Question

For the reaction at 500k , C+CO2=2CO the equilibrium mixture contained CO2 and CO at partial pressures of 6.6 and 3.2atm. Find the value of Kp

Answer 1

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: