1. Write the equilibrium reaction

On adding graphite, carbon reacts with carbon dioxide:

$CO_2(g) + C(s) \rightleftharpoons 2CO(g)$

Since graphite is a solid, it does not appear in the equilibrium constant expression.

2. Initial pressure and change

Initially, only $CO_2$ is present:

$P_{CO_2}^{\text{initial}} = 0.5\,\text{atm}, \qquad P_{CO}^{\text{initial}} = 0$

Let $x$ atm of $CO_2$ react.

Then at equilibrium:

• $CO_2$ decreases by $x$
• $CO$ increases by $2x$

So,

$P_{CO_2}=0.5-x$ $P_{CO}=2x$

3. Use total pressure

Given total equilibrium pressure is $0.8$ atm:

$P_{\text{total}} = P_{CO_2}+P_{CO}$

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

$0.8=0.5+x$

$x=0.3$

Thus,

$P_{CO_2}=0.5-0.3=0.2\,\text{atm}$ $P_{CO}=2(0.3)=0.6\,\text{atm}$

4. Write equilibrium constant in terms of pressure

For the reaction

$CO_2(g)+C(s)\rightleftharpoons 2CO(g)$

$K_p=\frac{(P_{CO})^2}{P_{CO_2}}$

Substitute the equilibrium pressures:

$K_p=\frac{(0.6)^2}{0.2}$

$K_p=\frac{0.36}{0.2}=1.8\,\text{atm}$

5. Match with options

$K = 1.8\,\text{atm}$

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.