Gas-Phase Dissociation Equilibria — O₂⇌2O and H₂+½O₂⇌H₂O: 21-Mat-A1 Thermodynamics, May 2015, Q6

Question 6: Gas-Phase Dissociation Equilibria — O₂⇌2O and H₂+½O₂⇌H₂O (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.

Given. (a) Pure O₂ tank, $P_{tot}=1$ atm, $\tfrac12\text{O}_2=\text{O}$, $\Delta G^\circ=187{,}800$ J/mol O at $T=1000$ K. (b) O₂/H₂/H₂O tank, $P_{tot}=1$ atm, $T=1750$ K, $\text{H}_2+\tfrac12\text{O}_2=\text{H}_2\text{O}$, $\Delta G^\circ=-246{,}000+54.84T$, and $P_{O_2}$ is held at $1\times10^{-10}$ atm.

Find. The equilibrium gas-phase composition (mole fractions / partial pressures) in each tank.

Approach. Write $K=\exp(-\Delta G^\circ/RT)$ for each reaction in terms of partial pressures (numerically equal to mole fractions since $P_{tot}=1$ atm), then close the system with the appropriate mass/pressure balance.

(a) Equilibrium constant. $K=\exp\!\left(\dfrac{-187{,}800}{(8.314)(1000)}\right)=1.549\times10^{-10}$, with $K=\dfrac{P_O}{\sqrt{P_{O_2}}}$.
Mass balance. Basis 1 mol O₂ initially; let $x$ mol O form (consuming $x/2$ mol O₂). Total moles $=1+x/2$, so $y_O=\dfrac{x}{1+x/2}$, $y_{O_2}=\dfrac{1-x/2}{1+x/2}$, and with $P_{tot}=1$ atm, $P_i=y_i$. Solving $y_O/\sqrt{y_{O_2}}=K$ (numerically, since $K$ is tiny the correction from $1+x/2\approx1$ is negligible): $$\boxed{y_O\approx1.55\times10^{-10}},\qquad \boxed{y_{O_2}\approx1-1.55\times10^{-10}\approx1.0000000000}.$$ Oxygen is essentially undissociated at 1000 K — only about 1.5 molecules in $10^{10}$ exist as monatomic O.
(b) Equilibrium constant at 1750 K. $\Delta G^\circ=-246{,}000+54.84(1750)=-150{,}030$ J, so $$K=\exp\!\left(\frac{150{,}030}{(8.314)(1750)}\right)=3.008\times10^{4},\qquad K=\frac{P_{H_2O}}{P_{H_2}\sqrt{P_{O_2}}}.$$
Ratio of H₂O to H₂. With $P_{O_2}=1\times10^{-10}$ atm fixed, $$\frac{P_{H_2O}}{P_{H_2}}=K\sqrt{P_{O_2}}=(3.008\times10^{4})\sqrt{1\times10^{-10}}=3.008\times10^{4}\times10^{-5}=\boxed{0.3008}.$$
Close with the total-pressure balance. $P_{O_2}$ is negligible next to 1 atm, so $P_{H_2}+P_{H_2O}\approx1$ atm. With $P_{H_2O}=0.3008\,P_{H_2}$: $$P_{H_2}(1+0.3008)=1\ \Rightarrow\ \boxed{P_{H_2}=0.769\ \text{atm}},\qquad \boxed{P_{H_2O}=0.231\ \text{atm}},\qquad P_{O_2}=1\times10^{-10}\ \text{atm (given)}.$$

Part	Species	Mole fraction / P_i (atm)
(a) 1000 K	O₂	≈ 1.0000000000
(a) 1000 K	O	1.55 × 10⁻¹⁰
(b) 1750 K	H₂	0.769
	H₂O	0.231
	O₂	1 × 10⁻¹⁰ (fixed)

Concept & Theory

Topic. Gas-phase dissociation/formation equilibria and the equilibrium constant $K=\exp(-\Delta G^\circ/RT)$.

Key relations. $\Delta G^\circ=-RT\ln K$; for reaction $\tfrac12\text{O}_2=\text{O}$, $K=P_O/\sqrt{P_{O_2}}$; for $\text{H}_2+\tfrac12\text{O}_2=\text{H}_2\text{O}$, $K=P_{H_2O}/(P_{H_2}\sqrt{P_{O_2}})$; Dalton's law $\sum P_i=P_{tot}$.

Why this works. At equilibrium and $P_{tot}=1$ atm, partial pressures equal mole fractions, so the mass-action expression can be solved directly for composition once one extra constraint (an initial-mole basis in (a), a fixed $P_{O_2}$ in (b)) closes the system algebraically.

Common pitfall. Forgetting the $\sqrt{\cdot}$ on $P_{O_2}$ from its ½ stoichiometric coefficient, or, in (a), not recognising that a very small $K$ means the total-mole correction ($1+x/2$) can be dropped without materially changing the five-figure answer.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 7 (chemical equilibrium in gas mixtures, the equilibrium constant).

Question 6 of 7: Gas-Phase Dissociation Equilibria — O₂⇌2O and H₂+½O₂⇌H₂O

Question 6: Gas-Phase Dissociation Equilibria — O₂⇌2O and H₂+½O₂⇌H₂O (20 marks)