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

The equilibrium that takes place after adding graphite is
$$\text{CO}_2(g) + C(s) \rightleftharpoons 2\text{CO}(g)$$

C(s) is a solid, so it will not appear in the equilibrium-constant expression.

Let $$x$$ be the fall in $$\text{CO}_2$$ pressure (in atm) when equilibrium is reached.
Initial partial pressures (atm):
  $$P_{\text{CO}_2}^{\text{initial}} = 0.5$$, $$P_{\text{CO}}^{\text{initial}} = 0$$

Change on reaching equilibrium:
  $$P_{\text{CO}_2}: 0.5 - x$$
  $$P_{\text{CO}}: 0 + 2x$$ (because 1 mol $$\text{CO}_2$$ produces 2 mol $$\text{CO}$$)

Total pressure at equilibrium is given to be $$0.8 \,\text{atm}$$:
$$\bigl(0.5 - x\bigr) + 2x = 0.8$$
$$0.5 + x = 0.8 \;\;\Rightarrow\;\; x = 0.3$$

Equilibrium partial pressures:
$$P_{\text{CO}_2} = 0.5 - 0.3 = 0.2 \,\text{atm}$$
$$P_{\text{CO}} = 2x = 0.6 \,\text{atm}$$

The equilibrium constant in terms of pressure for the reaction
$$\text{CO}_2(g) + C(s) \rightleftharpoons 2\text{CO}(g)$$ is
$$K_p = \frac{(P_{\text{CO}})^2}{P_{\text{CO}_2}}$$

Substituting the calculated values:
$$K_p = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \,\text{atm}$$

Therefore the equilibrium constant is $$1.8 \,\text{atm}$$.

Option D which is: $$1.8 \,\text{atm}$$