Gas-Phase Decomposition Equilibrium in a Rigid Vessel: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2016, Q6

Question 6: Gas-Phase Decomposition Equilibrium in a Rigid Vessel (Part B — 25 marks)

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. Rigid (constant-$V$) vessel, pure A charged at $T_0=300$ K, $P_0=760$ mmHg (no dissociation yet, $\alpha_0=0$). Heating drives A(g) → B(g) + C(g), which increases the total moles. Measured $P=1114$ mmHg at 400 K and 1584 mmHg at 500 K. Ideal gas, so $n_0R/V=P_0/T_0=2.533$ mmHg/K is a fixed constant of the vessel.

Find. The total pressure at 600 K.

Figure 5 — van’t Hoff plot of the extracted $K_p$ values. $\ln K_p$ rises with $1/T$ decreasing (endothermic decomposition); linear extrapolation to 600 K gives $K_p\approx478$, from which the pressure follows.

Approach. At each temperature the measured pressure gives the dissociation fraction $\alpha$ (since one mole of A makes two moles of products), hence $K_p$; two $K_p$ values fix a van’t Hoff line, which is extrapolated to 600 K and solved back for $\alpha$ and the pressure.

Dissociation fraction from each pressure. Starting from $n_0$ moles of A, at fraction $\alpha$ the total moles are $n_0(1+\alpha)$, so $P=\dfrac{P_0}{T_0}(1+\alpha)T$ and $\alpha=\dfrac{P\,T_0}{P_0\,T}-1$: $$\alpha_{400}=\frac{1114}{2.533(400)}-1=0.099,\qquad \alpha_{500}=\frac{1584}{2.533(500)}-1=0.250.$$
Equilibrium constant at 400 and 500 K. For A → B + C, $K_p=\dfrac{p_Bp_C}{p_A P^\circ}$; with mole fractions $\tfrac{\alpha}{1+\alpha}$ (B, C) and $\tfrac{1-\alpha}{1+\alpha}$ (A) this reduces to $K_p=\dfrac{P\,\alpha^2}{(1-\alpha^2)}$ (here $K_p$ carries mmHg units, i.e. $P^\circ=1$ mmHg; the unit choice cancels in the van’t Hoff ratio and in the back-calculation): $$K_{p,400}=\frac{1114(0.099)^2}{1-0.099^2}=11.1,\qquad K_{p,500}=\frac{1584(0.250)^2}{1-0.250^2}=106.$$
van’t Hoff line and extrapolation to 600 K. $\ln\dfrac{K_{p,500}}{K_{p,400}}=-\dfrac{\Delta H}{R}\!\left(\dfrac1{500}-\dfrac1{400}\right)$ gives $-\Delta H/R=-4515$ K ($\Delta H\approx+37.5$ kJ/mol, endothermic). Extrapolating, $$\ln K_{p,600}=\ln K_{p,500}-4515\!\left(\tfrac1{600}-\tfrac1{500}\right)\;\Rightarrow\;K_{p,600}=478.$$
Pressure at 600 K. At 600 K the vessel pressure is $P=\dfrac{P_0}{T_0}(1+\alpha)(600)=1520(1+\alpha)$, and $K_p=\dfrac{P\alpha^2}{1-\alpha^2}=\dfrac{1520\,\alpha^2}{1-\alpha}$. Setting this to 478: $$\frac{1520\,\alpha^2}{1-\alpha}=478\;\Rightarrow\;\alpha=0.425,\qquad P_{600}=1520(1+0.425)=\boxed{2166\ \text{mmHg}\ (2.85\ \text{atm}).}$$

The pressure rise (760 → 2166 mmHg over 300 → 600 K) far outpaces simple thermal expansion (which alone would give 1520 mmHg) because the endothermic decomposition keeps generating extra moles as the temperature climbs — a nice illustration of Le Chatelier’s principle acting through both temperature and the mole-increasing reaction.

Temperature	Pressure (mmHg)	$\alpha$	$K_p$
300 K (charge)	760	0	—
400 K	1114	0.099	11.1
500 K	1584	0.250	106
600 K (estimate)	2166 (2.85 atm)	0.425	478

Concept & Theory

Topic. Homogeneous gas-phase reaction equilibrium at constant volume, extent of reaction from total pressure, and van’t Hoff extrapolation.

Key relations. Ideal gas $PV=n_{tot}RT$ with $n_{tot}=n_0(1+\alpha)$; $K_p=\dfrac{P\alpha^2}{1-\alpha^2}$ for A→B+C from a pure-A charge; van’t Hoff $\ln(K_2/K_1)=-\dfrac{\Delta H}{R}(1/T_2-1/T_1)$.

Why this works. Because the vessel is rigid and initially holds pure A with no dissociation, $n_0R/V$ is known from the 300 K charge; every later pressure is then a direct measure of the total moles, hence of $\alpha$ and $K_p$. Two temperatures fix the (assumed constant-$\Delta H$) van’t Hoff line, which is all that is needed to predict a third temperature.

Common pitfall. Assuming $\alpha=0$ at 400 K (the gas is already partly dissociated), or forgetting that heating alone raises the pressure even without reaction — the reaction contribution is the pressure above the $1520$ mmHg that pure expansion would give.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics, 8th ed., §13.4 (reaction coordinate, $K_p$ for gas-phase reactions) and §13.5 (temperature dependence).

Question 6 of 6: Gas-Phase Decomposition Equilibrium in a Rigid Vessel

Question 6: Gas-Phase Decomposition Equilibrium in a Rigid Vessel (Part B — 25 marks)