Gas-Phase $A\rightarrow 2B$ from Total-Pressure Data: 23-Chem-A4 Chemical Reactor Engineering, December 2013, Q4

Question 4: Gas-Phase $A\rightarrow 2B$ from Total-Pressure Data (20 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 and temperature (280°C). For $A\rightarrow 2B$ each mole of A consumed adds one mole of gas, so the total pressure rises: $P_{tot}=P_{A0}+2(P_{A0}-P_A)=2P_{A0}-P_A$, hence $P_A=2P_{A0}-P_{tot}$ (with $P_{A0}=0.01974$ atm). Partial pressure is a proxy for concentration at fixed $T,V$.

$t$ (s)	$P_{tot}$ (atm)	$P_A=2P_{A0}-P_{tot}$ (atm)
0.0000	0.01974	0.01974
0.1389	0.02487	0.01461
0.2222	0.02724	0.01224
0.3611	0.03026	0.00922
0.5000	0.03263	0.00685

Find. (a) demonstrate second order fails; (b) demonstrate first order fits; (c) mean $k$ with units; (d) activation energy from the $k$-ratio between 280°C and 255°C.

Figure 4 — $\ln(P_{A0}/P_A)$ vs. $t$ is linear through the origin (slope $=k$), confirming first-order kinetics; the $1/P_A$ (second-order) test curves upward instead.

Approach. Convert total pressure to $P_A$, then test the two integrated forms. Second order: $1/P_A$ vs. $t$ should be linear; first order: $\ln(P_{A0}/P_A)$ vs. $t$ should be linear. The straight one wins; its slope is $k$. Finally apply the two-temperature Arrhenius relation.

Part (a) — second-order test fails. For $1/P_A$ vs. $t$ the successive slopes climb steadily (128 → 159 → 193 → 270 atm⁻¹s⁻¹): the plot is concave-up, not a straight line, so$$\frac{1}{P_A}-\frac{1}{P_{A0}}=k\,t\quad\text{does NOT fit — second order rejected.}$$
Part (b) — first-order test fits. For $\ln(P_{A0}/P_A)$ vs. $t$ the per-point slopes are nearly constant ($\approx2.11$ s⁻¹) and a least-squares line has intercept $\approx0$ ($R^2>0.999$):$$\ln\frac{P_{A0}}{P_A}=k\,t\quad\text{fits — first order confirmed.}$$
Part (c) — mean rate constant. The regression slope is$$\boxed{k=2.11\ \text{s}^{-1}}\quad(\text{first order}\Rightarrow\text{units s}^{-1}).$$
Part (d) — activation energy. With $k(255\,{}^\circ\text{C})=\tfrac12 k(280\,{}^\circ\text{C})$, $T_1=553.15$ K, $T_2=528.15$ K, the Arrhenius two-point form gives$$E_a=\frac{R\ln(k_1/k_2)}{\tfrac{1}{T_2}-\tfrac{1}{T_1}}=\frac{(8.314)\ln 2}{\tfrac{1}{528.15}-\tfrac{1}{553.15}}=\boxed{67.3\ \text{kJ/mol}}.$$

Quantity	Result
(a) Second-order fit	$1/P_A$ vs. $t$ curves upward — fails
(b)/(c) First-order fit & $k$	Linear; $k=2.11$ s⁻¹
(d) Activation energy $E_a$	67.3 kJ/mol

Concept & Theory

Topic. Integral method of rate-data analysis for a constant-volume gas reaction using total pressure, plus a two-temperature Arrhenius estimate.

Key relations. Stoichiometric pressure map $P_A=2P_{A0}-P_{tot}$; first order $\ln(P_{A0}/P_A)=kt$; second order $1/P_A-1/P_{A0}=kt$; Arrhenius $\ln(k_1/k_2)=\dfrac{E_a}{R}\!\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$.

Why this works. At fixed $T$ and $V$, partial pressure is proportional to concentration, so pressure-based integrated laws have the same form as concentration-based ones. The known stoichiometry converts the single measured total pressure into $P_A$.

Common pitfall. Using total pressure directly in the rate law, or getting the pressure map backwards. Because $A\rightarrow2B$ makes net gas, $P_{tot}$ rises; a $A\rightarrow\tfrac12 B$-type reaction would make it fall. Always temperatures in kelvin for Arrhenius.

Source. Fogler §7.2 (integral method; total-pressure data) and §3.3 (Arrhenius); Levenspiel Ch. 3.

Question 4 of 5: Gas-Phase $A\rightarrow 2B$ from Total-Pressure Data

Question 4: Gas-Phase $A\rightarrow 2B$ from Total-Pressure Data (20 marks)