Question

The value of $K _{ p }$ in the reaction
$MgCO _{3}( s ) \rightleftharpoons MgO ( s )+ CO _{2}( g )$ is

• A. $K _{ p }= P \left( CO _{2}\right)$
• B. $K _{ p }=\frac{ P \left( MgCO _{3}\right)}{ P \left( CO _{2}\right) \times P ( MgO )}$
• C. $K _{ p }=\frac{ P \left( CO _{2}\right) \times P \left( CO _{2}\right) \times P ( MgO )}{ P ( MgCO )_{3}}$
• D. $K _{ p }=\frac{ P \left( CO _{2}\right) \times P ( MgO )}{ P ( MgCO )_{3}}$

Answer 1

Answer: (A)

As active mass of solid substance is always unity, they cannot be included in the equilibrium constant expression. Thus equilibrium constant in terms of partial pressure $\left( K _{ p }\right)= P _{\left( CO _{2}\right)}$