Part B — Chemical Thermodynamics: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2019, Q4

Question B1: Gas-Phase Decomposition Equilibrium in a Rigid Vessel (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.

Find. the total pressure in the rigid vessel at 600 K.

Figure B1 — A fixed charge of pure A in a rigid vessel is heated; the total pressure rises both from ideal-gas heating and from the mole increase of the decomposition, so each observed $P$ encodes the dissociation fraction $\alpha(T)$.

Approach. Use the rigid-vessel ideal-gas relation to turn each measured pressure into a dissociation fraction and then an equilibrium constant; fit van’t Hoff to the two points, extrapolate $K$ to 600 K, and solve the coupled equilibrium/pressure relation for the final pressure.

Anchor the vessel with the initial state. Pure A at 300 K (negligible dissociation) gives $n_0R/V=P_0/T_0=760/300$ mmHg/K. With A $\rightarrow$ B+C the total moles are $n_0(1+\alpha)$, so at any $T$ $$P=\frac{n_0R}{V}(1+\alpha)T=\frac{760}{300}(1+\alpha)T.$$
Extract the dissociation fraction at each temperature. Solving the above for $\alpha$, $$\alpha=\frac{P}{(760/300)\,T}-1:\quad \alpha(400)=\frac{1114}{1013.3}-1=0.099,\quad \alpha(500)=\frac{1584}{1266.7}-1=0.251.$$
Equilibrium constant at each point. With mole fractions $y_A=(1-\alpha)/(1+\alpha)$, $y_B=y_C=\alpha/(1+\alpha)$ and $P^\circ=760$ mmHg, $$K=\frac{y_By_C}{y_A}\frac{P}{P^\circ}=\frac{\alpha^2}{1-\alpha^2}\frac{P}{760}:\quad K_{400}=0.0146,\ K_{500}=0.1396.$$
van’t Hoff between the two points. Assuming a constant heat of reaction, $$\ln\frac{K_{500}}{K_{400}}=-\frac{\Delta H^\circ}{R}\!\left(\frac1{500}-\frac1{400}\right)\;\Rightarrow\; \Delta H^\circ=+37.5\ \text{kJ/mol},\quad \ln K=-\frac{4514}{T}+7.06.$$ The positive $\Delta H^\circ$ (endothermic) is why heating drives the decomposition forward.
Extrapolate to 600 K. $$\ln K_{600}=-\frac{4514}{600}+7.06=-0.465\;\Rightarrow\; K_{600}=0.628.$$
Close the coupled equilibrium–pressure relation. At 600 K, $P=\tfrac{760}{300}(600)(1+\alpha)=1520(1+\alpha)$; substituting into $K$ collapses it to $K_{600}=\dfrac{2\alpha^2}{1-\alpha}=0.628$, a quadratic giving $\alpha_{600}=0.425$. Hence $$P_{600}=1520(1+0.425)=\boxed{2166\ \text{mmHg}\ (2.85\ \text{atm})}.$$

Quantity	Result
$\alpha$ at 400 / 500 K	0.099 / 0.251
$K$ at 400 / 500 K	0.0146 / 0.1396
$\Delta H^\circ$ of decomposition	+37.5 kJ/mol (endothermic)
$\alpha$ at 600 K	0.425
Pressure at 600 K	2166 mmHg (2.85 atm)

Concept & Theory

Topic. Reaction equilibrium in a rigid vessel; van’t Hoff extrapolation.

Key relations. Ideal-gas $PV=n_{tot}RT$ at constant $V$; extent relations $y_i(\alpha)$; $K=\dfrac{\alpha^2}{1-\alpha^2}\dfrac{P}{P^\circ}$; van’t Hoff $\ln(K_2/K_1)=-\dfrac{\Delta H^\circ}{R}(1/T_2-1/T_1)$.

Why this works. Because the vessel is rigid, pressure is a direct proxy for total moles, and the pure-A initial state fixes $n_0R/V$ once and for all. The measured pressures therefore convert cleanly into $\alpha(T)$ and then $K(T)$; two $K$ values fix the van’t Hoff line, which extrapolates to 600 K, and the same rigid-vessel pressure relation closes the final state.

Common pitfall. Forgetting that $P$ at 600 K itself depends on $\alpha$ (treating $P$ as known) — the equilibrium and the ideal-gas pressure equations must be solved together. Also, $K$ must carry the $P/P^\circ$ factor since $\Delta n=+1$.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §13.1–13.4 (single-reaction equilibrium, $K$ from composition, temperature dependence of $K$).

Question 4 of 6: Part B — Chemical Thermodynamics

Part A — Process Mass & Energy Balances

Part B — Chemical Thermodynamics

Question B1: Gas-Phase Decomposition Equilibrium in a Rigid Vessel (Part B — equal value)