i} = nR\ln\frac{P_i}{P_f}$$

• n = moles, R = 8.314 J·K⁻¹·mol⁻¹, V = volume, P = pressure.

• The pressure form follows from Boyle’s law (PiVi = PfVf at constant T and n).

Derivation

• Ideal-gas internal energy depends only on T, so isothermal means ΔU = q + w = 0 and q_rev = −w_rev.

• Reversible work (negative when the gas expands):

$$w_{rev} = -nRT\ln\frac{V_f}{V_i}$$

• Therefore:

$$q_{rev} = +nRT\ln\frac{V_f}{V_i}\qquad \Delta S = \frac{q_{rev}}{T} = nR\ln\frac{V_f}{V_i}$$

Molecular interpretation

• Expansion gives more accessible positions, more microstates, and ΔS > 0.

• Compression gives fewer positions, so ΔS < 0 for the gas (the surroundings gain entropy, and ΔS(univ) is still ≥ 0).

Assumptions and pitfalls

• ΔS(sys) is path-independent, so the same value applies to a reversible or irreversible (e.g., free) expansion between the same two states. Only the calculation path must be reversible.

• The result holds only for ideal gases at constant T, and it does not depend on the value of T.

• Units of V or P cancel in the ratio, but use R in J·K⁻¹·mol⁻¹ so ΔS comes out in J·K⁻¹.

• Decreasing P gives Pi/Pf > 1 and a positive ΔS. Do not invert the ratio.

Worked example (Problem 9.9): 5.25 mol ideal gas, 24.252 L → 34.058 L

$$\Delta S = (5.25)(8.314)\ln\frac{34.058}{24.252} = (43.65)(0.3396) = +14.8\ \text{J·K}^{-1}$$

Worked example (Problem 9.11): 1.50 mol Ne, 15.0 atm → 0.500 atm

$$\Delta S = (1.50)(8.314)\ln\frac{15.0}{0.500} = (12.47)(3.401) = +42.4\ \text{J·K}^{-1}$$

Calculate the standard entropy of a phase changes

Key idea

• Phase changes at the transition temperature occur at constant pressure and at equilibrium, so they are naturally reversible.

• At constant P, q = ΔH, so q_rev = ΔH(trans), giving:

$$\Delta S_{trans} = \frac{\Delta H_{trans}}{T_{trans}}$$

• ΔH(trans) = molar enthalpy of transition (J·mol⁻¹); T(trans) = transition temperature (K).

• “Standard” means the transition at standard pressure (1 bar), i.e., at the normal melting or boiling point.

Sign rules

• Melting and vaporization absorb heat (ΔH > 0), so ΔS > 0 (more freedom of motion, more microstates).

• Freezing and condensation reverse the sign:

$$\Delta H_{freezing} = -\Delta H_{melting}\qquad \Delta S_{freezing} = -\frac{\Delta H_{melting}}{T_m}$$

• Typical magnitudes: vaporization (about 85–110 J·K⁻¹·mol⁻¹) is far larger than fusion (about 10–25), because a gas has vastly more accessible volume than a liquid. Hydrogen-bonded liquids such as water and ethanol sit above the typical (Trouton) value because H-bonding orders the liquid.

Why ΔS(univ) = 0 at the transition

• The surroundings supply or absorb exactly ΔH at the same T, so ΔS(surr) = −ΔH/T cancels ΔS(sys). This is the signature of equilibrium.

Pitfalls

• Convert kJ to J before dividing (entropy is quoted in J·K⁻¹).

• Use kelvin.

• Scale by moles: n = m/M if mass is given.

• The formula applies only at the transition temperature. Elsewhere you must build a path (heat to T(trans), transition, heat again).

Worked example (Problem 9.15b): vaporization of 50.0 g ethanol at 351.5 K (ΔH(vap) = 43.5 kJ·mol⁻¹, M = 46.07 g·mol⁻¹)

$n = \frac{50.0\ \text{g}}{46.07\ \text{g·mol}^{-1}} = 1.085\ \text{mol}$$\Delta S = \frac{(1.085\ \text{mol})(43\,500\ \text{J·mol}^{-1})}{351.5\ \text{K}} = +134\ \text{J·K}^{-1}$

Worked example (Problem 9.15a): freezing 1.00 mol H₂O(l) at 273.2 K

$$\Delta S = \frac{(1.00\ \text{mol})(-6010\ \text{J·mol}^{-1})}{273.2\ \text{K}} = -22.0\ \text{J·K}^{-1}$$

(Melting ice at 0 °C gives +22.0 J·K⁻¹·mol⁻¹.)

