Water–Gas Shift Equilibrium Composition at Two Temperatures: 21-Mat-A1 Thermodynamics, December 2017, Q5

Question 5: Water–Gas Shift Equilibrium Composition at Two Temperatures (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. Basis 1 mol total gas, initial mole fractions $y_{CO}=y_{CO_2}=y_{H_2}=y_{H_2O}=0.25$, total $P=1$ atm. $CO_2+H_2=CO+H_2O$; $\Delta G^\circ_{700\text{K}}=+14$ kJ; $\Delta G^\circ_{1500\text{K}}=-9.6$ kJ.

Find. Equilibrium mole fractions of CO, $CO_2$, $H_2$, $H_2O$ at (a) 700 K and (b) 1500 K.

Approach. Get $K=\exp(-\Delta G^\circ/RT)$ at each T, define extent of reaction $x$ (moles of $CO_2$+$H_2$ consumed), express $K$ in terms of $x$ (total moles are unchanged since $\Delta n_{gas}=0$), and solve the resulting linear equation for $x$.

Equilibrium table. With extent $x$: $n_{CO_2}=n_{H_2}=0.25-x$, $n_{CO}=n_{H_2O}=0.25+x$, and because the reaction has equal moles of gas on each side, mole fractions equal these numbers directly ($n_{tot}=1$ throughout). Since $P_{tot}=1$ atm, $K=\dfrac{y_{CO}\,y_{H_2O}}{y_{CO_2}\,y_{H_2}}=\left(\dfrac{0.25+x}{0.25-x}\right)^2.$
(a) 700 K equilibrium constant. $$K_{700}=\exp\!\left(\dfrac{-\Delta G^\circ}{RT}\right)=\exp\!\left(\dfrac{-14{,}000}{(8.314)(700)}\right)=\exp(-2.406)=\boxed{0.0902}.$$
(a) Solve for extent. $\sqrt{K_{700}}=0.3004=\dfrac{0.25+x}{0.25-x}\ \Rightarrow\ x=\dfrac{0.3004(0.25)-0.25}{1+0.3004}=\boxed{-0.1345}.$ The negative extent means the reaction runs in reverse (toward $CO_2+H_2$), consistent with $K_{700}<1$.
(a) Equilibrium composition at 700 K. $$y_{CO_2}=y_{H_2}=0.25-(-0.1345)=\boxed{0.3845},\qquad y_{CO}=y_{H_2O}=0.25+(-0.1345)=\boxed{0.1155}.$$ i.e. 38.45% $CO_2$, 38.45% $H_2$, 11.55% CO, 11.55% $H_2O$ (sums to 100.0%).
(b) 1500 K equilibrium constant. $$K_{1500}=\exp\!\left(\dfrac{-(-9600)}{(8.314)(1500)}\right)=\exp(0.7697)=\boxed{2.159}.$$
(b) Solve for extent. $\sqrt{K_{1500}}=1.4695=\dfrac{0.25+x}{0.25-x}\ \Rightarrow\ x=\dfrac{1.4695(0.25)-0.25}{1+1.4695}=\boxed{+0.0475}.$ Positive extent: at the higher T the equilibrium (now $K>1$) shifts toward $CO+H_2O$.
(b) Equilibrium composition at 1500 K. $$y_{CO_2}=y_{H_2}=0.25-0.0475=\boxed{0.2025},\qquad y_{CO}=y_{H_2O}=0.25+0.0475=\boxed{0.2975}.$$ i.e. 20.25% $CO_2$, 20.25% $H_2$, 29.75% CO, 29.75% $H_2O$ (sums to 100.0%).

T	K	$y_{CO_2}=y_{H_2}$	$y_{CO}=y_{H_2O}$
700 K	0.0902	38.45%	11.55%
1500 K	2.159	20.25%	29.75%

Concept & Theory

Topic. Gas-phase equilibrium (the water–gas shift reaction) from $\Delta G^\circ$ via the equilibrium constant.

Key relations. $\Delta G^\circ=-RT\ln K$; $K=\prod (y_iP)^{\nu_i}$; for a mole-conserving reaction ($\Delta n_{gas}=0$) mole fractions are independent of total pressure.

Why this works. Because $CO_2+H_2=CO+H_2O$ has equal moles of gas on each side, the total mole count stays at 1 throughout, so a single extent-of-reaction variable $x$ tracks the whole system and $K$ reduces to a simple ratio of two linear expressions in $x$ — no need for a higher-order polynomial as would arise if $\Delta n_{gas}\ne0$.

Common pitfall. Mis-assigning the sign of $\Delta G^\circ_{1500\text{K}}=-9.6$ kJ (it is negative, so $K>1$ and the shift favors products) — a sign slip here flips which species end up in excess.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 12 (reactions involving gases, equilibrium constant, extent of reaction).

Question 5 of 7: Water–Gas Shift Equilibrium Composition at Two Temperatures

Question 5: Water–Gas Shift Equilibrium Composition at Two Temperatures (20 marks)