6.3 Chemical Reaction Equilibrium
Key Takeaways
- The thermodynamic equilibrium constant K is fixed by standard Gibbs energy: ΔG° = −RT ln K (with K defined consistently with the standard state).
- Temperature shifts K through van ’t Hoff: exothermic reactions have K decreasing as T rises; endothermic reactions have K increasing with T.
- Pressure and inert dilution shift equilibrium conversion for gas-phase reactions with mole change via Le Chatelier’s principle, even when K itself is approximately P-independent for ideal gases.
- Equilibrium conversion is the thermodynamic ceiling; kinetics (rate) determine how fast you approach it—catalysts speed rates without changing K.
- UPDA items often combine qualitative Le Chatelier reasoning with simple K expressions in partial pressures or mole fractions.
6.3 Chemical Reaction Equilibrium
Quick Answer: (\Delta G^{\circ} = -RT \ln K) fixes the equilibrium constant. T changes K (van ’t Hoff / Le Chatelier). P and inerts change equilibrium conversion for gas reactions with (\Delta n \neq 0). Equilibrium ≠ kinetics: K sets the ceiling; rate laws and catalysts set the speed.
Phase equilibrium (Sections 6.1–6.2) equates fugacities across phases. Chemical reaction equilibrium equates chemical potentials through a stoichiometric reaction, fixing the composition a closed system approaches given enough time. Domain B tests the thermo side; Domain D (Chemical Reaction Engineering) adds rate laws and reactor types—this section keeps the boundary clear.
Equilibrium Constant and ΔG°
For a general reaction
[ \nu_A A + \nu_B B \rightleftharpoons \nu_C C + \nu_D D ]
the standard Gibbs energy change of reaction (\Delta G^{\circ}(T)) relates to the thermodynamic equilibrium constant (K):
[ \Delta G^{\circ}(T) = -RT \ln K ]
| (\Delta G^{\circ}) | K | Equilibrium position |
|---|---|---|
| Large negative | K ≫ 1 | Products strongly favored |
| Zero | K = 1 | Comparable product/reactant activities |
| Large positive | K ≪ 1 | Reactants favored; low equilibrium conversion |
For ideal gases with standard state 1 bar (or 1 atm in older texts), K is often expressed as (K_f) in fugacities; at low pressure:
[ K \approx K_p = \prod_i (p_i)^{\nu_i} = \prod_i (y_i P)^{\nu_i} = K_y , P^{\Delta n} ]
where (\Delta n = \sum \nu_{\mathrm{products}} - \sum |\nu_{\mathrm{reactants}}|) for the gas-phase stoichiometry as written (net change in moles of gas).
For the ammonia synthesis reaction written as:
[ \mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}, \quad \Delta n = 2 - 4 = -2 ]
[ K_p = \frac{p_{\mathrm{NH_3}}^2}{p_{\mathrm{N_2}} p_{\mathrm{H_2}}^3} = \frac{y_{\mathrm{NH_3}}^2}{y_{\mathrm{N_2}} y_{\mathrm{H_2}}^3} P^{-2} ]
So at fixed K and T, higher total P increases equilibrium mole fraction of NH₃—classic industrial high-pressure synthesis.
Liquid-phase or heterogeneous reactions use activities (mole fractions, molalities, or pure solid activity ≈ 1). On UPDA MCQs, read whether K is (K_c), (K_p), or in mole fractions, and keep units/standard states consistent with the given K.
Effect of Temperature on K
The van ’t Hoff relation (integrated form for constant (\Delta H^{\circ})):
[ \ln \frac{K_2}{K_1} = -\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) ]
| Reaction type | Raise T | Effect on K |
|---|---|---|
| Exothermic ((\Delta H^{\circ} < 0)) | → | K decreases; equilibrium shifts toward reactants |
| Endothermic ((\Delta H^{\circ} > 0)) | → | K increases; equilibrium shifts toward products |
Le Chatelier statement: adding heat to an exothermic equilibrium favors the endothermic reverse direction.
Industrial tension: ammonia and SO₃ oxidation want low T for high K but high T for fast rates—hence catalysts and optimized temperature profiles (Domain D). Equilibrium thermo alone says “cold favors NH₃,” but a cold reactor without catalyst is useless on a plant timescale.
Effect of Pressure and Inerts on Conversion
For ideal-gas reactions, K_p is a function of T only (not of total P). Still, equilibrium conversion depends on P when (\Delta n \neq 0):
| Gas-phase (\Delta n) | Increase total P at fixed T, feed | Equilibrium conversion of reactants |
|---|---|---|
| (\Delta n < 0) (moles decrease) | → | Increases |
| (\Delta n > 0) (moles increase) | → | Decreases |
| (\Delta n = 0) | → | Little/no effect via P for ideal gases |
Inert dilution at fixed total P lowers partial pressures of reactants/products. For reactions with (\Delta n < 0), inerts typically hurt equilibrium conversion (similar to lowering partial pressures). For (\Delta n > 0), dilution can help conversion. Always reason from the reaction as written and the expression for K.
Liquid systems are weakly compressible; pressure effects on K are usually negligible on exam qualitative items unless high-P specialty chemistry is mentioned.
