Chapter 5: Isothermal Reactor Design: Conversion

Rate Constants From Semilog Graphs

This example shows how to determine the rate constant for a reaction by finding the slope of a line on semilog paper. Please refer to Example 5.1 from the text. This example picks up from step 5 in the text (bottom of page 152).

\(-\frac{dC_A}{dt} = kC_A \quad (E4-1.3)\)

5. Evaluate. For isothermal operation, \(k\) is constant, so we can integrate this equation (E4-1.3).

\( -\int_{C_{A_0}}^{C_A} \frac{dC_A}{C_A} = \int_0^t k \,dt = k \int_0^T dt \)

Using the initial condition that when \(t = 0\), then \(C_A = C_{A_0}\). The initial concentration of A after mixing the two volumes together is 1.0 kmol/m³ (1 mol/L).

Integrating yields:

\( \ln \frac{C_{A_0}}{C_A} = kt \) (E4-1.4)

The concentration of ethylene oxide at any time \( t \) is:

\( C_A = C_{A_0} e^{-kt} \) (E4-1.5)

The concentration of ethylene glycol at any time \( t \) can be obtained from the reaction stoichiometry:

\( \text{A + B} \longrightarrow \text{C} \) \( N_C = N_{A_0} X = N_{A_0} - N_A \)

For liquid-phase reactions \( V = V_0 \):

\( C_C = \frac{N_C}{V} = \frac{N_C}{V_0} = C_{A_0} - C_A = C_{A_0} \left( 1 - e^{-kt} \right) \) (E4-1.6)

Rearranging and taking the logarithm of both sides yields:

\[ \ln \frac{C_{A_0} - C_C}{C_{A_0}} = -kt \quad \text{(E4-1.7)} \]

We see that a plot of \(\ln[(C_{A_0} - C_C)/C_{A_0}]\) as a function of \(t\) will be a straight line with a slope \(-k\). Calculating the quantity \((C_{A_0} - C_C)/C_{A_0}\) (Table E4-1.3) and then plot:

\( k = \frac{\ln 10}{t_2 - t_1} = \frac{2.3}{8.95 - 1.55} = 0.311 \, \text{min}^{-1} \quad \text{(E4-1.8)} \)

The rate law becomes:

\( \boxed{-r_A = (0.311 \, \text{min}^{-1}) C_A} \quad \text{(E4-1.9)} \)

This rate law can now be used in the design of an industrial CSTR.

Back to Chapter 5