The equilibrium constant the reaction CaCO-3-s-leftrightharpoons CaO-s-CO-2-left - g right - is-displaystyle K-c- displaystylefrac-left - CaO right -left - CO-2- right -left - CaCO-3- right -displaystyle K-c- displaystylefrac-left - CaCO-3- right -left - CaO right -left - CO-2- right -displaystyle K-c-left - CO-2- right -displaystyle K-c-displaystylefrac-1-left - CO-2- right -

Question

Toppr Admin · Accepted Answer

The equilibrium reaction is as follows-CaCO3-s-x21D4-CaO-s-CO2-g-The expression for the equilibrium constant is Kc-CaO-CO2-CaCO3-The concentration of solids is considered as unity-Substituting -CaCO3-1 and -CaO-1 in the expression for the equilibrium constant- we get-Kc-CO2-Hence- option C is the correct answer-

The equilibrium constant for the reaction CaCO3(s)⇋CaO(s)+CO2(g) is:Kc=[CaO][CO2][CaCO3]Kc=[CaCO3][CaO][CO2]Kc=[CO2]Kc=1[CO2]

The equilibrium constant for the reaction $C a C O_{3} (s) ⇋ C a O (s) + C O_{2} (g)$ is:
$K_{c} = \frac{[C a O] [C O_{2}]}{[C a C O_{3}]}$
$K_{c} = \frac{[C a C O_{3}]}{[C a O] [C O_{2}]}$
$K_{c} = [C O_{2}]$
$K_{c} = \frac{1}{[C O_{2}]}$