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23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2016

Question 5 of 6: Favourable T–P Region for Methanol Synthesis

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2016 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: two parts — Part A (Q1–Q3) Process Balances, Part B (Q4–Q6) Chemical Thermodynamics; the candidate answers two questions from each part (four constitute a complete paper, equal value). All six questions are solved below for completeness. Property data not printed on the paper (molar masses, critical constants, acentric factors, Rackett Zc) are stated explicitly in each Given block as open-book look-ups.

Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material balances, combustion/Orsat analysis, humidity; Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (8th ed.) — heats of reaction and Kirchhoff’s law; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — fugacity, generalized virial correlations, reaction equilibrium and VLE; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the NIST Chemistry WebBook.

Question 5: Favourable T–P Region for Methanol Synthesis (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. $\Delta G^\circ_{298}=-25.2$ kJ/mol, $\Delta H^\circ_{298}=-90.7$ kJ/mol, stoichiometric feed (CO:H₂ = 1:2), $\Delta n=-2$. Assume $\Delta H^\circ,\Delta S^\circ$ roughly constant over the range.

Find. the boundary curve $P_{\min}(T)$ above which equilibrium CO conversion $X\ge0.10$, and the favourable region on a $P$–$T$ plot.

Approach. Get $\Delta S^\circ$ from $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$, build $K(T)$; write the equilibrium condition as a conversion group $G(X)=K(P/P^\circ)^2$, evaluate $G$ at $X=0.10$, and invert for the pressure boundary $P_{\min}(T)=\sqrt{G(0.10)/K(T)}$.

298365432499566633700024487296120P_min(T)favourable: X >= 10%X < 10%Temperature T (K)Pressure P (atm)
Figure 2 — Boundary $P_{\min}(T)$ for 10% equilibrium conversion. The shaded region (low temperature, high pressure) satisfies $X\ge10\%$; high temperature demands impractically high pressure.
  1. Standard entropy and $K(T)$. $$\Delta S^\circ=\frac{\Delta H^\circ-\Delta G^\circ}{T}=\frac{(-90.7+25.2)\times10^3}{298}=-219.8\ \tfrac{\text{J}}{\text{mol K}},$$ $$\Delta G^\circ(T)=-90{,}700+219.8\,T,\qquad K(T)=\exp\!\Big[\!-\Delta G^\circ(T)/RT\Big].$$
  2. Conversion group. For stoichiometric feed (1 mol CO, 2 mol H₂), at conversion $X$ the totals give $y_{\text{CH}_3\text{OH}}=X/(3-2X)$, etc. With $\Delta n=-2$ the equilibrium relation rearranges to $$\underbrace{\frac{X(3-2X)^2}{4(1-X)^3}}_{G(X)}=K\left(\frac{P}{P^\circ}\right)^2.$$
  3. Evaluate at the 10% threshold. $$G(0.10)=\frac{0.10(2.8)^2}{4(0.9)^3}=0.2689.$$
  4. Pressure boundary. Solving for the minimum pressure that reaches 10%: $$\boxed{P_{\min}(T)=P^\circ\sqrt{\frac{0.2689}{K(T)}}}.$$ Because $G$ increases with $X$, the condition $X\ge10\%$ is equivalent to $P\ge P_{\min}(T)$.
  5. Tabulate the boundary. Using $K(T)$: $$\begin{array}{c|ccccc}T\,(\text{K}) & 400 & 450 & 500 & 550 & 600\\\hline P_{\min}\,(\text{atm}) & 0.34 & 1.55 & 5.22 & 14.1 & 32.2\end{array}$$ At 298 K, $K\approx2.6\times10^4$ so $P_{\min}\approx0.003$ atm (essentially any pressure works); by 600 K one needs $\sim$32 atm. (Strictly, $P^\circ$ is the 1-bar standard state, so these boundary values are in bar; 1 bar = 0.987 atm, a difference of about 1% that does not change the region.) The favourable region is low temperature and/or high pressure — consistent with the exothermic, mole-reducing reaction (Le Chatelier).
QuantityResult
$\Delta S^\circ$−219.8 J·mol⁻¹K⁻¹
10% conversion group $G(0.10)$0.2689
Boundary$P_{\min}(T)=\sqrt{0.2689/K(T)}$ atm
$P_{\min}$ at 400 / 500 / 600 K0.34 / 5.22 / 32.2 atm
Favourable regionlow $T$, high $P$