# How is equilibrium related to Gibbs' free energy?

Sep 12, 2015

Every reaction in a closed system is in some sort of equilibrium. Depending on the equilibrium constant $K$ or its variants (${K}_{a}$, ${K}_{b}$, ${K}_{\text{sp}}$, etc), the equilibrium is skewed to a certain extent towards either the products or reactants.

A common type of equilibrium is found in acid/base reactions. For acids and bases, there is a specific equation you can use to determine the extent of this equilibrium called the Henderson-Hasselbalch equation:

$p H = p K a + \log \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$
$p H = p K b + \log \left(\frac{\left[B {H}^{+}\right]}{\left[B\right]}\right)$

where $A$ means acid and $B$ means base, while ${A}^{-}$ means conjugate base and $B {H}^{+}$ means conjugate acid.

For example, consider the dissociation of acetic acid in water:

$C {H}_{3} C O O H + {H}_{2} O r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}^{-} + {H}_{3} {O}^{+}$

The $K a$ of $C {H}_{3} C O O H$ is about $1.8 \times {10}^{- 5}$. Let us suppose that creating a $\text{1 M}$ acetic acid solution in de-ionized water gives a pH of $4$.

The amount of acetic acid that dissociates is the amount of dissociated acetic acid that forms. Therefore:

$p H = p K a + \log \left(\frac{\left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]}\right)$

$4 = - \log \left(1.8 \times {10}^{- 5}\right) + \log \left(\frac{x}{1 - x}\right)$

$4 = 4.74 + \log \left(\frac{x}{1 - x}\right) \implies - 0.74 = \log \left(\frac{x}{1 - x}\right)$

${10}^{-} 0.74 = 0.182 = \frac{x}{1 - x}$

$0.182 - 0.182 x = x \implies 0.182 = 1.182 x$

Dissociated acetic acid:
$x = \text{0.154 M}$

Undissociated acetic acid:
$1 - x = \text{0.846 M}$

Since the ratio in the Henderson-Hasselbalch equation is:

$\frac{\left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]} = \frac{0.154}{0.846}$

The equilibrium is skewed $\text{5.49 to 1}$ (from $0.846 : 0.154$) in favor of the reactants side, the undissociated acetic acid. That is not extremely significant though. $\text{10 to 1}$ in favor of the products side is approximately where it becomes significant.

Even a reaction like this:

${H}_{2} S {O}_{4} \left(a q\right) r i g h t \le f t h a r p \infty n s H S {O}_{4}^{-} \left(a q\right) + {H}^{+} \left(a q\right)$

is simply extremely heavily skewed towards the products side.

The $p K a$ of sulfuric acid is about $- 3$, so the $K a$ is about $1000$. That is over 50 million times the $K a$ of acetic acid, so we can just say that for practical intents and purposes, it is to completion. In other words, you can assume that aqueous sulfuric acid is ionized.

Sep 12, 2015

Every reaction in a closed system is in some sort of equilibrium. Depending on the equilibrium constant $K$ or its variants (${K}_{a}$, ${K}_{b}$, ${K}_{\text{sp}}$, etc), the equilibrium is skewed to a certain extent towards either the products or reactants.

For a typical reaction with equilibrium constant $K$, you could also check the extent of the equilibrium using the Gibbs' free energy.

$\Delta G = \Delta {G}^{o} + R T \ln Q$

where $Q$ is the not-yet-equilibrium constant.

At equilibrium:

$0 = \Delta {G}^{o} + R T \ln K$

$\Delta {G}^{o} = - R T \ln K$

The value $\Delta {G}^{o}$ is the standard Gibbs' free energy, meaning at ${25}^{o} C$ and $\text{1 bar}$ of pressure, otherwise known as $\text{STP}$.

Let's say for some reaction, we had $\Delta {G}^{o} = - 50 \text{kJ/mol}$, and we were doing the reaction at $\text{300 K}$ ($\approx {27}^{o} C$).

$\Delta G = \Delta {G}^{o} + R T \ln Q$

$= - 50 + \left(8.314472 \cdot 300\right) \ln Q$

Now let's say we already calculated the quantity for $\Delta G$ to be $- 80 \text{kJ/mol}$ after doing an experiment using a calorimeter.

$- 30 = \left(8.314472 \cdot 300\right) \ln Q$

$\textcolor{b l u e}{Q} = {e}^{\text{-30/(8.314472"*"300)}} = \textcolor{b l u e}{0.988}$

For exact equilibrium, $Q = K = 1$, but here, $Q < K$, so we say that the reaction is skewed a little bit towards the reactants.