Calculation of the entropy change when the temperature of a substance is changed or when pressure or volume is changed.

Temperature change: derivation

• For an infinitesimal reversible heat transfer, dq_rev = C dT, where C = n·c(molar). Then:

$$dS = \frac{dq_{rev}}{T} = \frac{C\,dT}{T}$$

• Integrating for constant C:

$$\Delta S = \int_{T_1}^{T_2}\frac{C\,dT}{T} = C\ln\frac{T_2}{T_1}$$

• Constant volume uses C = n·C(V,m); constant pressure uses C = n·C(P,m):

$$\Delta S = nC_{V,m}\ln\frac{T_2}{T_1}\qquad\text{or}\qquad \Delta S = nC_{P,m}\ln\frac{T_2}{T_1}$$

• If C varies with T, integrate C(T)/T dT instead.

• Heating (T₂ > T₁) gives ΔS > 0 because more energy levels become thermally accessible.

Volume or pressure change (ideal gas, constant T)

$$\Delta S = nR\ln\frac{V_2}{V_1} = nR\ln\frac{P_1}{P_2}$$

Both change together (ideal gas)

• Entropy is a state function, so the two contributions are calculated separately and added:

$$\Delta S = nC_{V,m}\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}$$

• Equivalent pressure form (using C(P,m) − C(V,m) = R):

$$\Delta S = nC_{P,m}\ln\frac{T_2}{T_1} - nR\ln\frac{P_2}{P_1}$$

Heat absorbed by a large block at nearly constant T

• ΔS ≈ q/T. For 65 J: at 298 K, ΔS ≈ 0.218 J·K⁻¹. At 373 K, ΔS ≈ 0.174 J·K⁻¹.

• Why the difference? The same energy added to an already hotter (more disordered) system creates a smaller relative increase in accessible microstates.

Assumptions and pitfalls

• Choose the right heat capacity (C(V) for fixed volume, C(P) for fixed pressure).

• T must be in kelvin.

• A compression (V₂ < V₁) lowers entropy.

• Diatomic gases at moderate T: C(V,m) = (5/2)R (translation plus rotation; vibration frozen out).

Worked example A: 20.0 L N₂ at 5.00 kPa heated from 20.0 °C to 400. °C (C(V,m) = 20.81 J·K⁻¹·mol⁻¹, fixed volume)

$n = \frac{PV}{RT} = \frac{(5.00\ \text{kPa})(20.0\ \text{L})}{(8.3145\ \text{kPa·L·K}^{-1}\text{mol}^{-1})(293.15\ \text{K})} = 0.0410\ \text{mol}$$\Delta S = (0.0410)(20.81)\ln\frac{673.15}{293.15} = (0.854)(0.831) = +0.710\ \text{J·K}^{-1}$

Worked example B (Problem 9.13): 1 mol N₂ compressed reversibly from 3.00 L to 0.500 L while warming from 18.5 °C to 28.1 °C (C(V) = (5/2)R)

$\Delta S_V = (8.314)\ln\frac{0.500}{3.00} = -14.9\ \text{J·K}^{-1}$$\Delta S_T = \tfrac{5}{2}(8.314)\ln\frac{301.25}{291.65} = (20.79)(0.0324) = +0.67\ \text{J·K}^{-1}$$\Delta S_{total} = -14.9 + 0.67 = -14.2\ \text{J·K}^{-1}$

Compression dominates.

Calculate the standard reaction entropy from standard molar entropies (Example 9.9).

Formula

$$\Delta S^\circ_{rxn} = \sum \nu_p\, S^\circ_{products} - \sum \nu_r\, S^\circ_{reactants}$$

• S° = standard molar (absolute) entropy (J·K⁻¹·mol⁻¹) at 1 bar and the stated T (usually 298 K).

• ν = stoichiometric coefficient, including fractions.

• This works because entropy has an absolute scale (third law), so S° values are tabulated directly instead of as formation values.

Qualitative checks before calculating

• Moles of gas increase, so ΔS° > 0 (usually large).

• Moles of gas decrease, so ΔS° < 0.

• Larger, more complex molecules and gases have higher S° than small molecules and solids.

