Diphosgene Decomposition — Order and Activation Energy: 23-Chem-A4 Chemical Reactor Engineering, May 2013, Q1

Question 1: Diphosgene Decomposition — Order and Activation Energy (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, constant-temperature $(280\,{}^\circ\text{C})$ gas-phase batch reactor; $A\rightarrow 2B$ so total pressure rises as $A$ decomposes. Initial charge is pure $A$ at $P_{A0}=15$ torr.

Quantity	Value
$t$ (s)	0, 500, 800, 1300, 1800
$P_{tot}$ (torr)	15, 18.9, 20.7, 23, 24.8
Initial diphosgene pressure $P_{A0}$	15 torr
Second temperature	265 °C ($k$ = ½ of the 280 °C value)

Find. (a) that the 2nd-order model fails; (b) that a 1st-order law fits; (c) the mean rate constant; (d) the activation energy $E_a$.

Figure 1 — First-order diagnostic for diphosgene decomposition. The points $\ln P_A$ vs $t$ fall on a straight line (slope $=-k$), confirming first-order kinetics.

Figure 2 — Second-order diagnostic. $1/P_A$ vs $t$ is visibly concave-up (the local slope keeps increasing), so the second-order model does not fit the data.

Approach. Convert total pressure to diphosgene partial pressure via the $A\rightarrow2B$ stoichiometry, then diagnose the order from which linearization is straight and obtain $E_a$ from the Arrhenius two-point relation.

Relate the diphosgene partial pressure to total pressure. For the constant-volume, constant-temperature batch decomposition $A\rightarrow 2B$, each mole of $A$ that reacts produces two moles of $B$, so the moles (and pressure) rise by the extent of reaction. If $x$ (torr) of $A$ has decomposed, then $$P_{tot}=(P_{A0}-x)+2x = P_{A0}+x \;\Rightarrow\; P_A = P_{A0}-x = 2P_{A0}-P_{tot}.$$ With $P_{A0}=15$ torr this gives $P_A = \{15,\,11.1,\,9.3,\,7.0,\,5.2\}$ torr.
(a) Second-order test — plot $1/P_A$ vs $t$. A second-order rate law $-\mathrm dP_A/\mathrm dt = k'P_A^2$ integrates to $1/P_A = 1/P_{A0}+k't$, a straight line. The data give $1/P_A = \{0.0667,\,0.0901,\,0.1075,\,0.1429,\,0.1923\}$ torr$^{-1}$; the successive slopes rise from $4.7\times10^{-5}$ to $9.9\times10^{-5}$ torr$^{-1}$s$^{-1}$ (Figure 2 is concave-up). Because the slope is not constant, $$\boxed{\text{the second-order model does not fit the data.}}$$
(b) First-order differential test. Estimate the rate $-\mathrm dP_A/\mathrm dt$ by finite differences over each interval and evaluate it at the mean pressure of the interval; for a first-order law $-\mathrm dP_A/\mathrm dt=kP_A$ the ratio $k=(-\mathrm dP_A/\mathrm dt)/P_A$ should be constant: $$k = \frac{-\Delta P_A/\Delta t}{\bar P_A}=\{5.98,\,5.88,\,5.64,\,5.90\}\times10^{-4}\ \text{s}^{-1}.$$ The four values agree to within 6%, and equivalently $\ln P_A$ vs $t$ is a straight line (Figure 1). The data therefore obey a first-order rate law.
(c) Mean rate constant. Averaging the differential values (and the slope of the $\ln P_A$–$t$ regression, $-k$, which gives the same figure): $$k = -\frac{\mathrm d(\ln P_A)}{\mathrm dt}=\boxed{5.86\times10^{-4}\ \text{s}^{-1}}\quad(\text{first order}).$$
(d) Activation energy. Given $k(265\,{}^\circ\text{C})=\tfrac12\,k(280\,{}^\circ\text{C})$, apply the two-temperature Arrhenius relation with $T_1=553.15$ K, $T_2=538.15$ K: $$E_a = \frac{R\,\ln\!\big(k_1/k_2\big)}{\left(\tfrac1{T_2}-\tfrac1{T_1}\right)} =\frac{(8.314)\ln 2}{5.039\times10^{-5}}=1.14\times10^{5}\ \text{J/mol}=\boxed{114\ \text{kJ/mol}}.$$

Quantity	Result
(a) Second-order fit	Fails — $1/P_A$ vs $t$ is curved
(b)–(c) Rate law & constant	First order, $k = 5.86\times10^{-4}$ s$^{-1}$
(d) Activation energy	$E_a = 114$ kJ/mol

Concept & Theory

Topic. Constant-volume batch kinetics from total-pressure data; integral vs. differential order determination; Arrhenius temperature dependence.

Key relations. Stoichiometric pressure bookkeeping $P_A=2P_{A0}-P_{tot}$ for $A\to2B$; first-order integral $\ln(P_{A0}/P_A)=kt$; second-order integral $1/P_A-1/P_{A0}=k't$; Arrhenius $\ln(k_1/k_2)=-\tfrac{E_a}{R}\big(\tfrac1{T_1}-\tfrac1{T_2}\big)$.

Why this works. In a rigid isothermal vessel, pressure is a direct proxy for moles, so the measured $P_{tot}$ maps one-to-one onto the reaction extent. The order is found by testing which integrated form linearizes the data; the first-order plot is straight while the second-order plot curves, so first order is confirmed and its slope is $k$. A single $k$-ratio at two temperatures fixes $E_a$ without needing the pre-exponential factor.

Common pitfall. Forgetting that $A\to2B$ makes the total pressure rise—students sometimes set $P_A=P_{tot}$ or use $P_A=P_{A0}-(P_{tot}-P_{A0})$ with the wrong sign. Convert temperatures to kelvin before the Arrhenius step.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 5 (collection and analysis of rate data; integral & differential methods) and Ch. 3 (Arrhenius equation).

Question 1 of 5: Diphosgene Decomposition — Order and Activation Energy

Question 1: Diphosgene Decomposition — Order and Activation Energy (20 marks)