Testing a Dual-Site LHHW Rate Law for DME Synthesis: 23-Chem-A4 Chemical Reactor Engineering, Undated paper, Q2

Question 2: Testing a Dual-Site LHHW Rate Law for DME Synthesis (25 marks: a 20, b 5)

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. Point rates of DME formation against methanol partial pressure; feed methanol + N₂ only, 375 K, 1340 kPa.

$p_M$ (kPa)	30	40	80	120	160	240
$r_{DME}$ (mol/kg·h)	0.155	0.156	0.200	0.217	0.219	0.220

Find. (a) Whether the dual-site rate law is consistent with the data; (b) best-fit $k_1$ and $K_M$.

Assumption

The feed contains no butanol, so $p_B=0$ and the $K_Bp_B$ adsorption term drops out. The tabulated rates are treated as point (differential) rates at the stated $p_M$. The data give no information on the reverse reaction, so the rate is used as printed (no reverse term).

Approach. With $p_B=0$ the law becomes $r_{DME}=k_1\left[\dfrac{K_Mp_M}{1+K_Mp_M}\right]^2$. Taking the square root and inverting gives a straight line in $p_M$ (the standard LHHW linearization, Fogler Ch. 10). Collinear transformed data answer part (a), and the slope and intercept give part (b). The fit is then checked back on the untransformed rates.

Linearize (part a). $\sqrt{r_{DME}}=\sqrt{k_1}\,\dfrac{K_Mp_M}{1+K_Mp_M}$, so $$\frac{p_M}{\sqrt{r_{DME}}}=\frac{1}{\sqrt{k_1}\,K_M}+\frac{1}{\sqrt{k_1}}\,p_M.$$ If the mechanism is right, a plot of $p_M/\sqrt{r_{DME}}$ against $p_M$ is a straight line.
Transformed data. $p_M/\sqrt{r_{DME}}=76.2,\ 101.3,\ 178.9,\ 257.6,\ 341.9,\ 511.7$ for the six points.
Least-squares line. Intercept $=14.82$, slope $=2.058$, coefficient of determination $R^2=0.9996$ (Figure 2a). The six points lie on a straight line.
Back-check on the rates. With the fitted constants the model predicts $r_{DME}=0.154,\ 0.170,\ 0.199,\ 0.210,\ 0.216,\ 0.223$. Five of the six points agree within 0.007 mol/kg·h (about 3%). The exception is the 40 kPa point: 0.156 measured against 0.170 predicted, or 8.7% low. On the rates themselves $R^2=0.948$ (Figure 2b).
Conclusion for (a). Yes. The data rise and then level off toward a plateau, as the squared-Langmuir form requires, and the linearized plot is straight, so the postulated rate equation fits. The one outlier (40 kPa, almost equal to the 30 kPa rate) is best read as experimental scatter. All the data lie in the near-saturation region ($K_Mp_M\ge4$), so they confirm the saturating form but cannot sharply separate this dual-site law from other saturating forms. Rates below about 10 kPa would be needed for that.
Parameters (part b). $\sqrt{k_1}=1/\text{slope}$ and $K_M=\text{slope}/\text{intercept}$: $$k_1=\frac{1}{2.058^2}=0.236,\qquad K_M=\frac{2.058}{14.82}=0.139$$ $$\boxed{k_1=0.236\ \text{mol DME/(kg cat}\cdot\text{h)},\qquad K_M=0.139\ \text{kPa}^{-1}}$$ Because $K_M^2p_M^2$ is dimensionless, $k_1$ carries the rate units and equals the saturation rate, just above the measured plateau of 0.220.

Figure 2a — Linearized test: $p_M/\sqrt{r_{DME}}$ against $p_M$ is a straight line ($R^2=0.9996$), confirming the dual-site LHHW form.

Figure 2b — Fitted rate law against the measured rates. The curve saturates at $k_1=0.236$; only the 40 kPa point falls noticeably below it.

Check

The linearization weights the points unevenly. A direct non-linear least-squares fit to the rates gives $k_1=0.241$ mol/kg·h and $K_M=0.123$ kPa⁻¹, which agrees within 2% on $k_1$ and 12% on $K_M$. $K_M$ is the less well-determined constant because every point is close to saturation.

Quantity	Result
(a) Linearized fit	straight line, $R^2=0.9996$: the rate law fits
(b) $k_1$	0.236 mol/(kg cat·h)
(b) $K_M$	0.139 kPa⁻¹
Largest deviation	40 kPa point, −8.7%

Concept & Theory

Topic. Discriminating and fitting Langmuir–Hinshelwood–Hougen–Watson (LHHW) rate laws.

Key relations. Dual-site surface reaction between two adsorbed methanol molecules gives $r=k_1K_M^2p_M^2/(1+K_Mp_M)^2$; linearization $p_M/\sqrt r=1/(\sqrt{k_1}K_M)+p_M/\sqrt{k_1}$; saturation rate $r\to k_1$ as $p_M\to\infty$.

Why this works. The squared denominator appears when the rate-limiting step needs two adjacent occupied sites. The square root removes that exponent, and inverting turns the Langmuir fraction into a straight line. At low pressure the law is second order in $p_M$; at high pressure the surface is full and the rate becomes independent of $p_M$. The data show only that second, zero-order end clearly.

Common pitfall. Treating $k_1$ as the initial slope, when it is the plateau rate here because $K_M^2$ has been written into the numerator. Another is carrying the $K_Bp_B$ term into the fit when the feed contains no butanol.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 10 (rate-limiting step, LHHW rate-law analysis and linear regression).

Question 2 of 5: Testing a Dual-Site LHHW Rate Law for DME Synthesis

Question 2: Testing a Dual-Site LHHW Rate Law for DME Synthesis (25 marks: a 20, b 5)