Pitfalls

• Elements in their standard states do not have S° = 0. That is true only for ΔfH° and ΔfG°. Elemental entropies at 298 K are positive.

• Multiply S° by every coefficient, including fractions like the ½ in ½N₂.

• S° is in J, while enthalpies are in kJ. Convert before combining in ΔG.

• The slide’s notation “S_f°” is misleading, because S° is an absolute entropy, not a formation quantity.

• The result is per mole of reaction as written.

Worked example 1: ½N₂(g) + O₂(g) → NO₂(g), with S° = 191.5, 205.0, 240.0 J·K⁻¹·mol⁻¹ respectively

$$\Delta S^\circ = 240.0 - \left(\frac{191.5}{2} + 205.0\right) = 240.0 - 300.75 = -60.8\ \text{J·K}^{-1}\text{mol}^{-1}$$

Gas moles fall from 1.5 to 1, so a negative value is expected.

Worked example 2: 2SO₂(g) + O₂(g) → 2SO₃(g), with S° = 248, 205, 257 J·K⁻¹·mol⁻¹

$$\Delta S^\circ = 2(257) - \left[2(248) + 205\right] = 514 - 701 = -187\ \text{J·K}^{-1}\text{mol}^{-1}$$

Three moles of gas become two, so a negative value is expected.

Gibbs free energy as the criterion for spontaneity (four possible cases with respect to signs of enthalpy and entropy)

Definition

• G is a state function:

$$G = H - TS$$

• At constant T and P:

$$\Delta G = \Delta H - T\Delta S$$

• ΔH (J or kJ) is the heat at constant P and reflects the energy-minimization tendency.

• TΔS is the entropy-maximization tendency, weighted by temperature.

• ΔG < 0 means spontaneous, ΔG = 0 means equilibrium, and ΔG > 0 means non-spontaneous (the reverse is spontaneous).

The four sign cases

ΔH	ΔS	ΔG	Outcome
−	+	always −	Spontaneous at all T
+	−	always +	Non-spontaneous at all T
+	+	− only at high T	Entropy-driven; spontaneous above T = ΔH/ΔS
−	−	− only at low T	Enthalpy-driven; spontaneous below T = ΔH/ΔS

Why T matters: In the mixed cases ΔH and ΔS “pull in opposite directions,” and T sets the relative weight of the entropy term.

Conditions and pitfalls

• ΔG applies at constant T and P with only P–V work.

• Spontaneous does not mean fast. ΔG gives direction only; rate belongs to kinetics.

• Spontaneous does not mean exothermic. Endothermic processes can be spontaneous if TΔS is large enough.

• Do not confuse ΔG (actual conditions) with ΔG° (standard conditions).

• Convert J·K⁻¹ entropy to kJ·K⁻¹ before combining with ΔH in kJ.

• Slide correction: a more negative ΔG° (not “higher”) means a more favorable reaction.

Worked example: is melting ice spontaneous at +10 °C and at −10 °C? (ΔH = 6010 J·mol⁻¹, ΔS = 22.0 J·K⁻¹·mol⁻¹, assumed independent of T)

$T = 283.15\ \text{K}:\quad \Delta G = 6010 - (283.15)(22.0) = -219\ \text{J·mol}^{-1}\ \text{(spontaneous)}$ $T = 263.15\ \text{K}:\quad \Delta G = 6010 - (263.15)(22.0) = +221\ \text{J·mol}^{-1}\ \text{(not spontaneous; water freezes)}$ $T_{crossover} = \frac{6010}{22.0} = 273\ \text{K}\quad\text{(the melting point)}$

Relationship of change in Gibbs free energy of system and change in entropy of universe. Relationship to 2nd law.

Second law

$$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} \ge 0$$

• ΔS(univ) > 0 for an irreversible (spontaneous) process.

• ΔS(univ) = 0 for a reversible process (equilibrium).

• Entropy increases until a maximum is reached at equilibrium, where concentrations are constant and forward rate equals backward rate.

