Reversible Pseudo-First-Order Esterification: 23-Chem-A4 Chemical Reactor Engineering, December 2013, Q2

Question 2: Reversible Pseudo-First-Order Esterification (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. Titration measures total acid = organic acid + HCl, so $[A]=(\text{total})-0.01934$. The HCl catalyst concentration is constant. The initial and equilibrium organic-acid concentrations are $[A]^\circ=0.0677$ M and $[A]_e=0.04858-0.01934=0.02924$ M.

$t$ (min)	Total acid (mol/L)	$[A]=$ total $-0.01934$
0	0.08704	0.06770
50	0.08080	0.06146
100	0.07550	0.05616
160	0.07020	0.05086
290	0.06218	0.04284
∞	0.04858	0.02924 = $[A]_e$

Find. (a) confirm linearity of $\ln\{([A]-[A]_e)/([A]^\circ-[A]_e)\}$ vs. $t$; (b) $k_f$ and $k_r$; (c) $K=k_f/k_r$.

Figure 2 — The linearized rate expression plots as a straight line through the origin (slope $=-(k_f+k_r)$), confirming reversible pseudo-first-order behaviour.

Approach. Subtract the constant HCl to recover $[A]$, form the dimensionless group, take logs, and least-squares fit against $t$: a straight line (slope $-(k_f+k_r)$, near-zero intercept) proves the model. The equilibrium constant $K=[E]_e/[A]_e$ then splits the sum into $k_f$ and $k_r$.

Part (a) — linearize and test. With $[A]^\circ-[A]_e=0.03846$, form $y=\ln\dfrac{[A]-[A]_e}{[A]^\circ-[A]_e}$ at each time (0, −0.177, −0.357, −0.576, −1.039 at $t=0,50,100,160,290$). A least-squares line gives a correlation $R^2>0.999$ with intercept $\approx 0$, so$$\ln\frac{[A]-[A]_e}{[A]^\circ-[A]_e}=-(k_f+k_r)\,t\quad\text{holds — reversible pseudo-first-order confirmed.}$$
Part (b) — sum of rate constants from the slope. The fitted slope is $-(k_f+k_r)$:$$k_f+k_r=3.59\times10^{-3}\ \text{min}^{-1}.$$
Equilibrium constant ties the ratio. Every mole of A consumed makes one mole of E, so $[E]_e=[A]^\circ-[A]_e=0.03846$ M and$$K=\frac{[E]_e}{[A]_e}=\frac{k_f}{k_r}=\frac{0.03846}{0.02924}=\boxed{1.32}\quad\text{(part c)}.$$
Split the sum. From $k_f=Kk_r$ and $k_f+k_r=k_r(1+K)$:$$k_r=\frac{3.59\times10^{-3}}{1+1.32}=\boxed{1.55\times10^{-3}\ \text{min}^{-1}},\quad k_f=Kk_r=\boxed{2.04\times10^{-3}\ \text{min}^{-1}}.$$

Quantity	Result
(a) Model	Straight line, $R^2>0.999$, intercept ≈ 0 — confirmed
(b) $k_f+k_r$	$3.59\times10^{-3}$ min⁻¹
(b) $k_f$ / $k_r$	$2.04\times10^{-3}$ / $1.55\times10^{-3}$ min⁻¹
(c) $K=k_f/k_r$	1.32

Concept & Theory

Topic. Integrated rate law for a reversible first-order reaction and extraction of forward/reverse rate constants by linear regression.

Key relations. $\ln\dfrac{[A]-[A]_e}{[A]^\circ-[A]_e}=-(k_f+k_r)t$; equilibrium $K=k_f/k_r=[E]_e/[A]_e$; mass conservation $[E]=[A]^\circ-[A]$.

Why this works. A reversible first-order reaction relaxes exponentially toward equilibrium with a single time constant $1/(k_f+k_r)$. The slope of the semi-log plot delivers that sum; equilibrium delivers the ratio; two equations, two unknowns.

Common pitfall. Plotting total acid instead of $[A]$ (the constant HCl must be subtracted first), or using $\ln[A]$ vs. $t$ (valid only for an irreversible reaction — here the driving force is $[A]-[A]_e$, not $[A]$).

Source. Fogler §3.3 (reversible reactions) and §7 (integral method of rate-data analysis); Levenspiel Ch. 3.

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

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