Question

The expression of $K_C$ for the following reaction is :
$CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$

• A. $K_c=\frac{\left[CaO\right]\times \left[C{O}_2\right]}{\left[CaC{O}_3\right]}$
• B. $K_c=\frac{[C{O}_2{]}^2}{\left[CaC{O}_3\right]}$
• C. $K_c=\left[C{O}_2\right]$
• D. None of the above

Answer 1

The correct option is C $K_c=\left[C{O}_2\right]$

This is an example of hetrogenous chemical reaction.
So, the active masses of pure solids and liquids are considered as 1 and they do not appear in the $K_c$ expression.

$CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)K_c=\frac{\left[C{O}_2\right]\times \left[1\right]}{\left[1\right]}=\left[C{O}_2\right]$




Theory:

Equilibrium constant :
Equilibrium constant is the ratio of rate of the forward reaction and rate of the backward reaction.

The equilibrium constant for a general reaction is written as (using law of mass action):

$aA+bB\rightleftharpoons cC+dD$

$K_{eq}=\frac{K_f}{K_b}=\frac{[C{]}_{eq}^c\times [D{]}_{eq}^d}{[A{]}_{eq}^a\times [B{]}_{eq}^b}$

where $[C{]}_{eq}^c,[D{]}_{eq}^d,\left[A{]}_{eq}^{a}and\right[B{]}_{eq}^b$ are the equilibrium concentrations of the reactants and products.

Types of equilibrium constant :
$K_p$ ( It is the equilibrium constant calculated from the partial pressures of a reaction equation.)
$K_c$ (A constant in terms of concentration, where all the concentrations are at equilibrium and are expressed in $mol{L}^{-1})$

For ex :

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

$K_c=\frac{\left[PC{l}_3{]}_{eq}\right[C{l}_2{]}_{eq}}{[PC{l}_5{]}_{eq}}$