Derivation of ΔG(sys) = −TΔS(univ)

1. Place the system in a large heat bath, so the surroundings’ T is unchanged and q(surr) = −q(sys).

2. At constant P, q(sys) = ΔH(sys), so q(surr) = −ΔH(sys).

3. Using ΔS = q/T:

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

4. Start from ΔG(sys) = ΔH(sys) − TΔS(sys) and divide by −T:

$$-\frac{\Delta G_{sys}}{T} = -\frac{\Delta H_{sys}}{T} + \Delta S_{sys} = \Delta S_{surr} + \Delta S_{sys} = \Delta S_{univ}$$

5. Multiply by −T:

$$\Delta G_{sys} = -T\,\Delta S_{univ}$$

Interpretation

• ΔS(univ) > 0 is equivalent to ΔG(sys) < 0.

• At equilibrium, ΔS(univ) = 0 and ΔG = 0.

• ΔG is the second law rewritten using only the system’s properties. This is useful because the surroundings are never measured directly.

• Physically, −ΔH(sys)/T is the entropy that the system’s released heat adds to the surroundings.

Assumptions and pitfalls

• The surroundings are a huge reservoir at constant T and P.

• ΔG < 0 does not mean the system’s entropy rises. A system can become more ordered if the surroundings gain more entropy than the system loses (e.g., freezing below 0 °C).

• Watch signs: the surroundings gain what the system loses.

Worked example (Problem 9.5): 40.0 kJ flows from a reservoir at 800. K to one at 200. K

$\Delta S_{hot} = \frac{-40\,000\ \text{J}}{800\ \text{K}} = -50\ \text{J·K}^{-1}\qquad \Delta S_{cold} = \frac{+40\,000\ \text{J}}{200\ \text{K}} = +200\ \text{J·K}^{-1}$$\Delta S_{univ} = -50 + 200 = +150\ \text{J·K}^{-1} > 0\quad\Rightarrow\quad\text{spontaneous}$

Worked example: freezing water at −10 °C (ΔH = −6010 J·mol⁻¹, ΔS(sys) ≈ −22.0 J·K⁻¹·mol⁻¹)

$\Delta S_{surr} = \frac{+6010}{263.15} = +22.84\ \text{J·K}^{-1}\qquad \Delta S_{univ} = -22.0 + 22.84 = +0.84\ \text{J·K}^{-1}$$\Delta G_{sys} = -(263.15)(0.84) \approx -221\ \text{J·mol}^{-1}$

This matches the result in Card 7.

Calculate a standard Gibbs free energy of formation from enthalpy and entropy data

Definitions

• ΔfG° is the Gibbs energy change for forming 1 mol of a compound from its elements in their standard states (1 bar, specified T). Elements in their standard states have ΔfG° = 0.

• Apply the Gibbs equation to the formation reaction:

$$\Delta_f G^\circ = \Delta_f H^\circ - T\,\Delta_f S^\circ$$

• The formation entropy is not the same as S° of the compound:

$$\Delta_f S^\circ = S^\circ(\text{compound}) - \sum \nu\, S^\circ(\text{elements})$$

Procedure

1. Write the balanced formation equation for exactly 1 mol of product (fractional coefficients are fine).

2. Calculate ΔfS° from S° of the compound and of each element.

3. Convert ΔfS° from J·K⁻¹ to kJ·K⁻¹.

4. Evaluate ΔfH° − TΔfS° at the given T (e.g., 298.15 K).

Pitfalls

• Using S° of the compound instead of ΔfS°.

• Forgetting that elemental S° values are not zero.

• Mixing kJ and J.

• Assuming ΔfH° and ΔfS° are independent of T (only a good approximation over modest ranges).

Worked example: ΔfG° of H₂O(g) at 298.15 K
(Data are standard table values, not from your slides: ΔfH° = −241.8 kJ·mol⁻¹; S° = 188.8 for H₂O(g), 130.7 for H₂(g), 205.0 for O₂(g), all in J·K⁻¹·mol⁻¹.)

Reaction: H₂(g) + ½O₂(g) → H₂O(g)

$\Delta_f S^\circ = 188.8 - \left[130.7 + \tfrac{1}{2}(205.0)\right] = -44.4\ \text{J·K}^{-1}\text{mol}^{-1}$$\Delta_f G^\circ = -241.8\ \text{kJ·mol}^{-1} - (298.15\ \text{K})(-0.0444\ \text{kJ·K}^{-1}\text{mol}^{-1}) = -228.6\ \text{kJ·mol}^{-1}$

This agrees with the −228.57 kJ·mol⁻¹ used in Card 11.

Use concept of absolute entropies and how they arise. Connection with 3rd law.

