Order and Rate Constant from Total-Pressure Data: 23-Chem-A4 Chemical Reactor Engineering, December 2017, Q5

Question 5: Order and Rate Constant 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. $2A\rightarrow A_2$, constant $V$ and $T=325$ °C; total pressure vs time (pure A initially, so $p_{A0}=P_{tot}(0)=84.25$ kPa):

$t$ (min)	0	10	20	30	40	50	60	70	80
$P_{tot}$ (kPa)	84.25	78.91	74.45	71.12	68.52	66.25	64.52	63.05	61.72

Find. the reaction order $n$ and rate constant $k$.

Approach. Convert total pressure to the partial pressure of A through the stoichiometry, then apply the integral method — test $n=2$ by plotting $1/p_A$ against $t$ — and confirm with the differential (log–log) slope.

Relate $p_A$ to the total pressure. For $2A\rightarrow A_2$ at constant $V,T$, consuming $(p_{A0}-p_A)$ of A forms $\tfrac12(p_{A0}-p_A)$ of $A_2$, so $P_{tot} = p_A + \tfrac12(p_{A0}-p_A) = \tfrac12(p_{A0}+p_A),$ giving $$p_A = 2P_{tot} - p_{A0} = 2P_{tot} - 84.25.$$ This yields $p_A = 84.25, 73.57, 64.65, 57.99, 52.79, 48.25, 44.79, 41.85, 39.19$ kPa.
Integral test for second order. If $n=2$, $-dp_A/dt = k p_A^2$ integrates to $\dfrac{1}{p_A} - \dfrac{1}{p_{A0}} = k\,t,$ so a plot of $1/p_A$ versus $t$ should be a straight line of slope $k$.
Regress and read the constant. The $1/p_A$–$t$ data are highly linear (coefficient of determination $R^2 = 0.999$), with slope $$k = \boxed{1.71\times10^{-4}\ \text{kPa}^{-1}\,\text{min}^{-1}}, \qquad \boxed{n = 2.}$$
Confirm the order. A first-order plot ($\ln p_A$ vs $t$) is visibly curved ($R^2 = 0.983$, systematically bowed), and the differential method — the log–log slope of $-dp_A/dt$ against $p_A$ from central differences — gives $n\approx2.3$ (differencing amplifies the scatter, so the integral test is the better estimate). Both corroborate second-order kinetics, as expected for the bimolecular step $2A\rightarrow A_2$.

Figure 5.1 — Second-order integral test: 1/p_A is linear in time (R² = 0.999). The slope is the rate constant k = 1.71×10⁻⁴ kPa⁻¹ min⁻¹.

Quantity	Value
Reaction order, $n$	2 (second order)
Rate constant, $k$	$1.71\times10^{-4}$ kPa$^{-1}$·min$^{-1}$

Concept & Theory

Topic. Determining reaction order and rate constant from constant-volume total-pressure data (integral and differential methods).

Key relations. Stoichiometric pressure link $P_{tot} = \tfrac12(p_{A0}+p_A)$; integrated second order $\tfrac{1}{p_A}-\tfrac{1}{p_{A0}} = kt$; differential order from $\ln(-dp_A/dt) = \ln k + n\ln p_A$.

Why this works. At constant $V,T$ pressure is proportional to moles, so the measured total pressure maps linearly onto the partial pressure of A via the stoichiometry. Testing a candidate order by checking which integrated form linearises the data is robust to scatter, and the differential slope independently fixes the order.

Common pitfall. Using $P_{tot}$ directly as $p_A$ (it is not — product formation lowers the total less than reactant depletion would), and mishandling the mole-change sign, which flips the $P_{tot}$–$p_A$ relation.

Source. Levenspiel, Chemical Reaction Engineering, 3rd ed., Ch. 3 (interpretation of batch reactor data); Fogler, 5th ed., Ch. 7.

Question 5 of 5: Order and Rate Constant from Total-Pressure Data

Question 5: Order and Rate Constant from Total-Pressure Data (25 marks)