Methane–Steam Reforming — Equilibrium Composition at 1000 K: 23-Chem-A1 Process Balances and Chemical Thermodynamics, Undated paper, Q6

Question B3: Methane–Steam Reforming — Equilibrium Composition at 1000 K (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. A closed reaction of 2 mol CH₄ + 3 mol H₂O at $T=1000$ K, $P=1$ atm, reaching equilibrium among five species over three elements (C, H, O).

Species	$\Delta G^\circ_f$ at 1000 K (kJ/mol)
CH₄	+19.3
H₂O	−192.72
CO	−200.715
CO₂	−396.11
H₂	0

Find. the equilibrium mole numbers (and mole fractions) of all five species.

Figure 3 — The five-species system spans two independent reactions: strongly-favoured steam reforming (R1, mole-increasing, aided by the low 1 atm pressure) followed by the mild water-gas shift (R2). Their equilibrium constants close the composition.

Approach. Five species minus three elements leaves two independent reactions — take steam reforming and the water-gas shift. Evaluate each equilibrium constant from the Gibbs energies, write every species in terms of the two extents, and solve the coupled equilibrium equations.

Independent reactions. A convenient independent set is $$\text{(R1) reforming: } \mathrm{CH_4+H_2O\rightleftharpoons CO+3H_2}\ (\Delta n=+2),\qquad \text{(R2) shift: } \mathrm{CO+H_2O\rightleftharpoons CO_2+H_2}\ (\Delta n=0).$$
Equilibrium constants from $\Delta G^\circ$. $\Delta G_1^\circ=\Delta G_{f,CO}^\circ+3\Delta G_{f,H_2}^\circ-\Delta G_{f,CH_4}^\circ-\Delta G_{f,H_2O}^\circ=-200.715-19.3+192.72=-27.30$ kJ/mol and $\Delta G_2^\circ=-396.11+200.715+192.72=-2.68$ kJ/mol, so with $K=\exp(-\Delta G^\circ/RT)$ $$\boxed{K_1=26.7,\qquad K_2=1.38}\quad(RT=8.314\times10^{-3}\ \text{kJ/mol}\cdot\text{K}\times1000\ \text{K}=8.314\ \text{kJ/mol}).$$
Extents and mole table. Let $\xi_1,\xi_2$ be the extents of R1, R2. Then $$n_{CH_4}=2-\xi_1,\ n_{H_2O}=3-\xi_1-\xi_2,\ n_{CO}=\xi_1-\xi_2,\ n_{CO_2}=\xi_2,\ n_{H_2}=3\xi_1+\xi_2,$$ with total $n_t=5+2\xi_1$.
Equilibrium relations at $P=1$ atm. Since $P=1$, $$K_1=\frac{n_{CO}\,n_{H_2}^3}{n_{CH_4}\,n_{H_2O}}\Big(\frac{1}{n_t}\Big)^2,\qquad K_2=\frac{n_{CO_2}\,n_{H_2}}{n_{CO}\,n_{H_2O}}.$$
Solve. Solving the pair simultaneously gives $\xi_1=1.828$ and $\xi_2=0.311$ — a methane conversion of $\xi_1/2=$ 91.4%. The equilibrium composition is $$\mathrm{CH_4}\,0.17,\ \mathrm{H_2O}\,0.86,\ \mathrm{CO}\,1.52,\ \mathrm{CO_2}\,0.31,\ \mathrm{H_2}\,5.80\ \text{mol}\ (n_t=8.66),$$ i.e. mole fractions CH₄ 2.0%, H₂O 9.9%, CO 17.5%, CO₂ 3.6%, H₂ 67.0%.

Species	Equilibrium moles	Mole %
CH₄	0.17	2.0
H₂O	0.86	9.9
CO	1.52	17.5
CO₂	0.31	3.6
H₂	5.80	67.0
Total	8.66	100 (91.4% CH₄ conversion)

Concept & Theory

Topic. Multiple-reaction gas-phase equilibrium; independent-reaction counting and the extent method.

Key relations. Number of independent reactions $=$ species $-$ elements; $K_j=\exp(-\Delta G_j^\circ/RT)$; $K_j=\prod_i(y_i)^{\nu_{ij}}\,(P/P^\circ)^{\Delta n_j}$ with $y_i=n_i/n_t$.

Why this works. With five species built from only three elements, any two conservation-independent reactions span all possible equilibria; reforming and the shift are the natural choice. Each $\Delta G^\circ$ fixes one $K$, and expressing all mole numbers through the two extents turns the problem into two algebraic equations. Reforming is strongly favoured at 1000 K (large positive $\Delta n$ helped by the low 1 atm pressure), driving methane conversion above 90%; the mildly exothermic shift only lightly converts CO to CO₂ at this temperature.

Common pitfall. Reading the methane datum as $-19.3$ kJ/mol. The paper prints it positive, and it is: methane's $\Delta G^\circ_f$ turns positive above about 800 K. The wrong sign would give $K_1<1$ and almost no reforming. Another pitfall is dropping the $(P/P^\circ)^{\Delta n}$ pressure factor on R1 — because $\Delta n=+2$, that factor is essential, whereas R2 with $\Delta n=0$ needs none.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §13.6–13.9 (equilibrium for multiple reactions, $K$ from $\Delta G^\circ$, the extent method).

Question 6 of 6: Methane–Steam Reforming — Equilibrium Composition at 1000 K

Part A — Process Mass and Energy Balances

Question B3: Methane–Steam Reforming — Equilibrium Composition at 1000 K (equal value)