For the reaction C_{(s)} + CO_{2_{(g)}} right | vedclass

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(A) For the reaction,$C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the equilibrium constant $K_p$ is defined as the ratio of the square of the

Vedclass · Accepted Answer

$16$

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(A) For the reaction,$C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the equilibrium constant $K_p$ is defined as the ratio of the square of the partial pressure of the product to the partial pressure of the gaseous reactant.$K_p = \frac{(P_{CO})^2}{P_{CO_2}}$Given,$P_{CO} = 8 \ atm$ and $P_{CO_2} = 4 \ atm$.Substituting these values into the expression:$K_p = \frac{(8)^2}{4} = \frac{64}{4} = 16$Thus,the $K_p$ of the reaction is $16$.Therefore,the correct option is $A$.

For the reaction $C_{(s)} + CO_{2_{(g)}} \rightleftharpoons 2CO_{(g)}$,the partial pressure of $CO_2$ and $CO$ at equilibrium are $4 \ atm$ and $8 \ atm$ respectively. $K_p$ for the reaction is

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