Standard Entropy and Free Energy Changes in Chemical Reactions

Entropy and Free Energy in Chemical Reactions

Standard State and Standard Molar Entropy

The standard state of a substance is defined as its pure form at 1 bar (or 1 atm, historically) and 25°C (298 K). The standard molar entropy, S°, is the absolute entropy of 1 mole of a pure substance in its standard state at 25°C. Unlike standard enthalpy of formation (ΔH°f), which is zero for elements in their standard state, S° is always positive for all substances, including elements.

• Standard State (Gas): Pure gas at exactly 1 bar (or 1 atm) pressure and 25°C.

• Standard State (Liquid/Solid): Pure substance in its most stable form at 1 bar and 25°C.

• Standard State (Solution): 1 M concentration at 25°C.

• Units: J/mol·K (joules per mole per kelvin).

The Third Law of Thermodynamics

The Third Law of Thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is zero: S(0 K, perfect crystal) = 0. This provides an absolute reference point for entropy, allowing us to assign absolute entropy values to all substances.

• At 0 K, a perfect crystal has only one possible arrangement (W = 1), so S = k ln(1) = 0.

• As temperature increases, energy disperses and the number of microstates (W) increases, so S increases.

• No real substance at room temperature has S = 0; all have S > 0.

Calculating Standard Entropy Change (ΔS°rxn)

The standard entropy change for a reaction is calculated using the standard molar entropies of the reactants and products:

$\Delta S^\circ_{\mathrm{rxn}} = \sum n_p S^\circ(\text{products}) - \sum n_r S^\circ(\text{reactants})$

• All stoichiometric coefficients must be included as multipliers.

• Unlike enthalpy, absolute entropy values can be measured directly.

Trends in Standard Molar Entropy

Standard molar entropy values show clear trends based on physical state, molar mass, and molecular complexity:

• Physical State: Gases have the highest S°, followed by liquids, then solids.

• Molar Mass: Heavier atoms and molecules generally have higher S° due to more closely spaced energy levels and more microstates.

• Molecular Complexity: More complex molecules (more atoms, more vibrational modes) have higher S°.

• Dissolution: Dissolving a solid or liquid in water usually increases entropy.

Example: Calculating ΔS°rxn

For the reaction: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

$\Delta S^\circ_{\mathrm{rxn}} = [4S^\circ(\mathrm{NO}) + 6S^\circ(\mathrm{H_2O})] - [4S^\circ(\mathrm{NH_3}) + 5S^\circ(\mathrm{O_2})]$

Predicting the Sign of ΔS°rxn

The sign of ΔS°rxn can often be predicted by comparing the number of moles of gas on each side of the reaction:

• More moles of gas on the product side → ΔS°rxn is positive.

• Fewer moles of gas on the product side → ΔS°rxn is negative.

Free Energy Changes in Chemical Reactions: Calculating ΔG°rxn

Three Methods to Calculate ΔG°rxn

1. ΔG°rxn = ΔH°rxn − TΔS°rxn Calculate ΔH°rxn using standard enthalpies of formation, ΔS°rxn using standard entropies, then combine at the specified temperature (usually 298 K).

2. ΔG°rxn = ΣnpΔG°f(products) − ΣnrΔG°f(reactants) Use tabulated standard free energies of formation (ΔG°f). ΔG°f = 0 for elements in their standard state.

3. Hess's Law for Free Energy Combine free energy changes from known reactions stepwise. ΔG is a state function, so the total change is path-independent.

Example: Calculating ΔG°rxn Using ΔH°rxn and ΔS°rxn

For the reaction: SO2(g) + 1/2 O2(g) → SO3(g)

$\Delta G^\circ_{\mathrm{rxn}} = \Delta H^\circ_{\mathrm{rxn}} - T\Delta S^\circ_{\mathrm{rxn}}$

At 25°C (298 K):

$\Delta G^\circ_{\mathrm{rxn}} = -98.9 \times 10^3 \; \mathrm{J} - 298 \; \mathrm{K} \times (-94.0 \; \mathrm{J/K}) = -70.9 \; \mathrm{kJ}$

Example: Calculating ΔG°rxn Using Standard Free Energies of Formation

For the reaction: CH4(g) + 8 O2(g) → CO2(g) + 2 H2O(g) + 4 O3(g)

$\Delta G^\circ_{\mathrm{rxn}} = [\Delta G^\circ_{f,\mathrm{CO_2}} + 2\Delta G^\circ_{f,\mathrm{H_2O}} + 4\Delta G^\circ_{f,\mathrm{O_3}}] - [\Delta G^\circ_{f,\mathrm{CH_4}} + 8\Delta G^\circ_{f,\mathrm{O_2}}]$

Using tabulated values, ΔG°rxn = −148.3 kJ.

Key Takeaways

• The Third Law provides an absolute zero for entropy, allowing calculation of absolute S° values.

• S° is always positive; elements in their standard state have S° ≠ 0.

• ΔS°rxn is calculated using standard molar entropies and stoichiometric coefficients.

• Predict the sign of ΔS°rxn by comparing moles of gas on each side of the equation.

• Three methods exist for calculating ΔG°rxn: (1) ΔH°rxn − TΔS°rxn, (2) ΣΔG°f, (3) Hess's Law.

• Always check units: ΔH in kJ, TΔS in J (convert as needed); temperature must be in Kelvin.