Question 4 of 5: Transesterification Kinetics and Selectivity

Question 4: Transesterification Kinetics and Selectivity (25 marks: a 12, b 3, c 3, d 7)

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. Batch series-reaction data at 60 °C; $C_{TA0} = 2.33$, $C_{MeOH0} = 13.44\ \text{mol/L}$; each of the three steps consumes one MeOH and produces one MeAc.

Find. (a) whether $\ln(C_{TA0}/C_{TA})$ is linear in $t$; (b) $C_{MeOH}(60)/C_{MeOH0}$; (c) selectivity and yield of glycerol on a TA basis at 60 min; (d) a graphical procedure to test the two-concentration rate law.

Approach. (a) apply the pseudo-first-order linear test; (b) close a methanol balance through the total MeAc produced; (c) form selectivity and yield from the glycerol and consumed/initial TA; (d) recast the bimolecular rate law into a straight-line plot.

1. (a) Pseudo-first-order test. If methanol were effectively constant, $-\tfrac{dC_{TA}}{dt} = k'C_{TA}$ with $k' = kC_{MeOH0}$, giving a straight line $\ln(C_{TA0}/C_{TA}) = k't$. A least-squares line through all nine points (Fig. 4.1) gives $$\boxed{k' \approx 0.025\ \text{min}^{-1}\ (R^2 \approx 0.95),}$$ but its intercept is 0.26 rather than the zero the model requires, and the residuals are systematic. The local slope is about $0.080\ \text{min}^{-1}$ over the first 5 min and $0.032\ \text{min}^{-1}$ at 10–15 min, then stays nearly constant at $0.018$–$0.023\ \text{min}^{-1}$. From 10 to 60 min the points are almost exactly linear ($k' \approx 0.021\ \text{min}^{-1}$, $R^2 = 0.997$). The pseudo-first-order model therefore does not fit the whole run. The misfit is concentrated in the rapid initial conversion (first 5–10 min), while the later data follow a first-order line with a smaller constant.
2. (b) Methanol depletion. Every mole of MeAc formed consumes one mole of MeOH, so $C_{MeOH}(t) = C_{MeOH0}-C_{MeAc}(t)$. At 60 min $$C_{MeOH}(60) = 13.44 - 3.89 = 9.55\ \text{mol/L},\qquad \boxed{\frac{C_{MeOH}(60)}{C_{MeOH0}} = \frac{9.55}{13.44} = 0.711.}$$ Methanol falls by nearly 29%, so treating it as constant is only roughly valid. This depletion would lower $k' = kC_{MeOH}$ steadily, by about 29% at 60 min, yet the part-(a) slope does not decline after 15 min. Methanol consumption alone therefore does not explain the curvature, which sits in the first few minutes; the two-concentration test of part (d) accounts for the depletion explicitly.
3. (c) Selectivity and yield of glycerol (TA basis). Glycerol produced is $C_G = 0.43\ \text{mol/L}$; TA consumed is $C_{TA0}-C_{TA}(60) = 2.33-0.43 = 1.90\ \text{mol/L}$. Hence $$\boxed{S_{G/TA} = \frac{C_G}{C_{TA0}-C_{TA}} = \frac{0.43}{1.90} = 0.226,}\qquad \boxed{Y_{G/TA} = \frac{C_G}{C_{TA0}} = \frac{0.43}{2.33} = 0.185.}$$ Only about 23% of the reacted TA has proceeded all the way to glycerol; the balance is held up as the diacetin/monoacetin intermediates.
4. (d) Graphical test of $-r_{TA} = k\,C_{TA}C_{MeOH}$. Compute the instantaneous rate $-r_{TA} = -\,dC_{TA}/dt$ from the slope of the $C_{TA}$–$t$ curve at each time (finite differences), and pair it with the product $C_{TA}\,C_{MeOH}$ evaluated at the same instant (using $C_{MeOH} = 13.44 - C_{MeAc}$). Plotting $-r_{TA}$ against $C_{TA}C_{MeOH}$ should give a straight line through the origin whose slope is $k$; systematic curvature would reject the second-order-overall form. (Illustration only — not evaluated.)