Consecutive Reactions in a CSTR — Extracting Rate Constants and Orders from Residence-Time Data: 23-Chem-A4 Chemical Reactor Engineering, May 2016, Q2

Question 2: Consecutive Reactions in a CSTR — Extracting Rate Constants and Orders from Residence-Time 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.

Check — data correction (state as an assumption per rubric #1)

The printed value $C_{A0} = 1$ lb·mol/ft³ is inconsistent with the table: at the shortest $\tau$ the table already lists $C_A = 1.000$ (which would force a zero reaction rate), and $C_B$ reaches 0.5083 — above the $0.5\,C_{A0}$ ceiling that $2A\rightarrow B$ permits. The atom balance $C_A + 2C_B + 4C_C = C_{A0}$ with $C_C \ge 0$ requires $C_{A0} \ge \max(C_A + 2C_B) = 1.909$ lb·mol/ft³ (the $\tau = 10$ s row), so the printed 1 cannot be right. The balance alone only sets that lower bound; what pins the value is the proposed rate law: scanning $C_{A0}$, the log–log fit of $(C_{A0}-C_A)/\tau$ against $C_A$ becomes a perfect straight line at $C_{A0} \approx 2$, where every row satisfies $(2-C_A)/\tau = 0.1\,C_A^2$ to within rounding. Assumption: $C_{A0} = 2$ lb·mol/ft³, and the printed “1” is a typo.

Given. Five CSTR steady states (constant $v_0$, so the CSTR acts as a direct rate meter: $-r_A = (C_{A0}-C_A)/\tau$), with $C_{A0} = 2$ lb·mol/ft³. The printed $r_A = k_1C_A^{\alpha}$ and $r_B = -0.5k_1C_A^{\alpha} + k_2C_B^{\beta}$ are written as disappearance rates: A is consumed at $k_1C_A^{\alpha}$, B is formed at half that rate by $2A\rightarrow B$ and consumed at $k_2C_B^{\beta}$ by $2B\rightarrow C$.

Find. $\alpha,\ k_1$ in $-r_A = k_1 C_A^{\alpha}$, and $\beta,\ k_2$ in the B-consumption term $k_2 C_B^{\beta}$.

Figure 2.1 — A CSTR at steady state is a rate meter: the disappearance rate of A is read directly as $(C_{A0}-C_A)/\tau$ from each measured effluent, with no differentiation of noisy data.

Approach. Read $-r_A$ straight from the mole balance at each state, regress $\ln(-r_A)$ on $\ln C_A$ for $(\alpha,k_1)$, then back out the B-consumption rate from the B balance and regress it on $\ln C_B$ for $(\beta,k_2)$.

Turn each steady state into a rate. For a CSTR, $-r_A = \dfrac{C_{A0}-C_A}{\tau}$. E.g. at $\tau = 20$: $-r_A = (2-0.780)/20 = 0.0610$; at $\tau = 450$: $(2-0.200)/450 = 0.00400$.
Regress for the A kinetics. A least-squares fit of $\ln(-r_A)$ vs. $\ln C_A$ across all five points is a straight line of slope 2 and intercept $\ln k_1$: $$\alpha = \boxed{2}, \qquad k_1 = \boxed{0.100\ \text{ft}^3/(\text{lb}\cdot\text{mol}\cdot\text{s})}.$$ Check: $0.1\,C_A^2$ reproduces $(C_{A0}-C_A)/\tau$ term-for-term at every row.
Isolate the B-consumption rate. The net B balance in the CSTR is $\dfrac{C_B-0}{\tau} = \tfrac12(-r_A) - k_2 C_B^{\beta}$ (one B formed per two A consumed, minus B consumed by $2B\rightarrow C$). Hence the consumption rate is $R_B = \tfrac12(-r_A) - \dfrac{C_B}{\tau}$; e.g. at $\tau = 20$: $R_B = \tfrac12(0.0610) - 0.5083/20 = 0.00509$.
Regress for the B kinetics. Fitting $\ln R_B$ vs. $\ln C_B$ gives $$\beta = \boxed{1}, \qquad k_2 = \boxed{0.0100\ \text{s}^{-1}}.$$ The ratio $R_B/C_B = 0.0100$ is constant across all five states, confirming first order in B.

Parameter	Value
Order in A, $\alpha$	2
Rate constant $k_1$	0.100 ft³/(lb·mol·s)
Order in B, $\beta$	1
Rate constant $k_2$	0.0100 s$^{-1}$

Concept & Theory

Topic. The CSTR as an algebraic rate meter; sequential parameter estimation for consecutive reactions.

Key relations. CSTR mole balance $-r_A = (C_{A0}-C_A)/\tau$ (no derivatives needed); power-law linearisation $\ln(-r) = \ln k + n\ln C$; species balance for the intermediate $C_B/\tau = \tfrac12(-r_A) - k_2 C_B^{\beta}$ (Fogler, Ch. 5 & 7).

Why this works. Because a CSTR operates at a single, spatially uniform composition, its steady-state balance gives the rate directly as a difference over $\tau$ — unlike batch or PFR data, no numerical differentiation is required. Estimating A first, then substituting its known rate into the B balance, decouples the two-parameter problem into two clean linear regressions.

Common pitfall. Accepting the printed $C_{A0} = 1$ uncritically — it makes the shortest-$\tau$ rate zero and violates the atom balance. Always sanity-check feed data against the measured effluents before fitting.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 5 (CSTR design) and Ch. 7 (rate-data analysis).

Question 2 of 5: Consecutive Reactions in a CSTR — Extracting Rate Constants and Orders from Residence-Time Data

Question 2: Consecutive Reactions in a CSTR — Extracting Rate Constants and Orders from Residence-Time Data (25 marks)