Ideal-Gas Batch Reaction $2A\rightarrow B+2C$ — Instantaneous Rates of Change at Constant $V$ and Constant $P$: 23-Chem-A4 Chemical Reactor Engineering, May 2016, Q1

Question 1: Ideal-Gas Batch Reaction $2A\rightarrow B+2C$ — Instantaneous Rates of Change at Constant $V$ and Constant $P$ (25 marks: a 12.5, b 12.5)

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. A rigid- or moving-boundary batch charge of pure A reacting by $2A\rightarrow B+2C$ (2 moles consumed produce 3 moles ⇒ expansion factor $\varepsilon = y_{A0}\,\delta = 1\cdot\tfrac{3-2}{2} = 0.5$), evaluated at the instant $-r_A = 1.6$.

Quantity	Value
Temperature $T$	170 °C = 443.15 K
Initial pressure / charge	$P_0 = 10$ atm, $n_{A0} = 2.25$ kmol pure A
Rate law	$-r_A = 1.122\times10^{4}\,C_A^{2}$ kmol/(m³·hr)
Evaluation instant	$-r_A = 1.6$ kmol/(m³·hr)
Expansion factor	$\varepsilon = 0.5$

Find. The six instantaneous derivatives $dn_A/dt,\ dN_A/dt,\ dp_A/dt,\ d\pi/dt,\ dV/dt,\ dX_A/dt$ for (a) constant-volume and (b) constant-pressure operation.

Figure 1.1 — The same charge run two ways: a rigid constant-volume vessel (pressure rises as moles increase) versus a constant-pressure moving boundary (volume expands). Every extensive rate is larger in the constant-pressure case because the vessel holds more moles at the fixed evaluation concentration.

Approach. The rate law fixes the concentration $C_A$ at the evaluation instant (identical in both cases); the stoichiometric table then converts that single rate into each requested derivative through the constant-$V$ or constant-$P$ mole/volume relations.

Fix the base state from the ideal-gas law. $C_{A0} = \dfrac{P_0}{RT} = \dfrac{10}{(0.082057)(443.15)} = 0.2750$ kmol/m³, and $V_0 = \dfrac{n_{A0}}{C_{A0}} = \dfrac{2.25}{0.2750} = 8.182$ m³. Also $RT = 36.36\ \text{m}^3\cdot\text{atm}\cdot\text{kmol}^{-1}$.
Invert the rate law for the concentration at the instant. Since $-r_A = kC_A^2$, $$C_A = \sqrt{\tfrac{-r_A}{k}} = \sqrt{\tfrac{1.6}{1.122\times10^4}} = \boxed{0.01194\ \text{kmol/m}^3}.$$ This same $C_A$ holds whether $V$ or $P$ is held constant, because the rate law is written in intensive terms.
(a) Constant volume — conversion and mole rate. With $C_A = C_{A0}(1-X)$, $X_a = 1 - C_A/C_{A0} = 1 - 0.01194/0.2750 = 0.9566$. The mole balance $\tfrac{dn_A}{dt} = -(-r_A)V_0$ (fixed $V$) gives $\tfrac{dn_A}{dt} = -(1.6)(8.182) = -13.09$ kmol/hr, hence $\tfrac{dX_A}{dt} = -\tfrac{1}{n_{A0}}\tfrac{dn_A}{dt} = 13.09/2.25 = 5.82\ \text{hr}^{-1}$.
(a) Constant volume — pressures and mole fraction. $\tfrac{dp_A}{dt} = RT\tfrac{dC_A}{dt} = RT\tfrac{1}{V_0}\tfrac{dn_A}{dt} = (36.36)(-1.6) = -58.18$ atm/hr. Total pressure rises with conversion, $\pi = P_0(1+\varepsilon X)$, so $\tfrac{d\pi}{dt} = P_0\varepsilon\tfrac{dX}{dt} = (10)(0.5)(5.82) = +29.09$ atm/hr. For the mole fraction $N_A = \dfrac{1-X}{1+\varepsilon X}$, $\dfrac{dN_A}{dX} = -\dfrac{1+\varepsilon}{(1+\varepsilon X)^2} = -\dfrac{1.5}{(1+0.5X)^2}$; at $X_a$ this is $-0.686$, so $\tfrac{dN_A}{dt} = -0.686(5.82) = -3.99\ \text{hr}^{-1}$. And $dV/dt = 0$ by definition.
(b) Constant pressure — the vessel now expands. At fixed $P$, $C_A = C_{A0}\dfrac{1-X}{1+\varepsilon X}$; solving $C_A/C_{A0} = 0.04343$ for $X$ gives $X_b = \dfrac{1-0.04343}{1+0.5(0.04343)} = 0.9363$, and the volume has grown to $V_b = V_0(1+\varepsilon X_b) = 8.182(1.468) = 12.01$ m³.
(b) Constant pressure — the six derivatives. The mole balance uses the current volume: $\tfrac{dn_A}{dt} = -(-r_A)V_b = -(1.6)(12.01) = -19.22$ kmol/hr, a larger draw-down than the constant-$V$ case because the vessel holds more gas at the same $C_A$. Then $\tfrac{dX_A}{dt} = 19.22/2.25 = 8.54\ \text{hr}^{-1}$, and the boundary moves at $\tfrac{dV}{dt} = V_0\varepsilon\tfrac{dX}{dt} = 8.182(0.5)(8.54) = +34.94$ m³/hr. With $\tfrac{dN_A}{dX} = -1.5/(1+0.5X_b)^2 = -0.696$, $\tfrac{dN_A}{dt} = -0.696(8.54) = -5.94\ \text{hr}^{-1}$. Since $C_A = C_{A0}N_A$ at fixed $P$, $\tfrac{dp_A}{dt} = RT\,C_{A0}\tfrac{dN_A}{dt} = (36.36)(0.2750)(-5.94) = -59.44$ atm/hr. Total pressure is held, so $\tfrac{d\pi}{dt} = 0$.

