Reversible Pseudo-First-Order Esterification of Formic Acid: 23-Chem-A4 Chemical Reactor Engineering, May 2013, Q2

Question 2: Reversible Pseudo-First-Order Esterification of Formic Acid (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. Isothermal (25 °C) batch reactor; alcohol and water in constant excess make the esterification behave as a reversible pseudo-first-order reaction $A\rightleftharpoons B$. Starting acid $A_0=0.0677$ mol/L; equilibrium $A_e=0.0292$ mol/L.

Quantity	Value
$t$ (min)	0, 50, 100, 160, 290
$[\text{HCOOH}]$ (mol/L)	0.0677, 0.0615, 0.0562, 0.0509, 0.0428
Equilibrium $A_e$	0.0292 mol/L
Integrated model	$\ln[(A_0-A_e)/(A-A_e)]=(k_f+k_r)t$

Find. (a) confirmation of the reversible pseudo-first-order model; (b) $k_f$ and $k_r$; (c) $K=k_f/k_r$.

Figure 3 — Reversible pseudo-first-order diagnostic for the formic-acid esterification: $\ln\!\frac{A_0-A_e}{A-A_e}$ vs $t$ is linear (slope $=k_f+k_r$), confirming the $A\rightleftharpoons B$ model.

Approach. Linearize the supplied integrated form to get $k_f+k_r$, use the equilibrium position to get $K=k_f/k_r$, then solve the two together for the individual constants.

(a) Establish the reversible pseudo-first-order model. With ethanol and water in large, effectively constant excess, the esterification reduces to $A\rightleftharpoons B$ with pseudo-first-order forward and reverse steps. The supplied integrated form is $$\ln\!\frac{A_0-A_e}{A-A_e} = (k_f+k_r)\,t,$$ where $A_e$ is the equilibrium (t → ∞) formic-acid concentration. Taking $A_e = 0.0292$ mol/L (the final tabulated value) and $A_0=0.0677$ mol/L, the left-hand side evaluated at each time is $\{0,\,0.176,\,0.355,\,0.573,\,1.041\}$. Plotted against $t$ these fall on a straight line through the origin (Figure 3), so the model holds: $$\boxed{\ln\!\frac{A_0-A_e}{A-A_e}\ \text{is linear in } t.}$$
Slope gives the sum of rate constants. The least-squares slope of the line is $$k_f+k_r = \boxed{3.59\times10^{-3}\ \text{min}^{-1}}.$$
(c) Pseudo-equilibrium constant. At equilibrium the forward and reverse rates balance, $k_fA_e=k_r(A_0-A_e)$, so $$K=\frac{k_f}{k_r}=\frac{A_0-A_e}{A_e}=\frac{0.0677-0.0292}{0.0292}=\boxed{1.32}.$$
(b) Individual rate constants. Combine the sum and the ratio. From $k_f=K\,k_r$ and $k_f+k_r=3.59\times10^{-3}$, $$k_r=\frac{k_f+k_r}{1+K}=\frac{3.59\times10^{-3}}{2.32}=\boxed{1.55\times10^{-3}\ \text{min}^{-1}},\qquad k_f=(k_f+k_r)-k_r=\boxed{2.04\times10^{-3}\ \text{min}^{-1}}.$$

Quantity	Result
(a) Model	Reversible pseudo-1st-order — plot is linear
$k_f+k_r$	$3.59\times10^{-3}$ min$^{-1}$
(b) $k_f$	$2.04\times10^{-3}$ min$^{-1}$
(b) $k_r$	$1.55\times10^{-3}$ min$^{-1}$
(c) $K=k_f/k_r$	1.32

Concept & Theory

Topic. Reversible (opposed) first-order reactions; approach to equilibrium; separating forward and reverse rate constants.

Key relations. Rate $-\mathrm dA/\mathrm dt=k_fA-k_r B$; integrated approach to equilibrium $\ln\frac{A_0-A_e}{A-A_e}=(k_f+k_r)t$; equilibrium condition $K=k_f/k_r=(A_0-A_e)/A_e$.

Why this works. A reversible first-order reaction relaxes exponentially toward equilibrium with a single time constant $1/(k_f+k_r)$, so the linearized plot yields only the sum of the constants. The equilibrium concentration supplies the second equation $(K=k_f/k_r)$, and the two together determine $k_f$ and $k_r$ individually.

Common pitfall. Using the initial concentration in place of $A_e$, or forgetting that the slope is $k_f+k_r$ (not $k_f$ alone). The equilibrium value $A_e$ must be the $t\to\infty$ concentration—here the tabulated $t\to\infty$ entry (0.0292 mol/L).

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 3 (reversible reactions and equilibrium) and Ch. 5 (analysis of rate data).

Question 2 of 5: Reversible Pseudo-First-Order Esterification of Formic Acid

Question 2: Reversible Pseudo-First-Order Esterification of Formic Acid (20 marks)