Chemical Equilibrium — Gibbs Energy, Temperature and Chemical Potentials
By Prof. Marco Ruzzi.
Theoretical background
At fixed temperature and pressure, the sign of the reaction Gibbs energy determines the direction of thermodynamic progress:
\[\Delta_rG=\Delta_rG^\circ+RT\ln Q.\] \[\Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ.\]At equilibrium Q=K and ΔrG=0, hence:
\[\Delta_rG^\circ=-RT\ln K.\] \[K=\exp\left(-\frac{\Delta_rG^\circ}{RT}\right).\]Exercise 6 — Exothermic oxidation and temperature
Discuss how temperature affects the spontaneity, from standard conditions, of the following exothermic reaction:
\[2\mathrm{NO}(g)+\mathrm{O}_2(g)\rightleftharpoons2\mathrm{NO}_2(g).\]Solution
We compare the reaction under standard conditions, where Q=1. ΔrH°<0 because the reaction is exothermic. ΔrS°<0 for this reaction, consistently with the reduction from three moles of gaseous reactants to two moles of gaseous products.
High temperatures. In the constant-ΔH°, constant-ΔS° approximation, sufficiently high T makes the positive contribution −TΔrS° dominate:
\[\Delta_rG^\circ\simeq-T\Delta_rS^\circ>0.\]The forward reaction is then unfavorable from standard conditions; raising T makes it less favorable.
Low temperatures. At sufficiently low T the negative enthalpic contribution dominates:
\[\Delta_rG^\circ\simeq\Delta_rH^\circ<0.\]The forward reaction is favorable from standard conditions when the magnitude of the negative enthalpy contribution exceeds the opposing entropy term:
\[|\Delta_rH^\circ|>|T\Delta_rS^\circ|.\]Within this approximation, that inequality is more marked at lower temperature.
Observations
If oxygen were liquid, the sign of the reaction entropy could not be inferred merely from the phases indicated in the reaction equation. To calculate ΔrS° and establish its sign, the standard molar entropies of all participating substances would be required: NO(g), NO₂(g) and O₂(l).
Likewise, if the exercise did not state that the reaction is exothermic, its enthalpy change would have to be calculated from the formation enthalpies of NO(g) and NO₂(g). For O₂(g) in its elemental standard reference state, the standard formation enthalpy is zero.
Exercise 7 — Copper carbonate and chemical potentials
The reaction:
\[\mathrm{CuCO}_3(s)\rightleftharpoons\mathrm{CuO}(s)+\mathrm{CO}_2(g)\]has Kp=50. A reaction mixture has a CO₂ pressure of 50 bar. Determine whether it is at equilibrium and write the resulting relation between the chemical potentials of its components.
Define the chemical-potential difference:
\[D=\mu_{\mathrm{CuO}(s)}+\mu_{\mathrm{CO}_2(g)}-\mu_{\mathrm{CuCO}_3(s)}.\]Solution
For pure solid phases, their activities are one. Using the pressure-based ideal-gas expression, with p°=1 bar:
\[Q_p=\frac{p_{\mathrm{CO}_2}}{p^\circ}=50=K_p.\]The mixture is therefore at equilibrium. In general, the equilibrium condition on chemical potentials is:
\[\begin{aligned} \Delta_rG={}&\sum_{\mathrm{products}}\nu_i\mu_i\\ &-\sum_{\mathrm{reactants}}\nu_i\mu_i=0. \end{aligned}\]For this reaction:
\[\mu_{\mathrm{CuO}(s)}+\mu_{\mathrm{CO}_2(g)}-\mu_{\mathrm{CuCO}_3(s)}=0.\]Problem — NO oxidation and its equilibrium constant
Consider:
\[2\mathrm{NO}(g)+\mathrm{O}_2(g)\rightleftharpoons2\mathrm{NO}_2(g).\]At T=298.15 K, the formation enthalpies are +90.2 kJ mol⁻¹ for NO(g) and +33.2 kJ mol⁻¹ for NO₂(g). The standard reaction entropy is −145.0 J mol⁻¹ K⁻¹.
- Write the expression for Kp.
- Calculate the equilibrium constant at 298.15 K.
- Starting from standard conditions, determine the direction in which the reaction moves toward equilibrium.
- What is ΔrG at equilibrium?
- Describe the effect of increasing temperature on the equilibrium.
1. Reaction quotient and equilibrium constant
For ideal gases, normalize each partial pressure by p°=1 bar:
\[Q=\frac{(p_{\mathrm{NO}_2}/p^\circ)^2}{(p_{\mathrm{NO}}/p^\circ)^2(p_{\mathrm{O}_2}/p^\circ)}.\]Equivalently:
\[Q=\frac{p_{\mathrm{NO}_2}^2p^\circ}{p_{\mathrm{NO}}^2p_{\mathrm{O}_2}}.\]At equilibrium the partial pressures no longer change and Kp=Q evaluated at those pressures. The constant is dimensionless.
2. Numerical value of the equilibrium constant
At equilibrium:
\[0=\Delta_rG^\circ+RT\ln K_p.\]Thus ΔrG°=−RT ln Kp, and calculating Kp requires the standard reaction Gibbs energy.
First calculate the enthalpy from the formation data:
\[\begin{aligned} \Delta_rH^\circ={}&2\Delta_fH^\circ(\mathrm{NO}_2)\\ &-2\Delta_fH^\circ(\mathrm{NO})\\ &-\Delta_fH^\circ(\mathrm{O}_2). \end{aligned}\] \[\Delta_rH^\circ=2(33.2)-2(90.2)-0=-114\ \mathrm{kJ\,mol^{-1}}.\]The entropy is supplied directly: −145.0 J mol⁻¹ K⁻¹, or −0.1450 kJ mol⁻¹ K⁻¹. Hence:
\[\Delta_rG^\circ=-114-(298.15)(-0.1450)\ \mathrm{kJ\,mol^{-1}}.\] \[\Delta_rG^\circ\simeq-70.8\ \mathrm{kJ\,mol^{-1}}.\]Using R=8.314 J mol⁻¹ K⁻¹ and the rounded Gibbs energy:
\[-\frac{\Delta_rG^\circ}{RT}\simeq\frac{70.8\times10^3}{(8.314)(298.15)}\simeq28.56.\] \[K_p=e^{28.56}\simeq2.5\times10^{12}.\]3. Direction from standard conditions
The standard conditions give Q=1. Since ΔrG°≈−70.8 kJ mol⁻¹, the reaction proceeds toward the products.
4. Gibbs energy at equilibrium
The reaction Gibbs energy vanishes at equilibrium:
\[\Delta_rG=0.\]5. Effect of temperature
Here ΔrH°=−114 kJ mol⁻¹ and ΔrS°=−145.0 J mol⁻¹ K⁻¹. The term TΔrS° is negative; its magnitude grows with temperature, so its contribution −TΔrS° to the Gibbs energy is increasingly positive. Raising the temperature favors the reactants in this exothermic equilibrium.