Question

For the equilibrium, $MgC{O}_3\left(s\right)\iff MgO\left(s\right)+C{O}_2\left(g\right)$, which of the following expression is correct?

• A. $K_p=\frac{P_{MgO}+P_{C{O}_2}}{P_{MgC{O}_3}}$
• B. $K_p=\frac{\left[MgO\right]+\left[C{O}_2\right]}{\left[MgC{O}_3\right]}$
• C. $K_p=\frac{P_{MgO}.P_{C{O}_2}}{P_{MgC{O}_3}}$
• D. $K_p=P_{C{O}_2}$

Answer 1

The correct option is D $K_p=P_{C{O}_2}$

The equilibrium reaction is $MgC{O}_3\left(s\right)\iff MgO\left(s\right)+C{O}_2\left(g\right)$.

The expression for the equilibrium constant is $K_p=\frac{P_{MgO}.P_{C{O}_2}}{P_{MgC{O}_3}}$.

However, the partial pressures of solids are considered as unity. Thus, $P_{MgO}=1$ and $P_{MgC{O}_3}=1$.

Hence, the expression for the equilibrium constant becomes $K_p=P_{C{O}_2}$.

Hence, option D is correct.