Answer:
The expression for the equilibrium constant \( K_p \) for the given reaction is:
\[ K_p = \frac{{P_{\text{CO}}^2}}{{P_{\text{CO}_2}}} \]
Given that the equilibrium mixture contained CO2 and CO at partial pressures of 6.6 atm and 3.2 atm respectively, we can plug these values into the expression for \( K_p \):
\[ K_p = \frac{{(3.2 \, \text{atm})^2}}{{6.6 \, \text{atm}}} \]
Solving this:
\[ K_p = \frac{{10.24 \, \text{atm}^2}}{{6.6 \, \text{atm}}} \approx 1.5545 \, \text{atm} \]
So, the value of \( K_p \) for the reaction at 500 K is approximately \( 1.5545 \, \text{atm} \).
Step-by-step explanation: