Question

A vessel at 1000 K contains \( \text{CO}_2 \) with a pressure of 0.5 atm. Some of \( \text{CO}_2 \) is converted into \( \text{CO} \) on addition of graphite. If total pressure at equilibrium is 0.8 atm, then \( K_p \) is:

• A. 0.18 atm
• B. 1.8 atm
• C. 0.3 atm
• D. 3 atm

Hint:

To calculate \( K_p \) for a reaction, use the partial pressures of the products and reactants at equilibrium.

Answer 1

The Correct Option is B

The problem involves the equilibrium reaction \( \text{CO}_2 + C \rightleftharpoons 2\text{CO} \). Initially, the pressure of \( \text{CO}_2 \) is 0.5 atm. At equilibrium, the total pressure is 0.8 atm. Let's denote the change in pressure of \( \text{CO}_2 \) that reacts as \( x \), so \( x \) moles of \( \text{CO}_2 \) are consumed and \( 2x \) moles of \( \text{CO} \) are produced.

Initial moles:

Substance	\( \text{CO}_2 \)	\( C \)	\( \text{CO} \)
Initial Pressure (atm)	0.5	-	0
Change in Pressure (atm)	-x	-	+2x
Equilibrium Pressure (atm)	0.5 - x	-	2x

The total pressure at equilibrium is given by:

\((0.5-x) + 2x = 0.8\)

Solving for \( x \):

\[0.5 - x + 2x = 0.8\]

\[x = 0.3\]

Therefore, the equilibrium pressure of \( \text{CO}_2 \) is \( 0.5 - 0.3 = 0.2 \) atm, and the equilibrium pressure of \( \text{CO} \) is \( 2 \times 0.3 = 0.6 \) atm.

The equilibrium constant \( K_p \) is defined by:

\( K_p = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} \)

Plug in the values:

\[K_p = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8\, \text{atm}\]

Thus, \( K_p \) is 1.8 atm.