Zero-Order Gas Reaction — Volume Change in a Constant-Pressure Batch Reactor: 23-Chem-A4 Chemical Reactor Engineering, May 2017, Q1

Question 1: Zero-Order Gas Reaction — Volume Change in a Constant-Pressure Batch Reactor (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. A homogeneous zero-order gas reaction $A\rightarrow rR$ (stoichiometric number $r$ unknown). Reactor 1 — constant volume, 20% inerts, total pressure rises $1\rightarrow1.3$ atm in $t=2$ min. Reactor 2 — constant pressure $P_2=3$ atm, feed 40% inerts (so $y_{A0}=0.6$), same temperature and same rate constant.

Find. The fractional volume change $\Delta V/V_0$ in Reactor 2 after 4 minutes.

Figure 1.1 — The two batch experiments. Reactor 1 (fixed volume) yields the kinetic group from its pressure rise; Reactor 2 (fixed pressure) expands as moles are generated.

Approach. Diagnose the observable of a zero-order reaction in the fixed-volume run (a linear pressure rise) to pin the single kinetic group $(r-1)k'$, then feed that group straight into Levenspiel's zero-order, variable-volume batch integral for the constant-pressure run — the individual $r$, $k$ never have to be separated.

Set the zero-order rate in pressure units. For a zero-order reaction $-r_A=k$ (independent of concentration). In the constant-volume vessel $C_A=p_A/RT$, so $-\dfrac{dp_A}{dt}=kRT\equiv k'$ is constant: the partial pressure of A falls linearly with time.
Convert the total-pressure rise to the kinetic group. With inerts fixed and $p_R=r\,(p_{A0}-p_A)$, the total pressure is $p_\text{tot}=p_\text{in}+p_A+p_R=p_0+(r-1)(p_{A0}-p_A)=p_0+(r-1)k't.$ The measured rise of $0.3$ atm over 2 min gives $$(r-1)k'=\frac{1.3-1.0}{2}=\boxed{0.15\ \text{atm/min}.}$$ Note the 20% inert content of Reactor 1 cancels — only the reacting group survives.
Write the zero-order, variable-volume batch law for Reactor 2. At constant $P,T$ the volume tracks total moles, $V=V_0(1+\varepsilon_A X)$ with expansion factor $\varepsilon_A=y_{A0}(r-1)=0.6(r-1)$. Levenspiel's zero-order integral (Ch. 3) for a constant-pressure batch is $$\frac{1}{\varepsilon_A}\ln(1+\varepsilon_A X)=\frac{k}{C_{A0}}\,t.$$
Collapse to the known group. Since $C_{A0}=p_{A0}/RT$ with $p_{A0}=0.6\times3=1.8$ atm, the concentration-based constant is $\dfrac{k}{C_{A0}}=\dfrac{kRT}{p_{A0}}=\dfrac{k'}{1.8}.$ Hence $$\ln(1+\varepsilon_A X)=\varepsilon_A\frac{k}{C_{A0}}\,t=\frac{0.6(r-1)\,k'}{1.8}\,t=\frac{(r-1)k'}{3}\,t.$$ Substituting $(r-1)k'=0.15$ and $t=4$ min: $$\ln(1+\varepsilon_A X)=\frac{0.15}{3}(4)=\boxed{0.20.}$$
Read off the fractional volume change. Because $V/V_0=1+\varepsilon_A X$, $$\frac{V}{V_0}=e^{0.20}=1.221\qquad\Rightarrow\qquad \frac{\Delta V}{V_0}=\varepsilon_A X=e^{0.20}-1=\boxed{0.221\ (+22.1\%).}$$

Check: the result uses only the group $(r-1)k'$, so no assumption about the individual stoichiometric number $r$ is needed. It is valid provided A is not exhausted within 4 min; the depletion condition $e^{0.20}-1<0.6(r-1)$ holds for every $r\ge2$, so the answer stands for any integer stoichiometry.

Quantity	Value
Kinetic group $(r-1)k'$ (from Reactor 1)	0.15 atm/min
Batch integral $\ln(1+\varepsilon_A X)$ at $t=4$ min	0.20
Volume ratio $V/V_0$	1.221
Fractional volume change $\Delta V/V_0$	+0.221 (+22.1%)

Concept & Theory

Topic. Zero-order homogeneous gas kinetics; constant-volume vs. constant-pressure batch analysis with mole change.

Key relations. Zero order $-r_A=k$; total-pressure method $p_\text{tot}=p_0+(r-1)(p_{A0}-p_A)$; expansion factor $\varepsilon_A=y_{A0}\,\delta$; zero-order variable-volume batch $\frac{1}{\varepsilon_A}\ln(1+\varepsilon_A X)=\frac{k}{C_{A0}}t$; $V/V_0=1+\varepsilon_A X$.

Why this works. A zero-order reaction consumes A at a rate that is independent of how much A remains, so in a rigid vessel the partial pressure of A — and therefore the total pressure — changes strictly linearly in time. That single slope carries all the kinetic information needed. Transferring to the constant-pressure vessel, the mole-generating stoichiometry expands the gas; the Levenspiel integral bookkeeps that expansion through $\varepsilon_A$, and the pressure ratio conveniently cancels the unknown $k$ and $r$ into one measurable group.

Common pitfall. Treating the constant-pressure run as if the volume were fixed (forgetting $\varepsilon_A$), or trying to solve separately for $r$ and $k$ — the data only determine their product, which is all the question needs.

Source. Levenspiel, Chemical Reaction Engineering, Ch. 3 (interpretation of batch data; variable-volume systems).

Question 1 of 5: Zero-Order Gas Reaction — Volume Change in a Constant-Pressure Batch Reactor

Question 1: Zero-Order Gas Reaction — Volume Change in a Constant-Pressure Batch Reactor (25 marks)