A) \[{{K}_{p}}=\frac{{{P}_{Mgo}}K\times {{P}_{C{{O}_{2}}}}}{{{P}_{MgC{{O}_{3}}}}}\]
B) \[{{K}_{p}}=\frac{[MgO][C{{O}_{2}}]}{[MgC{{O}_{3}}]}\]
C) \[{{K}_{p}}=\frac{{{P}_{Mg}}_{O}\times {{P}_{C{{O}_{2}}}}}{{{P}_{Mgc{{o}_{3}}}}}\]
D) \[{{K}_{p}}={{P}_{C{{O}_{2}}}}\]
Correct Answer: D
Solution :
[d] \[K{{ & }_{p}}={{P}_{C{{O}_{2}}}}\] K = equilibrium constant: Product of the molar concentration of product raised to power equal to its stoichiometric coefficient divided by product of molar concentration of reaction each raised to the power equal to its stoichiometric coefficient is constant at constant temperature.
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