Why entropy can have an absolute value

• U, H, and G have no natural zero, so only differences are measurable.

• Entropy does have a natural zero, because S = kB ln W and W can equal 1.

Third law

• A perfectly ordered crystal at 0 K has exactly one microstate (W = 1), so:

$$S = k_B \ln 1 = 0$$

• Quantum zero-point energy does not contradict this. The law concerns the number of accessible arrangements, not zero energy.

How absolute entropies are obtained

$$S^\circ(T) = \int_0^{T}\frac{C_P(T')}{T'}\,dT' + \sum\frac{\Delta H_{trans}}{T_{trans}}$$

• Measure heat capacities from near 0 K up to T and integrate C(P)/T. Add ΔH(trans)/T(trans) for every phase change on the way.

• C(P) → 0 as T → 0, so the integral converges.

• Since S(0) = 0, ΔS = S for heating from 0 K. This is why tabulated S° values are always positive and can be combined directly:

$$\Delta S^\circ_{rxn} = \sum S^\circ_{products} - \sum S^\circ_{reactants}$$

Edge cases and pitfalls

• Residual entropy: crystals with frozen-in disorder have S > 0 at 0 K. For CO with two random orientations per molecule, W = 2^N and S₀ = R ln 2 = 5.76 J·K⁻¹·mol⁻¹. For ice, S₀ ≈ R ln(3/2) = 3.4 J·K⁻¹·mol⁻¹.

• The slide’s “any pure substance has S = 0 at 0 K” is shorthand for a perfect crystal. Imperfect crystals and glasses have S > 0.

• Absolute entropy increases with temperature, with the number of atoms, and in the order solid < liquid < gas.

Worked example: 1 mol of ice at 273.15 K warmed to liquid water at 298.15 K (C_P,m) of liquid water = 75.3 J·K⁻¹·mol⁻¹, a standard value not from your slides)

$$\Delta S=\underbrace{\frac{6010}{273.15}}_{\text{melting}}+\underbrace{75.3\ln\frac{298.15}{273.15}}_{\text{heating}}=22.0+6.6=\bm{28.6\ J\cdot K^{-1}}\text{·mol}\bm{^{-1}}$$

This is the kind of step-by-step addition used to build S° from 0 K.

Calculate the standard Gibbs free energy of reaction from standard Gibbs free energies of formation

Formula (Hess’s law applied to G)

$$\Delta G^\circ_{rxn} = \sum \nu_p\,\Delta_f G^\circ(\text{products}) - \sum \nu_r\,\Delta_f G^\circ(\text{reactants})$$

• ΔfG° is in kJ·mol⁻¹, ν is the stoichiometric coefficient, and elements in their standard states contribute 0.

• G is a state function, so ΔG is path-independent and obeys Hess’s law. Reactions can be added, reversed (sign flips), and scaled (value scales).

Assumptions and pitfalls

• Values refer to standard states and a specific T (usually 298 K). Do not use a 298 K table at other temperatures.

• Multiply by coefficients. The answer is per mole of reaction as written.

• Slide correction: the more negative ΔG° is, the more favorable the reaction.

• ΔG° tells you about K, not the rate.

Worked example 1: 2H₂(g) + O₂(g) → 2H₂O(g) at 298 K (ΔfG° of H₂O(g) = −228.57 kJ·mol⁻¹)

$$\Delta G^\circ = 2(-228.57) - [0 + 0] = -457.14\ \text{kJ·mol}^{-1}$$

Worked example 2 (Hess combination): find ΔG° for 2CO + O₂ → 2CO₂ given

• (1) 2CH₄ + 3O₂ → 2CO + 4H₂O(l), ΔG° = −1088 kJ

• (2) CH₄ + 2O₂ → CO₂ + 2H₂O(l), ΔG° = −801 kJ

Target = 2 × (2) − (1):

• Reverse (1): 2CO + 4H₂O → 2CH₄ + 3O₂, ΔG° = +1088 kJ

• Double (2): 2CH₄ + 4O₂ → 2CO₂ + 4H₂O, ΔG° = −1602 kJ

• Sum (CH₄ and H₂O cancel, and 4O₂ − 3O₂ = O₂):

$$\Delta G^\circ = +1088 + 2(-801) = -514\ \text{kJ}$$

Predict the minimum temperature at which an endothermic reaction can occur spontaneously

Reasoning

• For an endothermic reaction (ΔH° > 0) to be spontaneous, TΔS° must overcome ΔH°. This requires ΔS° > 0 (the +/+ case in Card 7).

