Dissociation Equilibrium Compositions in Two Gas Tanks: 21-Mat-A1 Thermodynamics, December 2018, Q4

Question 4: Dissociation Equilibrium Compositions in Two Gas Tanks (20 marks)

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.

(a) Pure oxygen tank at 1000 K

Given. $\tfrac12$O₂(g)$=$O(g), $\Delta G^{\circ}=187{,}800$ J/mol O at $T=1000$ K; total pressure $P_{O_2}+P_O=1$ atm.

Find. Equilibrium $P_{O_2}$ and $P_O$ (the tank's composition).

Equilibrium constant. $K=\dfrac{P_O}{\sqrt{P_{O_2}}}=\exp\!\left(\dfrac{-\Delta G^{\circ}}{RT}\right)=\exp\!\left(\dfrac{-187{,}800}{(8.314)(1000)}\right)=\boxed{1.549\times10^{-10}}.$
Solve the mass balance. With $P_{O_2}+P_O=1$ and $P_O=K\sqrt{P_{O_2}}$, $K$ is so small that $P_{O_2}\approx1$ to 9 significant figures, giving $$P_{O_2}\approx\boxed{0.99999999985\ \text{atm}\ (\approx1.000\ \text{atm})},\qquad P_O=K\sqrt{P_{O_2}}\approx\boxed{1.549\times10^{-10}\ \text{atm}}.$$

(b) Oxygen–hydrogen–water tank at 1750 K

Given. H₂(g)$+\tfrac12$O₂(g)$=$H₂O(g), $\Delta G^{\circ}=-246{,}000+54.84T$ J/mol; total pressure 1 atm; $P_{O_2}$ held fixed at $1\times10^{-10}$ atm; $T=1750$ K.

Find. Equilibrium $P_{H_2}$ and $P_{H_2O}$ (with $P_{O_2}=10^{-10}$ atm already fixed).

$\Delta G^{\circ}$ and $K$ at 1750 K. $$\Delta G^{\circ}=-246{,}000+54.84(1750)=\boxed{-150{,}030\ \text{J/mol}},\qquad K=\frac{P_{H_2O}}{P_{H_2}\sqrt{P_{O_2}}}=\exp\!\left(\frac{150{,}030}{(8.314)(1750)}\right)=\boxed{3.008\times10^{4}}.$$
Mass balance. $P_{H_2}+P_{H_2O}+P_{O_2}=1$ with $P_{H_2O}=K\sqrt{P_{O_2}}\,P_{H_2}=(3.008\times10^4)(1\times10^{-5})P_{H_2}=0.3008\,P_{H_2}$: $$P_{H_2}(1+0.3008)+P_{O_2}=1\ \Rightarrow\ P_{H_2}=\frac{1-10^{-10}}{1.3008}=\boxed{0.7687\ \text{atm}},\qquad P_{H_2O}=0.3008\,P_{H_2}=\boxed{0.2313\ \text{atm}}.$$

Quantity	Value
(a) $K$ (1000 K)	1.549×10⁻¹⁰
(a) $P_{O_2}$, $P_O$	0.999999999850 atm, 1.549×10⁻¹⁰ atm
(b) $\Delta G^{\circ}$, $K$ (1750 K)	−150,030 J/mol, 3.008×10⁴
(b) $P_{H_2}$, $P_{H_2O}$, $P_{O_2}$	0.7687 atm, 0.2313 atm, 1×10⁻¹⁰ atm

Concept & Theory

Topic. Gas-phase dissociation/formation equilibria from $\Delta G^{\circ}=-RT\ln K$, closed with a total-pressure mass balance.

Key relations. $K=\exp(-\Delta G^{\circ}/RT)$; for $\tfrac12$O₂$=$O, $K=P_O/\sqrt{P_{O_2}}$; for H₂$+\tfrac12$O₂$=$H₂O, $K=P_{H_2O}/(P_{H_2}\sqrt{P_{O_2}})$; $\sum P_i=P_{total}$.

Why this works. Each reaction's $\Delta G^{\circ}$ fixes $K$ at the given $T$; combining that single algebraic constraint with the physical requirement that the partial pressures sum to the total pressure gives exactly enough equations to solve for the unknown composition, even when one partial pressure (here $P_{O_2}$ in part b) is externally imposed rather than free.

Common pitfall. In part (b), forgetting that $P_{O_2}=10^{-10}$ atm is a FIXED boundary condition (not a free unknown solved from $K$) — only $P_{H_2}$ and $P_{H_2O}$ partition the remaining pressure, and the small $\sqrt{P_{O_2}}=10^{-5}$ factor is what keeps the mixture mostly H₂ despite a large $K$.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 12–13 (chemical equilibrium in gas mixtures, $\Delta G^{\circ}$–$K$ relation).

Question 4 of 7: Dissociation Equilibrium Compositions in Two Gas Tanks

Question 4: Dissociation Equilibrium Compositions in Two Gas Tanks (20 marks)

(a) Pure oxygen tank at 1000 K

(b) Oxygen–hydrogen–water tank at 1750 K