Rate Expression from Constant-Volume Total-Pressure Data: 23-Chem-A4 Chemical Reactor Engineering, December 2018, Q2

Question 2: Rate Expression from Constant-Volume Total-Pressure Data (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. Constant-volume, constant-temperature (373.15 K) batch data for $A \rightarrow 2B$. The charge is 76.94 mol% $A$ + 23.06 mol% inert. On dropping the sealed reactor into the 100 °C bath, the charge first heats from 287.15 K to 373.15 K, so the recorded $t=0$ pressure of 1.31 atm is purely thermal: $1\times\frac{373.15}{287.15} = 1.30$ atm ✓. Hence at reaction start $p_{A0} = 0.7694\times1.31 = 1.008$ atm and $p_{\text{inert}} = 0.302$ atm.

Find. The order and rate constant of a rate law $-r_A = k\,C_A^{n}$ that fits the data.

Figure 2. Order diagnostic. Plotting the 1.5-order integral group $2\!\left(p_A^{-1/2}-p_{A0}^{-1/2}\right)$ against $t$ gives a straight line ($R^2 = 0.999$), whereas first- and second-order plots curve. The slope is the pressure-basis rate constant $k_p$.

Approach. Convert total pressure to the partial pressure of $A$ by stoichiometric bookkeeping, diagnose the order by the differential method, then confirm it and extract $k$ with the matching integrated rate law.

Relate total pressure to $p_A$. For $A \rightarrow 2B$ at constant $V,T$, each mole of $A$ consumed adds one net mole of gas, so $P = p_{\text{inert}} + p_A + 2(p_{A0}-p_A) = P_\infty - p_A$, where $P_\infty = 1.31 + p_{A0} = 2.318$ atm is the pressure at complete conversion. Therefore $$p_A = P_\infty - P = 2.318 - P.$$
Diagnose the order (differential method). A log–log fit of $-\,dp_A/dt$ against $p_A$ has slope $1.5$, pointing to a 1.5-order rate law. (First order would give a constant $\frac{1}{t}\ln\frac{p_{A0}}{p_A}$, which instead drifts; second order over-curves.)
Confirm with the integrated 1.5-order law. For $-\,dp_A/dt = k_p\,p_A^{1.5}$, integration gives $2\!\left(p_A^{-1/2}-p_{A0}^{-1/2}\right) = k_p t$. As Figure 2 shows, this group is linear in $t$ with $R^2 = 0.999$ and slope $$\boxed{k_p = 0.409\ \text{atm}^{-1/2}\,\text{hr}^{-1}.}$$
Convert to a concentration basis. With $C_A = p_A/RT$ and $R = 0.082057\ \text{L}\,\text{atm}\,\text{mol}^{-1}\text{K}^{-1}$, $T = 373.15$ K, the concentration-basis constant is $k_c = k_p (RT)^{1/2} = 0.409\,(30.62)^{1/2} = 2.26\ (\text{L/mol})^{1/2}\text{hr}^{-1}$.
State the rate expression. $$\boxed{-r_A = 2.26\;C_A^{1.5}\quad \left[(\text{L/mol})^{1/2}\,\text{hr}^{-1}\right]\ \text{at }100\,{}^\circ\text{C},}$$ equivalently $-\,dp_A/dt = 0.409\,p_A^{1.5}$ on a pressure basis.

Quantity	Value
Initial $A$ partial pressure $p_{A0}$	1.008 atm
Reaction order $n$	1.5
Pressure-basis constant $k_p$	0.409 atm$^{-1/2}$ hr$^{-1}$
Concentration-basis constant $k_c$	2.26 (L/mol)$^{1/2}$ hr$^{-1}$

Concept & Theory

Topic. Kinetics from constant-volume batch total-pressure data with a change in mole number.

Key relations. Stoichiometric pressure bookkeeping $p_A = P_\infty - P$ (for $\Delta n = +1$ per mole $A$); differential order test $\ln(-dp_A/dt)=\ln k_p + n\ln p_A$; integrated $n$-order law $\frac{p_A^{1-n}-p_{A0}^{1-n}}{n-1}=k_p t$.

Why this works. In a rigid isothermal vessel, partial pressures are proportional to moles, so the measured total pressure maps linearly onto $p_A$ once the stoichiometry (here $+1$ net mole per mole $A$) is accounted for. The correct order is the one whose integrated group plots straight; the differential method fixes it independently as a cross-check.

Common pitfall. Forgetting the inert partial pressure or the initial thermal expansion, which shifts $p_{A0}$ and $P_\infty$ and corrupts every derived $p_A$; and assuming an integer order without testing — here the true order is fractional (1.5).

Source. Levenspiel, 3rd ed., Ch. 3 (interpretation of batch reactor data); Fogler, 5th ed., §7.2.

Question 2 of 5: Rate Expression from Constant-Volume Total-Pressure Data

Question 2: Rate Expression from Constant-Volume Total-Pressure Data (25 marks)