Question

\( \text{CO}_2 \) is taken in a closed container initially at a pressure of 0.6 atm at 1500 K. After the addition of solid C, some of \( \text{CO}_2 \) is converted to CO. The equilibrium pressure is then 0.9 atm. Calculate the value of \( K_p \) at 1500 K.

Hint:

Always remember to exclude pure solids and liquids from equilibrium constant expressions ($K_c$ and $K_p$) because their concentrations/activities remain constant.

Answer 1

Step 1: Write the Balanced Reaction
\[ \text{CO}_{2(g)} + \text{C}_{(s)} \rightleftharpoons 2\text{CO}_{(g)} \] (Note: Solid carbon does not appear in \( K_p \) expression.)

Step 2: Construct ICE Table (in atm)
\[ \begin{array}{c|c|c} & \text{CO}_2 & \text{CO} \\ \hline \text{Initial} & 0.6 & 0 \\ \text{Change} & -x & +2x \\ \text{Equilibrium} & 0.6 - x & 2x \\ \end{array} \]

Step 3: Use Total Pressure at Equilibrium
\[ P_{\text{total}} = P_{\text{CO}_2} + P_{\text{CO}} \] \[ 0.9 = (0.6 - x) + 2x \] \[ 0.9 = 0.6 + x \Rightarrow x = 0.3 \]

Step 4: Calculate Equilibrium Partial Pressures
\[ P_{\text{CO}_2} = 0.6 - 0.3 = 0.3 \text{ atm} \] \[ P_{\text{CO}} = 2x = 2(0.3) = 0.6 \text{ atm} \]

Step 5: Calculate \( K_p \)
\[ K_p = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} \] \[ K_p = \frac{0.36}{0.3} = 1.2 \text{ atm} \]

Step 6: Final Answer
\[ \boxed{K_p = 1.2 \text{ atm}} \]