# Equilibrium and Free Energy

$$\ce{CO2(g) + Ca^2+(aq) + H2O(aq) <=> CaCO3(s) + 2H^+(aq)}$$

Find the equilibrium constant using a free energy table.

Finding the free energy and plugging it into the equation $$\mathrm{G = -RT \ln (\mathit{K})}$$ is easy, but I am confused about what $$K$$ it is. Is it $$K_c$$ or $$K_p$$, or does it even matter?

This is important because I am supposed to do equilibrium calculations with it later. I guess I have to use $$K_c$$, because only the concentrations of the aqueous solutions are given, plus $$\ce{CO2}$$ is in ppm. I think $$\mathrm{G = -RT \ln(\mathit{K})}$$ gives me $$K_p$$, so do I have to convert that into $$K_c$$ afterwards if I want to do equilibrium calculations?

• I think it should be $K_c$ as the K in the above equation you mentioned represents equilibrium constant for the reaction. Original equation has reaction quotient (Q) and at equillibrium it is just K. – Physicsapproval Jun 7 '17 at 8:32

K is properly defined as:

$$K = \dfrac{(a_C)^{c}(a_D)^{d}}{(a_A)^{a}(a_B)^{b}}$$

Activities involve relation of concentrations or parcial presures so K is dimensionless.

Let's find activities.

Ideal gases

$$a_{gas} = \dfrac{P}{P^o}$$ $P^º$ is 1 atm.

Solute in very dilute solutions

$$a_{solute} = \dfrac{C}{C^º}$$ $C^o$ is 1 M

Activities for solvents and pure solids are considered unity. $\ce{H2O}$ and $\ce{CaCO3}$ have unitary activity.

$$K = \dfrac{\ce{(a_{CaCO{_3}})(a_{H{^+}})^{2}}}{{\ce{(a_{H{_2}O})(a_{CO{_2}{_g}})(a_{Ca{^{2+}}{_{aq}}})}}}$$

$a_{CaCO{_3}}, a_{H{_2}O}$ are unity. It rests to use ideal gas equation and get $C_{CO2}$. $a_{Ca{^{2+}}{_{aq}}}$ and $a_{H{^+}}$ are usual concentrations.

EDIT

The way $K_p$ is converted to $K_c$ is:

$$K_p = K_c(RT)^{\Delta{n}_{gas}}$$

R: gas constant (0.082 L atm mol$^{-1}$ K$^{-1}$). T: temperature in Kelvin (K) at which the equilibrium reaction takes place.

To find $\Delta n_{ gas}$ , simply subtract the sum of the coefficients of the reactants from the sum of the coefficients of the products. You can deduce that using $P=\frac{nRT}{V}$, $\frac{n}{V}$ is concentration.