Gas-Phase Equilibrium for Ethylene Synthesis: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2014, Q8

Question 8: Gas-Phase Equilibrium for Ethylene Synthesis (Part D — 30%)

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. Stoichiometric feed (2 mol CO$_2$, 6 mol H$_2$); $T=800$ K. Standard-state (298 K) formation data and heat-capacity coefficients from SVA Appendix C are used to project $K$ to 800 K. $\Delta\nu = (1+4)-(2+6) = -3$.

Species	$\Delta H^\circ_{f,298}$ (kJ/mol)	$\Delta G^\circ_{f,298}$ (kJ/mol)
CO$_2$(g)	−393.509	−394.359
H$_2$(g)	0	0
C$_2$H$_4$(g)	+52.510	+68.460
H$_2$O(g)	−241.818	−228.572

Find. $K(800\,\mathrm{K})$ and the CO$_2$ conversion at 1 bar and at 100 bar.

CO2 conversion at 800 K. The reaction reduces the mole count (Δn = −3), so by Le Chatelier raising the pressure from 1 bar to 100 bar drives the equilibrium strongly toward products — conversion rises from about 8.5% to 54%.

Approach

Compute $\Delta H^\circ$ and $\Delta G^\circ$ at 298 K, project $K$ to 800 K through the heat-capacity integrals, then relate $K$ to the conversion via mole fractions — at 1 bar with ideal gas, and at 100 bar with pure-component fugacity coefficients (ideal solution).

Standard properties at 298 K. $$\Delta H^\circ_{298} = [52.510+4(-241.818)] - [2(-393.509)] = -127.74\ \text{kJ},$$ $$\Delta G^\circ_{298} = [68.460+4(-228.572)] - [2(-394.359)] = -57.11\ \text{kJ}.$$ The reaction is exothermic, so $K$ falls as temperature rises.
Project $K$ to 800 K. Using $\Delta C_p^\circ/R = \Delta A+\Delta B\,T+\Delta C\,T^2+\Delta D\,T^{-2}$ with $\Delta A=-15.10,\ \Delta B=1.557\times10^{-2},\ \Delta C=-4.39\times10^{-6},\ \Delta D=2.30\times10^{5}$, integrate the enthalpy and entropy: $$\Delta H^\circ_{800} = -156.98\ \text{kJ},\qquad \Delta S^\circ_{800} = -296.7\ \text{J/K},$$ $$\Delta G^\circ_{800} = \Delta H^\circ_{800} - T\Delta S^\circ_{800} = +80.37\ \text{kJ}.$$
Equilibrium constant. $$K = \exp\!\left(-\frac{\Delta G^\circ_{800}}{RT}\right) = \exp\!\left(-\frac{80{,}370}{8.314\times800}\right) = 5.65\times10^{-6}.$$
$K(800\ \mathrm{K}) = 5.65\times10^{-6}$.
Conversion relation. With CO$_2$ conversion $X$ on the stoichiometric feed (total moles $=8-3X$), $$K_y \equiv \frac{y_{C_2H_4}\,y_{H_2O}^4}{y_{CO_2}^2\,y_{H_2}^6} = \frac{256\,X^5\,(8-3X)^3}{186{,}624\,(1-X)^8},\qquad K = K_y\,K_\phi\left(\tfrac{P}{P^\circ}\right)^{\Delta\nu}.$$
(a) At 1 bar (ideal gas, $K_\phi=1$). $(P/P^\circ)^{-3}=1$, so $K_y=K=5.65\times10^{-6}$; solving, $$X = 0.0847.$$
At 1 bar: CO$_2$ conversion $\approx 8.5\%$.
(b) At 100 bar (ideal solution). Pure-component fugacity coefficients from the generalized correlation at 800 K give $\phi_{CO_2}=1.012,\ \phi_{H_2}=1.016,\ \phi_{C_2H_4}=1.011,\ \phi_{H_2O}=0.931$, hence $K_\phi=\phi_{C_2H_4}\phi_{H_2O}^4/(\phi_{CO_2}^2\phi_{H_2}^6)=0.675$. Then $$K_y = \frac{K}{K_\phi (P/P^\circ)^{-3}} = \frac{5.65\times10^{-6}}{0.675\times10^{-6}} = 8.38,$$ and solving the conversion relation, $$X = 0.541.$$
At 100 bar: CO$_2$ conversion $\approx 54\%$.

Quantity	Value
$\Delta G^\circ_{800}$	+80.4 kJ/mol
Equilibrium constant $K(800\,\mathrm{K})$	5.65 × 10⁻⁶
(a) Conversion at 1 bar	8.5 %
(b) $K_\phi$ at 100 bar	0.675
(b) Conversion at 100 bar	54 %

Concept & Theory

Topic. Gas-phase chemical-reaction equilibrium with temperature and pressure effects.

Key relations. $K=\exp(-\Delta G^\circ/RT)$; temperature projection through $\Delta H^\circ(T),\Delta S^\circ(T)$ with $\Delta C_p^\circ$; $K = K_y K_\phi (P/P^\circ)^{\Delta\nu}$; ideal solution ⇒ $\hat\phi_i = \phi_i$ (pure).

Why this works. $K$ depends on temperature only; pressure and non-ideality act through $K_y$, $K_\phi$ and $(P/P^\circ)^{\Delta\nu}$. Since $\Delta\nu=-3$, the $(P/P^\circ)^{-3}$ term makes $K_y$ (hence conversion) rise sharply with pressure — the quantitative form of Le Chatelier’s principle for a mole-reducing reaction.

Common pitfall. Forgetting the $(P/P^\circ)^{\Delta\nu}$ factor or its sign, or omitting the $\Delta C_p^\circ$ correction (using $K$ at 298 K would badly overpredict conversion). The fugacity correction is modest here ($K_\phi=0.68$) but should be included when the problem says “ideal solution.”

Source. Smith, Van Ness & Abbott, ch. 13 (chemical-reaction equilibria; equilibrium-constant temperature dependence and the fugacity/pressure factors).

Species	\(\Delta H^\circ_{f,298}\) (kJ/mol)	\(\Delta G^\circ_{f,298}\) (kJ/mol)
CO\(_2\)(g)	−393.509	−394.359
H\(_2\)(g)	0	0
C\(_2\)H\(_4\)(g)	+52.510	+68.460
H\(_2\)O(g)	−241.818	−228.572

Quantity	Value
\(\Delta G^\circ_{800}\)	+80.4 kJ/mol
Equilibrium constant \(K(800\,\mathrm{K})\)	5.65 × 10⁻⁶
(a) Conversion at 1 bar	8.5 %
(b) \(K_\phi\) at 100 bar	0.675
(b) Conversion at 100 bar	54 %

Question 8 of 8: Gas-Phase Equilibrium for Ethylene Synthesis

Question 8: Gas-Phase Equilibrium for Ethylene Synthesis (Part D — 30%)

Approach