• Spontaneity requires ΔG° = ΔH° − TΔS° < 0, which gives:

$$T > T_{min} = \frac{\Delta H^\circ}{\Delta S^\circ}$$

• At T(min), ΔG° = 0 (K = 1).

• If ΔH° > 0 and ΔS° < 0, there is no temperature at which the reaction is spontaneous.

Phase-change connection

• At a transition temperature, ΔG = 0 and ΔH = TΔS. The “minimum temperature” for vaporization is the normal boiling point (at 1 bar).

Assumptions and pitfalls

• Assumes ΔH° and ΔS° are independent of T (a reasonable approximation).

• Units: ΔH° in J·mol⁻¹ with ΔS° in J·K⁻¹·mol⁻¹ (or both in kJ). The result is in kelvin.

• T(min) is a standard-state result. Under non-standard conditions (Q ≠ 1) the crossover shifts, because the true criterion is ΔG = ΔG° + RT ln Q < 0.

• A reaction that is thermodynamically allowed above T(min) can still be very slow.

Worked example: Br₂(l) → Br₂(g), ΔH° = 31 kJ·mol⁻¹, ΔS° = 93 J·K⁻¹·mol⁻¹

$$T_{min} = \frac{31\times10^{3}\ \text{J·mol}^{-1}}{93\ \text{J·K}^{-1}\text{mol}^{-1}} = 333\ \text{K}$$

Check at 400 K:

$$\Delta G^\circ = 31 - (400)(0.093) = -6.2\ \text{kJ·mol}^{-1}\ \text{(spontaneous)}$$

Check at 300 K:

$$\Delta G^\circ = 31 - (300)(0.093) = +3.1\ \text{kJ·mol}^{-1}\ \text{(not spontaneous)}$$

Connection between Gibbs free energy and equilibrium constant for any reaction

Core equations

$$\Delta G = \Delta G^\circ + RT\ln Q$$

• R = 8.314 J·K⁻¹·mol⁻¹, T in K, and Q is the reaction quotient at the current conditions.

• At equilibrium, ΔG = 0 and Q = K, so:

$0 = \Delta G^\circ + RT\ln K\quad\Rightarrow\quad \Delta G^\circ = -RT\ln K$$K = e^{-\Delta G^\circ / RT}$

• K is just one special case of Q (the value at equilibrium).

Defining Q and K properly

• Use activities: gases as p/p°, solutes as c/c°, and pure solids and liquids as 1 (omit them).

• Example, A ⇌ 2B:

$$Q = \frac{[B]^2}{[A]}\qquad K = \frac{[B]^2_{eq}}{[A]_{eq}}$$

• K is dimensionless and depends on how the reaction is written. Doubling the equation squares K and doubles ΔG°.

Interpretation

• ΔG° < 0 gives K > 1 (products favored); ΔG° > 0 gives K < 1; ΔG° = 0 gives K = 1.

• Q < K gives ΔG < 0 (forward); Q > K gives ΔG > 0 (reverse); Q = K is equilibrium.

• At 298 K, each 5.7 kJ·mol⁻¹ (that is, 2.303RT) of ΔG° changes K by a factor of 10.

• Equilibrium is the minimum of G along the reaction coordinate.

• Tabulated ΔfH° and S° give ΔG°, and hence K, for any reaction.

Pitfalls

• ΔG° > 0 does not mean “no reaction.” It means K < 1, so some product still forms.

• Use consistent units (ΔG° in J when R is in J).

• Do not use ΔG° to judge spontaneity under non-standard conditions. Use ΔG.

Worked example 1: ½H₂(g) + ½I₂(s) → HI(g), ΔG° = +1.70 kJ·mol⁻¹ at 298.15 K

$\ln K = -\frac{1700\ \text{J·mol}^{-1}}{(8.3145\ \text{J·K}^{-1}\text{mol}^{-1})(298.15\ \text{K})} = -0.686$$K = e^{-0.686} = 0.50$

Worked example 2: Kp for 2H₂ + O₂ → 2H₂O(g), ΔG° = −457.14 kJ·mol⁻¹

$$\ln K_p = \frac{457\,140}{(8.314)(298)} = 184.5\quad\Rightarrow\quad K_p = 1.4\times10^{80}$$

The reaction goes essentially to completion.

