Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 4

Let $G$ be Gibbs free energy, $H$ enthalpy and $T$ thermodynamic temperature for a fixed amount and composition of material, with the derivative taken at constant pressure. Apply the Gibbs–Helmholtz equation to the standard molar reaction quantities, obtained by summing product minus reactant contributions with their stoichiometric coefficients:Define $K(T)$ as the dimensionless thermodynamic equilibrium constant expressed using activities in specified standard states. Since ${\Delta }_r{G}^{\circ }=-RT\ln K$, where $R$ is the molar gas constant,If the reaction enthalpy is constant over the interval, integration givesThis is the requested Van't Hoff equation. For an exothermic reaction, increasing temperature decreases $K$. Although the name “isochore” is often used in this context, the displayed enthalpy form follows from standard Gibbs energies and the stated constant-pressure Gibbs–Helmholtz identity; the analogous ideal-gas concentration form involves internal energy when the temperature dependence of standard concentration is retained.

The two remaining partial pressures sum to $200-160=40$ bar. Their ratio therefore gives $p_{{\mathrm{N}}_2}=10$ bar and $p_{\mathrm{N}{\mathrm{H}}_3}=30$ bar. With the ideal-gas activity approximation $a_i=p_i/p^{\circ }$ and $p^{\circ }=1$ bar,The equilibrium constant is dimensionless: pressures enter as ratios to the standard pressure. Blindly substituting pressures in pascals into an unnormalized expression would give a different, incorrect thermodynamic $K$.

The standard reaction enthalpy is $2(-45.90)=-91.80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ per mole of reaction as written, because the elements are in their reference states. ThusThis calculation uses the supplied constant-enthalpy simplification. At 200 bar, an accurate thermodynamic treatment would use chemical fugacities rather than ideal partial pressures; no chemical fugacity coefficients are supplied. The iron catalyst changes the approach to equilibrium, not its thermodynamic value.

Using the equilibrium constant calculated without premature rounding,Then ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$ givesThe sign agrees with the reduction in gas stoichiometric amount. The standard reaction entropy is not the absolute entropy of ammonia: its stoichiometric coefficient is two. ThereforeThis is an absolute standard molar entropy inferred within the paper's thermodynamic approximations, not a claim that elemental gases have zero absolute entropy. Only their standard enthalpies of formation are zero in this calculation.

For a reversible temperature change at constant pressure and fixed composition, $\mathrm{d}{S}_m=\delta q_{\mathrm{rev},\mathrm{m}}/T$ and $\delta q_{\mathrm{rev},\mathrm{m}}=C_{p,m}\mathrm{d}T$. HenceWith $T_0=298\mathrm{K}$, termwise integration yieldsThe last sign follows from integrating $e{T}^{-3}$, not $e{T}^{-2}$. Every term has units $\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$; the logarithm has a dimensionless argument. Inserting 800 K gives an entropy increase of $41.733\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$, and thereforeThe standard pressure remains fixed during this integration; the 200-bar equilibrium mixture is not the standard state. Differentiating the integrated expression recovers $C_p/T$, providing a check on all five terms. Using the ammonia heat-capacity polynomial here does not retroactively replace the explicitly requested constant-reaction-enthalpy approximation in the preceding equilibrium calculation.