Chapter 10: Catalysis and Catalytic Reactors
Calculating Fractional Coverage
If KCO=1 atm-1 amd KNO=5 atm-1, what is the ratio of sites containing CO to those containing NO when the conversion is 20%? 99%? What fraction of sites are occupied by NO? by CO? The feed is equal molar in CO and NO with PO=2 atm.
If KCO=1 atm-1 amd KNO=5 atm-1, what is the ratio of sites containing CO to those containing NO when the conversion is 20%? 99%? What fraction of sites are occupied by NO? by CO? The feed is equal molar in CO and NO with PO=2 atm.
Solution
\( C_{\text{CO}\cdot s} = K_{\text{CO}} P_{\text{CO}} = K_{\text{CO}} P_{\text{CO},0}(1 - X) \)
\( C_{\text{NO}\cdot s} = K_{\text{NO}} P_{\text{CO},0}(\theta_{\text{NO}} - X) = K_{\text{NO}} P_{\text{CO},0}(1 - X) \)
\(\text{CO to NO sites} = \frac{C_{\text{CO}\cdot s}}{C_{\text{NO}\cdot s}} = \frac{K_{\text{CO}}}{K_{\text{NO}}} = \frac{1}{5} = 0.2 \)
Fraction of CO sites.
\( \frac{C_{\text{CO}\cdot s}}{C_T} = \frac{K_{\text{CO}} P_{\text{CO},0} (1 - X)} {1 + K_{\text{CO}} P_{\text{CO},0} (1 - X) + K_{\text{NO}} P_{\text{CO},0} (1 - X)} \)
For X=.2
\( \frac{C_{\text{CO}\cdot s}}{C_T} = \frac{(1\, \text{atm}^{-1})(2\, \text{atm})(1 - 0.2)}{1 + (1\, \text{atm}^{-1})(2\, \text{atm})(1 - 0.2) + (5\, \text{atm}^{-1})(2\, \text{atm})(1 - 0.2)} = \frac{1.6}{1 + 1.6 + 8} = 0.15 \)
\( \frac{C_v}{C_T} = \frac{1}{1 + (1\, \text{atm}^{-1})(2\, \text{atm})(1 - 0.2) + (5\, \text{atm}^{-1})(2\, \text{atm})(1 - 0.2)} = \frac{1}{1 + 1.6 + 8} = 0.094 \)
at 20% conversion 15% of the sites are occupied by CO and 9.4% are vacant.
For X=.99
\( \frac{C_{\text{CO}\cdot s}}{C_T} = \frac{(1\, \text{atm}^{-1})(2\, \text{atm})(1 - 0.99)}{1 + (1\, \text{atm}^{-1})(2\, \text{atm})(1 - 0.99) + (5\, \text{atm}^{-1})(2\, \text{atm})(1 - 0.99)} = \frac{0.02}{1 + 0.02 + 0.1} = 0.18 \)
\( \frac{C_v}{C_T} = \frac{1}{1 + 0.02 + 0.1} = 0.089 \)
at 99% conversion 1.8% of the sites are occupied by CO and 89% are vacant.