Derivative	(a) Constant $V$	(b) Constant $P$
$dn_A/dt$ (kmol/hr)	−13.09	−19.22
$dX_A/dt$ (hr$^{-1}$)	+5.82	+8.54
$dN_A/dt$ (hr$^{-1}$)	−3.99	−5.94
$dp_A/dt$ (atm/hr)	−58.18	−59.44
$d\pi/dt$ (atm/hr)	+29.09	0
$dV/dt$ (m$^3$/hr)	0	+34.94

Concept & Theory

Topic. Variable-volume (mole-changing) gas batch reactor; the stoichiometric table and expansion factor $\varepsilon$.

Key relations. Constant $V$: $C_A = C_{A0}(1-X)$, $\pi = P_0(1+\varepsilon X)$; constant $P$: $C_A = C_{A0}\tfrac{1-X}{1+\varepsilon X}$, $V = V_0(1+\varepsilon X)$; mole fraction $N_A = \tfrac{1-X}{1+\varepsilon X}$; batch mole balance $\tfrac{dn_A}{dt} = -(-r_A)V$ (Fogler, Ch. 3–4).

Why this works. Because the rate law is intensive, the evaluation concentration is common to both operating modes; only the bookkeeping that turns $C_A$ into extensive rates differs. At constant $P$ the vessel is larger (it has expanded), so at the same $C_A$ it contains more A — every extensive rate ($dn_A/dt,\ dX/dt,\ dV/dt$) is therefore larger than at constant $V$, while total pressure is pinned so $d\pi/dt = 0$.

Common pitfall. Forgetting that the constant-$P$ mole balance must use the current expanded volume $V_b$, not $V_0$; using $V_0$ understates every constant-$P$ rate. A second trap is sign confusion on $d\pi/dt$: pressure rises at constant $V$ (moles increase) but is fixed at constant $P$.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 3 (stoichiometry, $\varepsilon$) and Ch. 4 (batch design).

Question 1 of 5: Ideal-Gas Batch Reaction $2A\rightarrow B+2C$ — Instantaneous Rates of Change at Constant $V$ and Constant $P$

Question 1: Ideal-Gas Batch Reaction $2A\rightarrow B+2C$ — Instantaneous Rates of Change at Constant $V$ and Constant $P$ (25 marks: a 12.5, b 12.5)