For the reaction MgCO_{3(s)} rightleftharpoon | vedclass

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(A) For a heterogeneous equilibrium reaction,the concentration or partial pressure of pure solids and pure liquids is taken as unity $(1)$.Given the r

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$K_p = P_{CO_2}$

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(A) For a heterogeneous equilibrium reaction,the concentration or partial pressure of pure solids and pure liquids is taken as unity $(1)$.Given the reaction: $MgCO_{3(s)} ightleftharpoons MgO_{(s)} + CO_{2(g)}$.The expression for the equilibrium constant $K_p$ is defined as the ratio of the product of partial pressures of gaseous products to the product of partial pressures of gaseous reactants,each raised to the power of their stoichiometric coefficients.$K_p = \frac{P_{MgO} 	imes P_{CO_2}}{P_{MgCO_3}}$.Since $MgCO_{3(s)}$ and $MgO_{(s)}$ are pure solids,their partial pressures are considered as $1$.Therefore,$K_p = 1 	imes P_{CO_2} / 1 = P_{CO_2}$.

For the reaction $MgCO_{3(s)} \rightleftharpoons MgO_{(s)} + CO_{2(g)}$,the value of $K_p$ is:

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