6.3 Chemical Reaction Equilibrium

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):

$$\Delta G^{\circ}(T) = -RT \ln K$$

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 $\Delta H^{\circ}$):

$$\ln \frac{K_2}{K_1} = -\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

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 $\Delta n \neq 0$:

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

Critical exam facts:

1. 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).
2. Zero conversion can mean either K is tiny or the reaction is frozen kinetically—or the reactor is too small/cold.
3. 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.
4. 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.

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

1. Claiming a catalyst changes K or equilibrium yield in a closed system.
2. Saying pressure changes K_p for ideal gases (it changes conversion when $\Delta n \neq 0$, not K_p(T)).
3. Applying liquid Le Chatelier pressure logic as if liquids were ideal gases.
4. Confusing phase equilibrium K-values ((y/x)) with reaction K.
5. Using stoichiometric feed assumptions when the stem specifies excess air or recycle.
6. Forgetting that $\Delta G^{\circ}$ (standard) sets K, while $\Delta G$ at actual conditions is zero at equilibrium.

Domain B Wrap for Chapter 6

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.