Question 1 of 14: Reaction-Order Kinetics in Food Processing

Question 1: Reaction-Order Kinetics in Food Processing (12.5 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.

(a) Half-order conversion at 30 min

Given. A batch reaction follows a half-order rate law; a 10 min run converts 75% of the liquid reactant.

Find. The fractional conversion after a 30 min (half-hour) run.

Approach. Integrate the half-order batch rate law to get $C(t)$, use the 10 min datum to fix the rate constant, then evaluate at $t=30$ min — checking first whether the reactant is exhausted before 30 min, since an order below 1 reaches zero concentration in finite time (unlike first order).

1. Integrate the half-order rate law. For $-dC/dt = kC^{n}$ with $n=0.5$, separating and integrating from $C_0$ at $t=0$ gives $$C^{0.5} = C_0^{0.5} - \tfrac{1}{2}kt.$$
2. Fix $k$ from the 10 min datum. At $t=10$ min, $X=0.75$, so $C = 0.25C_0$ and $C^{0.5}=0.5C_0^{0.5}$. Substituting, $$0.5C_0^{0.5} = C_0^{0.5} - \tfrac12 k(10) \ \Rightarrow\ k = 0.1\,C_0^{0.5}\ \text{min}^{-1}.$$
3. Find the time to complete conversion. The half-order law reaches $C=0$ at a finite time $t^{*}$ (a power-law rate never asymptotes the way an exponential does), found by setting $C^{0.5}=0$: $$t^{*} = \frac{C_0^{0.5}}{\tfrac12 k} = \frac{C_0^{0.5}}{0.05\,C_0^{0.5}} = \boxed{20\ \text{min}}.$$
4. Evaluate at $t=30$ min. Since $30\ \text{min} > t^{*}=20\ \text{min}$, the reactant is already fully consumed by the time the run reaches 30 min — the concentration cannot go negative, so it simply stays at zero from $t^{*}$ onward. The conversion at 30 min is therefore $\boxed{X = 100\%}$, not a value obtained by naively extending the $C^{0.5}(t)$ line past its zero-crossing.

(b) Determining the rate equation from two conversion points

Given. Batch reactor, $C_{A0}=1$ mol/L; conversion is 80% at $t=8$ min and 90% at $t=18$ min.

Find. The order $n$ and rate constant $k$ of the rate equation $-dC_A/dt = kC_A^{\,n}$.

Approach. Test the integer-order integrated forms against both data points at once: the correct order is the one that returns the same $k$ from both points, since a single data point can always be forced to fit any assumed order.

1. Tabulate concentrations. $C_A = C_{A0}(1-X)$: at $t=8$ min, $C_A=0.20$ mol/L; at $t=18$ min, $C_A=0.10$ mol/L.
2. Test second order. The integrated second-order law is $1/C_A - 1/C_{A0} = kt$. From $t=8$ min: $$k = \frac{1/0.20 - 1/1}{8} = \frac{5-1}{8} = 0.5\ \text{L}\,\text{mol}^{-1}\,\text{min}^{-1}.$$
3. Check against the second point. From $t=18$ min: $$k = \frac{1/0.10-1/1}{18} = \frac{10-1}{18} = 0.5\ \text{L}\,\text{mol}^{-1}\,\text{min}^{-1}.$$ Both points return the identical $k$, confirming the order.

The reaction is therefore second order in A, with $$\boxed{-\frac{dC_A}{dt} = kC_A^{2},\qquad k = 0.5\ \text{L}\,\text{mol}^{-1}\,\text{min}^{-1}.}$$