Rate Equation from Total-Pressure Data: 23-Chem-A4 Chemical Reactor Engineering, December 2019, Q3

Question 3: Rate Equation from 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 gas reaction $2A \rightarrow R$; pure $A$ charged at 1 atm/298.15 K, then heated to 373.15 K; total pressure $\pi(t)$ tabulated; complete conversion at $t\to\infty$.

Find. The reaction order and rate constant, i.e. the complete rate equation for $-r_A$.

Figure 3.1 — $1/p_A$ is linear in $t$ ($R^2 = 0.9998$), confirming a second-order reaction; the slope is the pressure-basis rate constant.

Approach. Convert the measured total pressure to the partial pressure of $A$ through the stoichiometric mole balance, then test integrated order plots ($1/p_A$ straight $\Rightarrow$ second order) and convert the pressure-basis constant to concentration units.

Initial partial pressure after heating. Heating pure $A$ from 298.15 to 373.15 K (before any reaction) raises its pressure to $$p_{A0} = (1\ \text{atm})\frac{373.15}{298.15} = 1.2516\ \text{atm}.$$
Relate $p_A$ to total pressure. For $2A \rightarrow R$, forming $x$ atm of $R$ consumes $2x$ atm of $A$: $p_A = p_{A0}-2x$, $p_R = x$, so $\pi = p_A + p_R = p_{A0}-x$. Eliminating $x$, $$\boxed{p_A = 2\pi - p_{A0}.}$$ (Check: at $t\to\infty$, $p_A\to 0$ gives $\pi\to p_{A0}/2 = 0.626$ atm, consistent with the still-falling data and “no $A$ found”.)
Test the order. Forming $p_A = 2\pi - p_{A0}$ and plotting $1/p_A$ versus $t$ gives a straight line (Fig. 3.1, $R^2 = 0.9998$), while $\ln p_A$ is curved — so the reaction is second order, $-r_A = k\,C_A^2$. The slope is the pressure-basis constant $$k_p = 0.207\ \text{atm}^{-1}\text{min}^{-1}.$$
Convert to concentration units. With $C_A = p_A/RT$, a second-order pressure constant converts as $k = k_p\,RT$: $$\boxed{k = (0.207)(0.082057)(373.15) = 6.33\ \text{L}\,\text{mol}^{-1}\text{min}^{-1}.}$$ For reference $C_{A0} = p_{A0}/RT = 0.0409\ \text{mol/L}$.
State the rate equation. $$-r_A = 6.33\,C_A^{2}\quad[\text{mol}\,\text{L}^{-1}\text{min}^{-1}],\ C_A\ \text{in mol/L}.$$

Quantity	Value
Heated initial pressure $p_{A0}$	1.252 atm
Reaction order	Second order
Pressure-basis constant $k_p$	0.207 atm$^{-1}$min$^{-1}$
Rate constant $k$	6.33 L·mol$^{-1}$min$^{-1}$
Rate equation	$-r_A = 6.33\,C_A^{2}$

Concept & Theory

Topic. Extracting order and rate constant from constant-volume total-pressure measurements.

Key relations. Ideal gas at fixed $V,T$: partial pressures add to $\pi$; mole-change bookkeeping gives $p_A = a\pi + b$ set by stoichiometry (here $p_A = 2\pi - p_{A0}$). Second order $\Rightarrow 1/p_A = 1/p_{A0}+k_p t$; conversion $k = k_p RT$.

Why this works. Only total pressure is measured, but stoichiometry ties every product mole to a fixed number of reactant moles, so $p_A$ is a linear function of $\pi$. Once $p_A(t)$ is known, the usual integrated-rate linearisations apply; the plot that is straight identifies the order without differentiation. The isothermal charge (heat first, react second) is what makes $p_{A0}$ a clean, computable baseline.

Common pitfall. Forgetting the factor of two ($p_A = 2\pi - p_{A0}$, not $\pi - p_{A0}$) for the $2A\rightarrow R$ stoichiometry; and using the 1 atm charge pressure instead of the heated $p_{A0} = 1.25$ atm.

Source. Levenspiel, Chemical Reaction Engineering, 3rd ed., Ch. 3 (constant-volume batch, total-pressure method).

Question 3 of 5: Rate Equation from Total-Pressure Data

Question 3: Rate Equation from Total-Pressure Data (25 marks)