Maximum HCN Yield from Non-Ideal Nitrogenation of Acetylene: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2018, Q6

Question B3: Maximum HCN Yield from Non-Ideal Nitrogenation of Acetylene (Part B — 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. $N_2+C_2H_2\leftrightarrow2HCN$; stoichiometric feed ($y_{N_2}=y_{C_2H_2}$); $T=573$ K; $P=200$ bar; $\Delta G^\circ_{573}=30.1$ kJ/mol; ideal-solution (Lewis–Randall) mixing so $\hat\phi_i=\phi_i^{\text{pure}}$.

Species	$T_c$ (K)	$P_c$ (bar)	$\omega$
$N_2$	126.2	33.9	0.04
$C_2H_2$	308.3	61.4	0.184
HCN	456.7	49.6	0.4

Find. The maximum (equilibrium) mole fraction of HCN in the product stream at 573 K and 200 bar.

Approach. Compute $K$ from $\Delta G^\circ$; because $\Delta n_{\text{gas}}=0$ the pressure cancels and $K=K_y K_\phi$; evaluate each pure-component fugacity coefficient with the Peng–Robinson EOS (a reproducible stand-in for the Lee–Kesler charts, since $P_r\approx3$–6 is off the virial scale), then solve the stoichiometric-feed equilibrium for $y_{HCN}$.

Thermodynamic equilibrium constant. $$K=\exp\!\left(-\frac{\Delta G^\circ_{573}}{RT}\right)=\exp\!\left(-\frac{30{,}100}{8.314(573)}\right)=1.80\times10^{-3}.$$
Fugacity-coefficient ratio (Peng–Robinson, Lewis–Randall). Evaluating each pure species at 573 K, 200 bar gives $\phi_{N_2}=1.075$, $\phi_{C_2H_2}=0.925$, $\phi_{HCN}=0.595$ (near-critical HCN, $T_r=1.25$, is the strongly non-ideal one). Thus $$K_\phi=\frac{\phi_{HCN}^{2}}{\phi_{N_2}\,\phi_{C_2H_2}}=\frac{0.595^2}{(1.075)(0.925)}=0.356.$$
Split $K$ into composition and fugacity parts. Since $\Delta n_{\text{gas}}=2-2=0$, the total-pressure factor $(P/P^\circ)^{\Delta n}=1$ drops out and $$K=K_y\,K_\phi\;\Rightarrow\; K_y=\frac{K}{K_\phi}=\frac{1.80\times10^{-3}}{0.356}=5.06\times10^{-3}.$$ Because $K_\phi<1$, real-gas non-ideality raises $K_y$ above the ideal value — HCN (in the numerator) is the most compressed species, favouring the products.
Solve the stoichiometric-feed equilibrium. With $y_{N_2}=y_{C_2H_2}=(1-y_{HCN})/2$, $$K_y=\frac{y_{HCN}^{2}}{y_{N_2}\,y_{C_2H_2}}=\frac{4\,y_{HCN}^{2}}{(1-y_{HCN})^{2}}\;\Rightarrow\; y_{HCN}=\frac{\sqrt{K_y}}{2+\sqrt{K_y}}=\boxed{0.034.}$$ For comparison, the ideal-gas estimate ($K_\phi=1$) gives only $y_{HCN}=\sqrt{K}/(2+\sqrt{K})=0.021$, so accounting for non-ideality lifts the maximum HCN fraction from about 2.1% to 3.4%.

Check: the exam specifies an "ideal solution" but supplies no fugacity chart; the pure-component $\phi_i$ here are computed with Peng–Robinson from the given $T_c,P_c,\omega$, which reproduces Lee–Kesler chart values to within a few percent at these reduced conditions. A candidate reading $\phi_i$ from generalized charts would obtain the same $y_{HCN}\approx0.03$–0.035.

Quantity	Result
$K$ from $\Delta G^\circ_{573}$	$1.80\times10^{-3}$
$\phi_{N_2},\phi_{C_2H_2},\phi_{HCN}$ (PR)	1.075 / 0.925 / 0.595
$K_\phi$ and $K_y$	0.356 and $5.06\times10^{-3}$
Maximum HCN mole fraction	0.034 (vs. 0.021 ideal)

Concept & Theory

Topic. High-pressure reaction equilibrium with real-gas fugacity corrections (Lewis–Randall ideal solution).

Key relations. $K=\exp(-\Delta G^\circ/RT)$; $K=K_y K_\phi (P/P^\circ)^{\Delta n}$; Lewis–Randall $\hat\phi_i=\phi_i^{\text{pure}}$; Peng–Robinson $\ln\phi=Z-1-\ln(Z-B)-\frac{A}{2\sqrt2 B}\ln\frac{Z+(1+\sqrt2)B}{Z+(1-\sqrt2)B}$.

Why this works. With $\Delta n_{\text{gas}}=0$ the explicit pressure dependence vanishes, leaving only the fugacity ratio $K_\phi$ to carry non-ideality. The Lewis–Randall rule lets each $\hat\phi_i$ be replaced by the pure-species value at the mixture $T,P$, so three independent EOS evaluations suffice. HCN, closest to its critical point at 573 K, has the smallest $\phi$; since it sits in the numerator, $K_\phi<1$ and equilibrium shifts toward more HCN than an ideal-gas treatment predicts.

Common pitfall. Assuming pressure has no effect on yield because it cancels in $K$ — it still enters through the fugacity coefficients; also, using the truncated virial equation where $P_r\gg1$ (here up to $\sim6$) is out of range, so a cubic EOS or generalized charts are required.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §11.6–11.7 (fugacity, Lewis–Randall rule) and §13.6 (gas-phase reaction equilibria with $K_\phi$).

Question 6 of 6: Maximum HCN Yield from Non-Ideal Nitrogenation of Acetylene

Part A — Process Balances

Question B3: Maximum HCN Yield from Non-Ideal Nitrogenation of Acetylene (Part B — equal value)