Calculating Equilibrium Constant from Gibbs Free Energy – Haber Process Examples

Relationship Between Gibbs Free Energy (ΔG°) and the Equilibrium Constant (K)

  • Fundamental equation introduced/used in class:
    ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT \ln K
  • Rearranged for solving K:
    ln⁡K=−ΔG∘RT\ln K = -\dfrac{\Delta G^{\circ}}{RT}
    K=e−ΔG∘RTK = e^{-\frac{\Delta G^{\circ}}{RT}}
  • Meaning of symbols
    • ΔG∘\Delta G^{\circ} = standard Gibbs free-energy change (units: \text{kJ·mol}^{-1}, but must be converted to \text{J·mol}^{-1} when used with R).
    • R = ideal-gas constant = 8.314\ \text{J·mol}^{-1}\text{K}^{-1}.
    • T = absolute temperature in Kelvin.
    • K = equilibrium constant (unitless) representing (products)(reactants)\dfrac{(\text{products})}{(\text{reactants})} at equilibrium.

Worked Example 1 – Haber Process at 25 °C

  • Reaction: N<em>2(g)+3 H</em>2(g)⇌2 NH3(g)\text{N}<em>2(g) + 3\,\text{H}</em>2(g) \rightleftharpoons 2\,\text{NH}_3(g)
  • Given/previously calculated: \Delta G^{\circ}_{25^{\circ}\text{C}} = -33.3\ \text{kJ·mol}^{-1}.
  • Convert ΔG∘\Delta G^{\circ} to joules: -33.3\ \text{kJ·mol}^{-1}\times1000 = -3.33\times10^{4}\ \text{J·mol}^{-1}.
  • Insert into rearranged formula:
    ln⁡K=−−3.33×104(8.314)(298)=13.4\ln K = -\dfrac{-3.33\times10^{4}}{(8.314)(298)}=13.4
  • Exponentiate to isolate K:
    K=e13.4≈7×105K = e^{13.4}\approx7\times10^{5}
  • Interpretation:
    • A very large K (≫1) indicates the equilibrium position at 25 °C lies heavily toward products (NH₃).
    • Numerically, product concentration dominates; reactants are depleted.

Worked Example 2 – Haber Process at 500 °C

  • Temperature conversion: 500∘C=500+273=773 K500^{\circ}\text{C} = 500+273 = 773\ \text{K}.
  • Provided \Delta G^{\circ}_{773\,\text{K}} = +61\ \text{kJ·mol}^{-1} (extracted from earlier slide).
  • Convert to joules: +61\ \text{kJ·mol}^{-1}\times1000 = 6.10\times10^{4}\ \text{J·mol}^{-1}.
  • Calculate ln⁡K\ln K:
    ln⁡K=−6.10×104(8.314)(773)=−9.49\ln K = -\dfrac{6.10\times10^{4}}{(8.314)(773)} = -9.49
  • Solve for K:
    K=e−9.49≈7.45×10−5K = e^{-9.49}\approx7.45\times10^{-5}
  • Interpretation:
    • K ≪ 1 implies equilibrium now favors reactants (N₂ and H₂).
    • Demonstrates temperature’s dramatic effect on equilibrium position for the Haber reaction.

Theoretical & Practical Significance

  • Temperature Dependence:
    • As temperature increases from 298 K to 773 K, ΔG∘\Delta G^{\circ} shifts from negative to positive, flipping the favored side.
    • Consistent with Le Châtelier’s Principle for an exothermic formation of NH₃: raising T drives equilibrium toward reactants.
  • Industrial Relevance:
    • Although low temperatures favor NH₃ yield (large K), practical production balances kinetics (rate) and thermodynamics.
    • High pressure and optimized catalysts are employed in real Haber plants to offset the reduced K at elevated T.
  • Conceptual Link:
    • Connects back to prior lectures on spontaneity: \Delta G^{\circ}<0 ↔ spontaneous under standard conditions.
    • Shows direct quantitative bridge between thermodynamic driving force (free energy) and observable equilibrium composition.

Calculation Reminders & Common Pitfalls

  • Always convert kJ→J\text{kJ} \to \text{J} when pairing ΔG∘\Delta G^{\circ} with R.
  • Temperature must be Kelvin.
  • Sign discipline: negative signs in ΔG∘\Delta G^{\circ} or rearranged formulae can invert interpretation if mishandled.
  • Exponentials grow fast; check with orders-of-magnitude reasoning (e.g., ln⁡K=13\ln K=13 → K≈105K≈10^{5}).

Quick Reference Equations

  • Core identity: ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^{\circ} + RT\ln Q (specializes to Q=KQ=K at equilibrium).
  • Equilibrium expression for Haber process (all gases):
    K=[NH<em>3]2[N</em>2][H2]3K = \dfrac{[\text{NH}<em>3]^2}{[\text{N}</em>2][\text{H}_2]^3} (activities if not ideal).

Take-Home Messages

  • You can translate a measured/calculated ΔG∘\Delta G^{\circ} directly into an equilibrium constant at any temperature.
  • The sign and magnitude of ΔG∘\Delta G^{\circ} dictate how far a reaction proceeds under standard conditions.
  • For exothermic industrial syntheses like ammonia formation, optimizing T involves a trade-off between favorable K (low T) and acceptable reaction rates (higher T).