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The dimensionless ratio of a gas species' chemical fugacity to its pressure in the pure-fluid convention. Its thermodynamic activity relative to a gas reference pressure $P^{\circ }$ is $\varphi P/P^{\circ }$. For mixtures the pressure denominator uses the species' partial pressure. The ideal-gas limit has $\varphi =1$.

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.

For pure phases at the $1$ bar reference pressure, the equilibrium condition is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }=0$. ThusThe calculation uses the stipulated constant reaction enthalpy and entropy, with the fluid at its reference chemical fugacity.

Integrate the solids’ molar volume contribution from $P^{\circ }=1$ bar and use ideal-gas water chemical fugacity $f=P$. For pure solids,The logarithm is dimensionless, and $\Delta V_s=-2.28\mathrm{J}\mathrm{ba}{\mathrm{r}}^{-1}$ can be multiplied directly by pressure in bars. Therefore the dehydration equilibrium with ideal-gas fluid givesHence $T=1295.57$ K, or $1022.42^{\circ }$C. Do not also add a constant gas-volume pressure term: the water contribution has already been integrated through $RT\ln (P/P^{\circ })$. Ideal-gas behavior at $6$ kbar is a prescribed model approximation, not a realistic equation of state. The solid volume correction is distinct from it.

With water fugacity coefficient ${\varphi }_f=0.8$, use $f={\varphi }_{f}P=4800$ bar. Retain the same ideal mineral-site activities, givingHenceThe lower fluid chemical fugacity stabilizes the water-bearing products of the forward dehydration relative to the ideal-gas calculation and lowers the required temperature by about $13.19$ K. Here the stated fugacity coefficient is treated as a given constant for evaluating the equilibrium.

Use the pressure correction to a mineral reaction for the solids, and the gas thermodynamic activity $f/P_0$ for water. These are separate contributions: do not also add a gas-volume pressure correction after including $RT\ln (f/P_0)$. For this reaction,At $7000\mathrm{bar}$ with pure solids and an ideal gas, $f=P$ and the solid-activity ratio is one. ThereforeandThe pressure unit in the solid term is bar, matching the supplied $\mathrm{J}\mathrm{ba}{\mathrm{r}}^{-1}$. The gas logarithm is dimensionless because $P/P_0=7000$. High water chemical fugacity opposes dehydration and raises the equilibrium temperature substantially.

The fugacity coefficient $\varphi =f/P=0.7$ gives $f=3500\mathrm{bar}$. Keep the same solid thermodynamic activities and pressure correction to a mineral reaction as in part (c), changing only the water chemical fugacity:The lower water chemical fugacity makes the dehydrated assemblage more favorable, so the equilibrium temperature decreases by about $29.0\mathrm{K}$. This direction is a useful independent check on the logarithm's sign.

Apply the pressure correction to a mineral reaction to the solids and use the chemical fugacity of the water product. For pure solids and water vapour treated as an ideal gas,The gas contribution is already included by the logarithm and must not be counted again as an independent volume correction. With pressures in bars,

Use the pressure correction to a mineral reaction for the solids, with $P_0=1\mathrm{bar}$. The water contribution must instead use its chemical fugacity. For pure solid phases and water vapour treated as an ideal gas, $f_{{\mathrm{H}}_2\mathrm{O}}=P$, givingThe last term already accounts for the gas-pressure contribution; adding another gas-volume correction would double count it. The pressure ratio inside the logarithm is dimensionless.

With pressure expressed in bars, the solid correction is $-2.28(4500-1)=-10257.72\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. ThusUsing $4500$ rather than $4499$ for the solid pressure increment would silently shift the specified one-bar reference.

At fixed temperature, $dG=VdP$ gives the pressure correction for the participating solids. Constant reaction volume reduces the integral to $\Delta V_{\mathrm{solids}}(P-P_0)$. A gas product treated by chemical fugacity instead contributes $\nu RT\ln (f/P_0)$ relative to its reference state. Adding a separate gas-volume correction to that term would double count the same pressure dependence.