Question

The equilibrium expression, $K=\left[C{O}_2\right]$ represents the reaction :

• A. $C\left(s\right)+O_2\left(g\right)\rightleftharpoons C{O}_2\left(g\right)$
• B. $CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons C{O}_2\left(g\right)$
• C. $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$
• D. $C{O}_2\left(g\right)\rightleftharpoons C\left(s\right)+O_2\left(g\right)$
• E. $CaO\left(s\right)+C{O}_2\left(g\right)\rightleftharpoons CaC{O}_3\left(s\right)$

Answer 1

The correct option is C $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$

R.E.F image

In your case, the decomposition of calcium carbonate, $CaC{O}_3$,a solid will produce

calcium oxide, $CaO,$ also a solid, and carbon dioxide, $C{O}_2$, a gas.

As you know, solids and pure liquids are not included in the expression of the

equilibrium constants $K_c$ and $K_p$ because their concentrations are considered

constant.

In this example, you would have

$k_c=\left[C{O}_2\right]$ and $k_p=\left(C{O}_2\right)$

So just by looking at the expressions for the equilibrium constants that the

concentration of the two solids has no impact on the position of the equilibrium.

Moreover, what's interesting here is that you an only change the concentration of

carbon dioxide by changing the temperature at which the reaction takes place.