Synthesis-Gas Reaction Equilibria: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2013, Q7

Question 7: Synthesis-Gas Reaction Equilibria (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Reactions (1) CH₄ + H₂O ⇌ CO + 3H₂ ($\Delta n_{gas}=+2$) and (2) CO + H₂O ⇌ CO₂ + H₂ ($\Delta n_{gas}=0$). Standard Gibbs energies of formation (J/mol), as printed in the exam table:

Species	$\Delta G_f^\circ$ at 600 K	$\Delta G_f^\circ$ at 1300 K
CH₄	$-22{,}980$	$+52{,}325$
H₂O(g)	$-214{,}105$	$-175{,}890$
CO	$-164{,}755$	$-227{,}035$
CO₂	$-395{,}305$	$-396{,}310$
H₂	0	0

Find. (a) favourable pressure; (b) favourable temperature; (c) $K_1$ at those conditions; (d) the H₂/CO ratio; (e) $\Delta H^\circ_{800\,\text{K}}$ for reaction 1.

Figure 7 — The two coupled equilibria: reforming (1) makes syngas with a net gain of 2 gas moles; the shift (2) consumes CO with no change in moles.

Approach. Le Chatelier settles (a) and (b) by mole-change and endothermicity; then $K=\exp(-\Delta G^\circ/RT)$ gives (c); the ratio in (d) follows from stoichiometry with the shift shown to be weak; and a two-point $\Delta G^\circ(T)$ fit yields $\Delta H^\circ$ for (e).

(a) Pressure. Reaction 1 produces syngas with $\Delta n_{gas}=+2$ (1 mol CH₄ + 1 mol H₂O → 4 mol product). Le Chatelier: raising pressure shifts a mole-increasing reaction backward. Therefore 1.0 bar gives the higher syngas yield — low pressure favours reforming.
(b) Temperature. Reaction 1 is strongly endothermic, so higher $T$ drives it forward. This is confirmed by its standard Gibbs change, $\Delta G_1^\circ=\Delta G_f^\circ(\text{CO})+3\Delta G_f^\circ(\text{H}_2)-\Delta G_f^\circ(\text{CH}_4)-\Delta G_f^\circ(\text{H}_2\text{O})$: $$\Delta G_1^\circ(600\,\text{K})=+72{,}330\ \tfrac{\text{J}}{\text{mol}}\ (K\ll1),\qquad \Delta G_1^\circ(1300\,\text{K})=-103{,}470\ \tfrac{\text{J}}{\text{mol}}\ (K\gg1).$$ So 1300 K gives the higher yield. Preferred conditions: high $T$, low $P$ → 1300 K, 1.0 bar.
(c) Equilibrium constant at 1300 K, 1 bar. $$K_1=\exp\!\Big(\!-\frac{\Delta G_1^\circ}{RT}\Big)=\exp\!\Big(\frac{103{,}470}{8.314\times1300}\Big)=\exp(9.573)=\boxed{1.44\times10^{4}}.$$ The huge value means reforming runs essentially to completion at 1300 K.
(d) H₂/CO ratio. Check the shift at 1300 K: $\Delta G_2^\circ=\Delta G_f^\circ(\text{CO}_2)-\Delta G_f^\circ(\text{CO})-\Delta G_f^\circ(\text{H}_2\text{O})=+6{,}615\ \text{J/mol}$, so $K_2=\exp(-6615/(8.314\times1300))=0.54$. Since $K_2<1$, the shift barely proceeds and consumes little CO. Assuming reaction 1 dominates and reaction 2 is negligible, the products are 3 mol H₂ per 1 mol CO: $$\boxed{\text{H}_2/\text{CO}\approx 3}.$$
(e) $\Delta H^\circ_{800\,\text{K}}$ for reaction 1. Treat $\Delta H^\circ,\Delta S^\circ$ as roughly constant over 600–1300 K and fit $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ to the two points: $$\Delta S^\circ=\frac{\Delta G_1^\circ(600)-\Delta G_1^\circ(1300)}{1300-600}=\frac{72{,}330-(-103{,}470)}{700}=251.1\ \tfrac{\text{J}}{\text{mol}\cdot\text{K}},$$ $$\Delta H^\circ=\Delta G_1^\circ(600)+600\,\Delta S^\circ=72{,}330+150{,}686=\boxed{+223\ \text{kJ/mol}}.$$ Being nearly $T$-independent over this range, $\Delta H^\circ_{800\,\text{K}}\approx+223$ kJ/mol — the expected large endothermicity of steam reforming.

Quantity	Result
(a) Favourable pressure	1.0 bar (reaction 1 gains moles)
(b) Favourable temperature	1300 K (endothermic)
(c) $K_1$ at 1300 K	$1.44\times10^{4}$
(d) H₂/CO ratio	≈ 3 (shift weak, $K_2=0.54$)
(e) $\Delta H^\circ_{800\,\text{K}}$, reaction 1	$+223$ kJ/mol

Check: all $\Delta G_f^\circ$ values are read from the printed exam table and agree with JANAF/NIST data to within about 0.1 kJ/mol. A ±100 J/mol change in any entry moves $\Delta H^\circ$ by well under 0.1%; CO₂ enters only through the shift constant $K_2$.

Concept & Theory

Topic. Simultaneous gas-phase reaction equilibria; Le Chatelier reasoning and the $\Delta G^\circ$–$K$ relation.

Key relations. $\Delta G^\circ=-RT\ln K$; $K=\prod(\hat y_i P/P^\circ)^{\nu_i}$ so pressure enters through $\Delta n_{gas}$; van't Hoff / two-point fit $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$ to extract $\Delta H^\circ$; $\Delta G^\circ_{rxn}=\sum\nu_i\Delta G_{f,i}^\circ$.

Why this works. For reforming, $K\propto P^{\Delta n}$ with $\Delta n=+2$, so lowering $P$ raises the equilibrium conversion; and because $\Delta G^\circ$ swings sharply negative between 600 K and 1300 K, the strong temperature dependence signals a large positive $\Delta H^\circ$ (endothermic) — exactly what the $\Delta S^\circ$/$\Delta H^\circ$ fit quantifies. The water-gas shift, with $\Delta n=0$, is pressure-insensitive and only mildly favourable, so it perturbs the 3:1 stoichiometric H₂/CO ratio only slightly.

Common pitfall. Forgetting a species when summing $\Delta G_f^\circ$ (H₂'s value is zero but its stoichiometric coefficient 3 still matters in the mole-count and in $\Delta S^\circ$), or applying a pressure argument to the shift reaction where $\Delta n=0$ makes pressure irrelevant. Also, the two-point $\Delta H^\circ$ estimate assumes constant $\Delta C_p$≈0 — reasonable for an estimate, not exact.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §13.1–13.6 (single- and multi-reaction equilibrium; $T$ and $P$ effects).

Question 7 of 7: Synthesis-Gas Reaction Equilibria

Question 7: Synthesis-Gas Reaction Equilibria (equal value)