Equilibrium vs Kinetics
| Concept | What it answers | Tools |
|---|---|---|
| Equilibrium | How far can the reaction go? | (\Delta G^{\circ}), K, Le Chatelier, VLE if multiphase |
| Kinetics | How fast does it go? | Rate law, (k(T)), E_a, catalyst, reactor type |
Critical exam facts:
- A catalyst increases forward and reverse rates but does not change (\Delta G^{\circ}) or K; it shortens time to the same equilibrium composition (in a closed system).
- Zero conversion can mean either K is tiny or the reaction is frozen kinetically—or the reactor is too small/cold.
- Equilibrium conversion is the upper bound for a single reaction in a closed system at given T (and P, feed); real reactors may stop earlier.
- Multiple reactions need simultaneous equilibria or kinetic selectivity—do not force one K to describe two independent paths without data.
Worked Example 1: From ΔG° to K
Suppose at 500 K a reaction has (\Delta G^{\circ} = -8.314,\mathrm{kJ/mol} = -8314,\mathrm{J/mol}).
[ \ln K = -\frac{\Delta G^{\circ}}{RT} = -\frac{-8314}{(8.314)(500)} = \frac{8314}{4157} = 2.00 ]
[ K = e^{2.00} \approx 7.4 ]
Products are moderately favored. If the stem gave (\Delta G^{\circ} = +8314,\mathrm{J/mol}) at the same T, then (\ln K = -2) and (K \approx 0.135)—reactants favored.
Worked Example 2: Ideal-Gas Equilibrium Conversion
Reaction: (A \rightleftharpoons 2B) (ideal gas), (\Delta n = +1). Pure A feed. Isothermal, isobaric batch (or flow with long residence time). Let X = equilibrium conversion of A.
Moles: A: (1-X), B: (2X), total: (1+X).
[ y_A = \frac{1-X}{1+X}, \quad y_B = \frac{2X}{1+X} ]
[ K_p = \frac{p_B^2}{p_A} = \frac{(y_B P)^2}{y_A P} = \frac{y_B^2}{y_A} P = \frac{4X^2}{(1-X)(1+X)} P = \frac{4X^2}{1-X^2} P ]
If K_p = 0.5 bar and P = 1 bar:
[ 0.5 = \frac{4X^2}{1-X^2}(1) \implies 0.5(1-X^2) = 4X^2 \implies 0.5 = 4.5 X^2 \implies X^2 = \frac{0.5}{4.5} = 0.111 ]
[ X = \sqrt{0.111} \approx 0.33\ (33%) ]
Pressure effect: same K_p at P = 4 bar:
[ 0.5 = \frac{4X^2}{1-X^2}(4) \implies 0.5 = \frac{16X^2}{1-X^2} \implies 0.5 - 0.5X^2 = 16X^2 \implies 0.5 = 16.5 X^2 ]
[ X \approx \sqrt{0.0303} \approx 0.17\ (17%) ]
Higher P lowers X because (\Delta n > 0)—Le Chatelier matches the algebra.
| P (bar) | Approx. X_eq (this example) |
|---|---|
| 1 | 0.33 |
| 4 | 0.17 |
Worked Example 3: Exothermic Shift with T
SO₂ oxidation toward SO₃ is strongly exothermic. Qualitatively:
- High T ⇒ smaller K ⇒ lower equilibrium SO₃ fraction
- Low T ⇒ larger K ⇒ higher equilibrium conversion
- Industrial converters use catalysts and staged cooling to balance rate vs equilibrium
If an MCQ asks only equilibrium, pick the colder option for higher conversion of an exothermic reaction; if it asks rate or approach to equilibrium in a small reactor, high T or catalyst may dominate the practical answer—read the stem.
Common UPDA Traps
- Claiming a catalyst changes K or equilibrium yield in a closed system.
- Saying pressure changes K_p for ideal gases (it changes conversion when (\Delta n \neq 0), not K_p(T)).
- Applying liquid Le Chatelier pressure logic as if liquids were ideal gases.
- Confusing phase equilibrium K-values ((y/x)) with reaction K.
- Using stoichiometric feed assumptions when the stem specifies excess air or recycle.
- Forgetting that (\Delta G^{\circ}) (standard) sets K, while (\Delta G) at actual conditions is zero at equilibrium.
Domain B Wrap for Chapter 6
| Topic | Core exam skill |
|---|---|
| Phase rule | Count F = C − P + 2 |
| VLE | Raoult, Henry, activity idea, critical vs triple |
| Bubble/dew/flash | First bubble/drop; flash balances + K_i; α for separation |
| Reaction equilibrium | K ↔ ΔG°; T, P, inerts; equilibrium vs rate |
Together with Chapter 5 (thermo laws, EOS, H and S), you now have the Domain B toolkit for the ~15% thermodynamics and phase-equilibria share of UPDA/MMUP Chemical exam. Transport chapters next apply driving forces that vanish at the equilibria you just defined.
If ΔG° for a reaction at temperature T is large and positive, which statement is correct?
For the ideal-gas reaction N₂ + 3 H₂ ⇌ 2 NH₃ (Δn = −2), increasing total pressure at fixed temperature and fixed feed composition tends to:
A process engineer adds a catalyst to an exothermic reversible reaction in a closed batch reactor. At the same final temperature, the long-time equilibrium composition compared with the uncatalyzed case will: