What are these formations called when they point down?
Formation of stalactites (pointing down) and stalagmites (pointing up) is a complex process. Solutions of minerals drip down and absorb carbon dioxide as water flows through the cave. Calcium carbonate dissolves in this liquid and redeposits on the rock as the carbon dioxide is dissipated into the environment.
Equilibrium Constant and
ΔG
At equilibrium the
The variable
When



Description 
>1  positive  negative  Products are favored at equilibrium. 
1  0  0  Reactants and products are equally favored. 
<1  negative  positive  Reactants are favored at equilibrium. 
Knowledge of either the standard free energy change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.
Sample Problem: Gibbs Free Energy and the Equilibrium Constant
The formation of nitrogen monoxide from nitrogen and oxygen gases is a reaction that strongly favors the reactants at 25°C.
\begin{align*}\text{N}_2(g)+\text{O}_2(g) \rightleftarrows 2\text{NO}(g)\end{align*}
The actual concentrations of each gas would be difficult to measure, and so the _{ @$\begin{align*}K_{eq}\end{align*}@$ } for the reaction can more easily calculated from the @$\begin{align*}\Delta G^\circ\end{align*}@$ , which is equal to 173.4 kJ/mol.
Step 1: List the known values and plan the problem.
Known
 @$\begin{align*}\Delta G^\circ=+173.4 \ \text{kJ} / \text{mol}\end{align*}@$
 @$\begin{align*}R=8.314 \ \text{J} / \text{K} \cdot \text{mol}\end{align*}@$
 @$\begin{align*}T=25^\circ \text{C}=298 \ \text{K}\end{align*}@$
Unknown
 @$\begin{align*}K_{eq}=?\end{align*}@$
In order to make the units agree, the value of @$\begin{align*}\Delta G^\circ\end{align*}@$ will need to be converted to J/mol (173,400 J/mol). To solve for @$\begin{align*}K_{eq}\end{align*}@$ , the inverse of the natural logarithm, @$\begin{align*}e^x\end{align*}@$ , will be used.
Step 2: Solve .
@$$\begin{align*}\Delta G^\circ &=RT \ln K_{eq} \\ \ln K_{eq}&=\frac{\Delta G^\circ}{RT} \\ K_{eq} &=e^{\frac{\Delta G^\circ}{RT}}=e^{\frac{173, 400 \ \text{J} / \text{mol}}{8.314 \ \text{J} / \text{K} \cdot \text{mol}(298 \ \text{K})}}=4.0 \times 10^{31} \end{align*}@$$
Step 3: Think about your result.
The large positive free energy change leads to a _{ @$\begin{align*}K_{eq}\end{align*}@$ } value that is extremely small. Both lead to the conclusion that the reactants are highly favored and very few product molecules are present at equilibrium.
Sample Problem: Free Energy from @$\begin{align*}K_{sp}\end{align*}@$
The solubility product constant @$\begin{align*}(K_{sp}) \end{align*}@$ of lead(II) iodide is 1.4 × 10 ^{ 8 } at 25°C. Calculate @$\begin{align*}\Delta G^\circ\end{align*}@$ for the dissociation of lead(II) iodide in water.
@$$\begin{align*}\text{PbI}_2(s) \rightleftarrows \text{Pb}^{2+}(aq) + 2\text{I}^(aq)\end{align*}@$$
Step 1: List the known values and plan the problem .
Known
 @$\begin{align*}K_{eq}=K_{sp}=1.4 \times 10^{8}\end{align*}@$
 @$\begin{align*}R=8.314 \ \text{J} / \text{K} \cdot \text{mol}\end{align*}@$
 @$\begin{align*}T=25^\circ \text{C}=298 \ \text{K}\end{align*}@$
Unknown
 @$\begin{align*}\Delta G^\circ=? \ \text{kJ} / \text{mol}\end{align*}@$
The equation relating @$\begin{align*}\Delta G^\circ\end{align*}@$ to _{ @$\begin{align*}K_{eq}\end{align*}@$ } can be solved directly.
Step 2: Solve.
@$$\begin{align*}\Delta G^\circ&=RT \ln K_{eq} \\ &=8.314 \ \text{J} / \text{K} \cdot \text{mol}(298 \ \text{K}) \ln (1.4 \times 10^{8}) \\ &=45,000 \ \text{J} / \text{mol} \\ &=45 \ \text{kJ} / \text{mol}\end{align*}@$$
Step 3: Think about your result.
The large, positive @$\begin{align*}\Delta G^\circ\end{align*}@$ indicates that the solid lead(II) iodide is nearly insoluble and so very little of the solid is dissociated at equilibrium.
Summary
 The relationship between @$\begin{align*}\Delta G \end{align*}@$ and _{ @$\begin{align*}K_{eq}\end{align*}@$ } is described.
 Calculations involving these two parameters are shown.
Practice
Questions
Read the material at the link below and answer the following questions:
http://www.chem1.com/acad/webtext/thermeq/TE5.html
 What is the difference between @$\begin{align*}\Delta G \end{align*}@$ and @$\begin{align*}\Delta G^\circ\end{align*}@$ ?
 At equilibrium, why does the equation between free energy and equilibrium constant reduce to @$\begin{align*}\Delta G^\circ = RT \ln K_{eq}\end{align*}@$ ?
 What other equilibrium units could we use?
Review
Questions
 When _{ @$\begin{align*}K_{eq}\end{align*}@$ } is large, what will be the sign of @$\begin{align*}\Delta G \end{align*}@$ ?
 When _{ @$\begin{align*}K_{eq}\end{align*}@$ } is small, are reactants or products favored?
 What does @$\begin{align*}R\end{align*}@$ stand for?