Worked example 3: ΔG for N₂ + 3H₂ ⇌ 2NH₃ at 298 K with p(N₂) = 1.0, p(H₂) = 3.0, p(NH₃) = 4.0 atm and ΔG° = −33.32 kJ·mol⁻¹

$Q = \frac{p_{NH_3}^2}{p_{N_2}\,p_{H_2}^3} = \frac{4^2}{(1)(3^3)} = \frac{16}{27} = 0.593$$RT\ln Q = (8.314\times10^{-3}\ \text{kJ·K}^{-1}\text{mol}^{-1})(298\ \text{K})(-0.523) = -1.30\ \text{kJ·mol}^{-1}$$\Delta G = -33.32 - 1.30 = -34.6\ \text{kJ·mol}^{-1} \approx -35\ \text{kJ·mol}^{-1}\ \text{(2 s.f.)}$

Note: the slide prints −1.43 for the RT ln Q term. The correct value is −1.30. The 2 s.f. answer of −35 is the same either way.

Summary of three laws of thermodynamics

First law: conservation of energy

• Energy cannot be created or destroyed, only converted between heat and work:

$$\Delta E = q + w$$

• Sign convention: q > 0 for heat absorbed by the system, and w > 0 for work done on the system.

• It says energy is conserved if a process occurs, but not whether it will occur.

Second law: direction of change

$$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} \ge 0$$

• Reversible: ΔS(univ) = 0. Irreversible (spontaneous): ΔS(univ) > 0.

• Microscopically, systems evolve toward the macrostate with the most microstates (S = kB ln W).

• For the system at constant T and P:

$$\Delta G_{sys} = -T\,\Delta S_{univ}$$

so spontaneous means ΔG < 0.

• ΔS(sys) alone can decrease (e.g., freezing, compression) as long as ΔS(univ) ≥ 0.

Third law: absolute entropy

• A perfect crystal at 0 K has S = 0 (W = 1), which allows absolute entropies and:

$$\Delta S^\circ_{rxn} = \sum S^\circ_{products} - \sum S^\circ_{reactants}$$

• A related consequence is that 0 K cannot be reached in a finite number of steps.

Master relationships

$\Delta G = \Delta H - T\Delta S$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -RT\ln K$$\Delta G = \Delta G^\circ + RT\ln Q$$\Delta S = \frac{q_{rev}}{T}\qquad \Delta S_{trans} = \frac{\Delta H_{trans}}{T_{trans}}$$\Delta S = nC_V\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}$

Biological extension: coupled reactions

• ΔG values add (Hess’s law), so a non-spontaneous reaction can be driven by a spontaneous one:

  • Glucose + Pi → glucose-6-phosphate: ΔG = +14.3 kJ (non-spontaneous)

  • ATP + H₂O → ADP + Pi: ΔG = −31.0 kJ (spontaneous)

  • Coupled: glucose + ATP → glucose-6-phosphate + ADP:

$$\Delta G = +14.3 + (-31.0) = -16.7\ \text{kJ}\quad\text{(spontaneous)}$$

• This is why ATP is called the energy currency of the cell.

Common pitfalls

• Applying ΔS(sys) ≥ 0 as if it were the second law. It holds for the universe, or for an isolated system.

• Confusing spontaneity (thermodynamics) with speed (kinetics).

• Treating ΔG° and ΔG as interchangeable.

Worked example linking all three laws: 2SO₂(g) + O₂(g) → 2SO₃(g) at 298 K (ΔH° = −198 kJ, ΔS° = −187 J·K⁻¹)

Third law: absolute S° values give ΔS(sys) = −187 J·K⁻¹.

First law (constant P): q(sys) = ΔH° = −198 kJ, so

$$\Delta S_{surr} = \frac{+198\,000\ \text{J}}{298\ \text{K}} = +664\ \text{J·K}^{-1}$$

Second law:

$$\Delta S_{univ} = -187 + 664 = +477\ \text{J·K}^{-1} > 0\quad\Rightarrow\quad\text{spontaneous}$$

Equivalent check:

$\Delta G^\circ = -T\Delta S_{univ} = -(298)(477) = -1.42\times10^{5}\ \text{J} = -142\ \text{kJ}$$K = e^{142\,000/(8.314\cdot298)} \approx 10^{25}$