The value of K p in the reaction MgCO 3( s ) leftharpoons MgO ( s )+ CO 2( g ) is

Question

The value of K p in the reaction MgCO 3( s ) leftharpoons MgO ( s )+ CO 2( g ) is (A) K p = P ( CO 2) (B) K p =( P ( MgCO 3)/ P ( CO 2) × P ( MgO )) (C) K p =( P ( CO 2) × P ( CO 2) × P ( MgO )/ P ( MgCO )3) (D) K p =( P ( CO 2) × P ( MgO )/ P ( MgCO )3)

Tardigrade Admin · Accepted Answer

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 ( K p )= P ( CO 2)

Q. The value of $K_{p}$ in the reaction
$M g C O_{3} (s) ⇌ M g O (s) + C O_{2} (g)$ is

$K_{p} = P (C O_{2})$

$K_{p} = \frac{P ( M g C O _{3} )}{P ( C O _{2} ) \times P ( M g O )}$

$K_{p} = \frac{P ( C O _{2} ) \times P ( C O _{2} ) \times P ( M g O )}{P ( M g CO ) _{3}}$

$K_{p} = \frac{P ( C O _{2} ) \times P ( M g O )}{P ( M g CO ) _{3}}$

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 $(K_{p}) = P_{(C O_{2})}$

Q. The value of Kp​ in the reaction MgCO3​(s)⇌MgO(s)+CO2​(g) is

Kp​=P(CO2​)

Kp​=P(CO2​)×P(MgO)P(MgCO3​)​

Kp​=P(MgCO)3​P(CO2​)×P(CO2​)×P(MgO)​

Kp​=P(MgCO)3​P(CO2​)×P(MgO)​