23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2015
Question 4 of 6: Thermodynamics of Calcium-Oxalate Dissociation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: six questions in two parts — Part A (Q1–Q3, Process Mass & Energy Balances) and Part B (Q4–Q6, Chemical Thermodynamics). Candidates answer two from Part A and two from Part B; four equally-weighted questions (25 marks each) constitute a complete paper. All six are solved below for completeness. Property data are stated explicitly in each Given block.
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — combustion stoichiometry, humidity, recycle/purge and reactive material balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — reaction equilibrium, van’t Hoff analysis, VLE with ideal solutions and excess-property/heat-of-mixing energy balances; supporting data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the NIST Chemistry WebBook.
Question 4: Thermodynamics of Calcium-Oxalate Dissociation (Part B — 25 marks)
Given. A solid-decomposition equilibrium with a single gaseous product, CO. Because both calcium compounds are pure solids (activity 1), the equilibrium constant is just the CO pressure referenced to the standard state $P^\circ=100$ kPa: $K=p_{CO}/P^\circ$.
T (°C)
T (K)
$p_{CO}$ (kPa)
$K=p/P^\circ$
375
648.15
1.09
0.0109
388
661.15
4.00
0.0400
403
676.15
17.86
0.1786
410
683.15
33.33
0.3333
416
689.15
78.25
0.7825
418
691.15
91.18
0.9118
Find. $\Delta G^\circ$ at each temperature, and $\Delta H^\circ$, $\Delta S^\circ$ from the temperature dependence.
Figure 4 — van’t Hoff plot ln p vs 1/T; the straight-line slope gives $\Delta H^\circ=-R\,d(\ln K)/d(1/T)$.
Approach. Compute $\Delta G^\circ=-RT\ln K$ at each temperature; obtain $\Delta H^\circ$ from the slope of a van’t Hoff plot ($\ln p$ vs $1/T$, assuming $\Delta H^\circ$ constant over this narrow range); then $\Delta S^\circ=(\Delta H^\circ-\Delta G^\circ)/T$ at each temperature.
Gibbs energy at each temperature. With $K=p/P^\circ$ and $\Delta G^\circ=-RT\ln K$, e.g. at 375 °C:
$$\Delta G^\circ=-8.314(648.15)\ln(0.0109)=+24.35\ \text{kJ/mol.}$$
Repeating for all six temperatures gives the falling series tabulated below — $\Delta G^\circ$ drops toward zero as the CO pressure approaches 1 bar.
Enthalpy from the van’t Hoff slope. A least-squares fit of $\ln p$ against $1/T$ has slope $-\Delta H^\circ/R=-4.60\times10^{4}$ K, so
$$\Delta H^\circ=-R\,(\text{slope})=\boxed{+383\ \text{kJ/mol}\;(\text{endothermic}).}$$
Entropy at each temperature. From $\Delta S^\circ=(\Delta H^\circ-\Delta G^\circ)/T$, e.g. at 375 °C: $(383{,}000-24{,}350)/648.15=553$ J/mol·K. The value is essentially constant (551–553 J/mol·K) across the set, confirming $\Delta H^\circ$ and $\Delta S^\circ$ are sensibly temperature-independent here:
$$\Delta S^\circ\approx\boxed{+552\ \text{J/mol}\cdot\text{K.}}$$
Check
The tabulated pressures rise very steeply (a factor of ~84 over only 43 K), which forces an unusually large van’t Hoff slope and hence a high $\Delta H^\circ\approx383$ kJ/mol and $\Delta S^\circ\approx552$ J/mol·K. These follow directly and self-consistently from the data as printed; if a textbook value is expected, re-examine the pressure column for transcription. The standard state is taken as $P^\circ=100$ kPa (1 bar); using 1 atm shifts $\Delta G^\circ$ and $\Delta S^\circ$ by $R\ln(101.325/100)$ but leaves $\Delta H^\circ$ unchanged.
