Chapter 5: Isothermal Reactor Design: Conversion
Gas Phase Reactions
\( 2\text{NOCl} \longrightarrow 2\text{NO} + \text{Cl}_2 \)
\( 2\text{A} \longrightarrow 2\text{B} + \text{C} \)
\( \text{A} \longrightarrow \text{B} + \frac{1}{2}\text{C} \)
1. Mole Balance/Design Equation
\(\frac{dX}{dV} = \frac{-r_A}{F_{A0}}\)
2. Rate Law
\(-r_A = kC_A^2, \quad k = 0.29 \, \frac{\mathrm{dm}^3}{\mathrm{mol} \cdot \mathrm{s}}\)
3. Stoichiometry
Gas \( C_A = \frac{F_A}{\nu} = \frac{F_A}{\nu_0 (1 + \varepsilon X)} \frac{P}{P_0} \frac{T_0}{T} = \left( \frac{F_{A0}}{\nu_0} \right) \frac{(1 - X)}{(1 + \varepsilon X)} \frac{P}{P_0} \frac{T_0}{T} \)
\(C_A = \frac{C_{A0}(1 - X)}{(1 + \varepsilon X)}\)
\(\varepsilon = y_{A0}\delta = 1\left(1 + \frac{1}{2} - 1\right) = \frac{1}{2}\)
\(T = T_0, \quad P = P_0\)
4. Combine
\(-r_A = kC_{A0}^2 \frac{(1 - X)^2}{(1 + \varepsilon X)^2}\)
\(\frac{dX}{dV} = \frac{kC_{A0}^2 (1 - X)^2}{\nu_0 C_{A0} (1 + \varepsilon X)^2}\)
\(\int_0^X \frac{(1 + \varepsilon X)^2}{(1 - X)^2} dX = \frac{kC_{A0}}{\nu_0} V = kC_{A0} \tau\)
\(\tau kC_{A0} = \int_0^X \frac{(1 + \varepsilon X)^2}{(1 - X)^2} dX = 2 \varepsilon (1 + \varepsilon) \ln(1 - X) + \varepsilon^2 X + \frac{(1 + \varepsilon)^2 X}{1 - X}\)
5. Evaluate
\(\tau kC_{A0} = 2 \left( \frac{1}{2} \right) \left( 1 + \frac{1}{2} \right) \ln(1 - 0.9) + \frac{0.9}{4} + \frac{(1.5)^2 (0.9)}{0.1} = 17.075\)
\(\tau = \frac{17.075}{(0.29)(0.2)} = 294.4 \, \text{s}\)
If \(\nu_0 = 10 \, \text{m}^3/\text{s}\), \( V = 2944 \, \text{dm}^3\)
(b), (c), (d) No answer will be given.