A vessel at 1000 K contains CO_2 with a pres | vedclass

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(A) The chemical equation for the reaction is: $CO_2(g) + C(s) \rightleftharpoons 2CO(g)$Initially,the pressure of $CO_2$ is $0.5 \ atm$ and $CO$ is $

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$1.8$

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(A) The chemical equation for the reaction is: $CO_2(g) + C(s) ightleftharpoons 2CO(g)$Initially,the pressure of $CO_2$ is $0.5 \ atm$ and $CO$ is $0 \ atm$.Let $x$ be the decrease in pressure of $CO_2$ at equilibrium.At equilibrium,the partial pressures are: $P_{CO_2} = (0.5 - x) \ atm$ and $P_{CO} = 2x \ atm$.The total pressure at equilibrium is given as $0.8 \ atm$.$P_{	ext{total}} = P_{CO_2} + P_{CO} = (0.5 - x) + 2x = 0.5 + x$$0.5 + x = 0.8 \implies x = 0.3 \ atm$.Now,calculate the equilibrium constant $K_p$:$K_p = \frac{(P_{CO})^2}{P_{CO_2}} = \frac{(2x)^2}{(0.5 - x)}$$K_p = \frac{(2 	imes 0.3)^2}{(0.5 - 0.3)} = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \ atm$.

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

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