A vessel at 1000 K contains $$CO_2$$ with a pressure of 0.5 atm . Some of $$CO_2$$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm , then Kp is :

We need to find $$K_p$$ for the reaction $$CO_2(g) + C(s) \rightleftharpoons 2CO(g)$$ at 1000 K.

Initially, the partial pressure of $$CO_2$$ is 0.5 atm, while the partial pressure of $$CO$$ is zero. If x atm of $$CO_2$$ decomposes, its pressure decreases by x and the pressure of $$CO$$ increases by 2x, giving equilibrium pressures of $$0.5 - x$$ for $$CO_2$$ and $$2x$$ for $$CO$$.

Since the total pressure at equilibrium is 0.8 atm, we set up the equation $$(0.5 - x) + 2x = 0.5 + x = 0.8$$, from which it follows that $$x = 0.3$$ atm.

Substituting this value of x back into the expressions for the equilibrium pressures yields $$P_{CO_2} = 0.5 - 0.3 = 0.2$$ atm and $$P_{CO} = 2(0.3) = 0.6$$ atm.

The equilibrium constant is then given by $$K_p = \frac{P_{CO}^2}{P_{CO_2}} = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8$$ atm.

The correct answer is Option 1: 1.8 atm.