How is equilibrium related to Gibbs' free energy?

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How is equilibrium related to Gibbs' free energy?

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Answer 1 · 2015-09-12T18:46:02

Every reaction in a closed system is in some sort of equilibrium. Depending on the equilibrium constant $K$ $K$ or its variants ( $K_{a}$ $K_a$ , $K_{b}$ $K_b$ , $K_{sp}$ $K_"sp"$ , 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:

$p H = p K a + log (\frac{[A^{-}]}{[H A]})$ $pH = pKa + log (([A^(-)])/([HA]))$
$p H = p K b + log (\frac{[B H^{+}]}{[B]})$ $pH = pKb + log (([BH^(+)])/([B]))$

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

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

$C H_{3} C O O H + H_{2} O ⇌ C H_{3} C O O^{-} + H_{3} O^{+}$ $CH_3COOH + H_2O rightleftharpoons CH_3COO^(-) + H_3O^(+)$

The $K a$ $Ka$ of $C H_{3} C O O H$ $CH_3COOH$ is about $1.8 \times 10^{- 5}$ $1.8xx10^(-5)$ . Let us suppose that creating a $1 M$ $"1 M"$ acetic acid solution in de-ionized water gives a pH of $4$ $4$ .

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

$p H = p K a + log (\frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O H]})$ $pH = pKa + log (([CH_3COO^(-)])/([CH_3COOH]))$

$4 = - log (1.8 \times 10^{- 5}) + log (\frac{x}{1 - x})$ $4 = -log(1.8xx10^(-5)) + log (x/(1 - x))$

$4 = 4.74 + log (\frac{x}{1 - x}) \Rightarrow - 0.74 = log (\frac{x}{1 - x})$ $4 = 4.74 + log (x/(1-x)) => -0.74 = log (x/(1-x))$

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

$0.182 - 0.182 x = x \Rightarrow 0.182 = 1.182 x$ $0.182 - 0.182x = x => 0.182 = 1.182x$

Dissociated acetic acid:
$x = 0.154 M$ $x = "0.154 M"$

Undissociated acetic acid:
$1 - x = 0.846 M$ $1-x = "0.846 M"$

Since the ratio in the Henderson-Hasselbalch equation is:

$\frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O H]} = \frac{0.154}{0.846}$ $[[CH_3COO^(-)]]/[[CH_3COOH]] = 0.154/0.846$

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

Even a reaction like this:

$H_{2} S O_{4} (a q) ⇌ H S O_{4}^{-} (a q) + H^{+} (a q)$ $H_2SO_4(aq) rightleftharpoons HSO_4^(-)(aq) + H^(+)(aq)$

is simply extremely heavily skewed towards the products side.

The $p K a$ $pKa$ of sulfuric acid is about $- 3$ $-3$ , so the $K a$ $Ka$ is about $1000$ $1000$ . That is over 50 million times the $K a$ $Ka$ 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.

Answer 2 · 2015-09-12T19:09:26

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

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

$DeltaG = DeltaG^o + RTlnQ$

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

At equilibrium:

$0 = DeltaG^o + RTlnK$

$DeltaG^o = -RTlnK$

The value $DeltaG^o$ is the standard Gibbs' free energy, meaning at $25^oC$ and $"1 bar"$ of pressure, otherwise known as $"STP"$ .

Let's say for some reaction, we had $DeltaG^o = -50 "kJ/mol"$ , and we were doing the reaction at $"300 K"$ ( $~~27^oC$ ).

$DeltaG = DeltaG^o + RTlnQ$

$= -50 + (8.314472*300)lnQ$

Now let's say we already calculated the quantity for $DeltaG$ to be $-80 "kJ/mol"$ after doing an experiment using a calorimeter.

$-30 = (8.314472*300)lnQ$

$color(blue)(Q) = e^("-30/(8.314472"*"300)") = color(blue)(0.988)$

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

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How is equilibrium related to Gibbs' free energy?

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