Question

How is equilibrium related to Gibbs' free energy?

Answer 1

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_"sp"`, etc), the equilibrium is skewed to a certain extent towards either the products or reactants.

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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:

`pH = pKa + log (([A^(-)])/([HA]))`
`pH = pKb + log (([BH^(+)])/([B]))`

where `A` means acid and `B` means base, while `A^(-)` means conjugate base and `BH^(+)` means conjugate acid.

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

`CH_3COOH + H_2O rightleftharpoons CH_3COO^(-) + H_3O^(+)`

The `Ka` of `CH_3COOH` is about `1.8xx10^(-5)`. Let us suppose that creating a `"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:

`pH = pKa + log (([CH_3COO^(-)])/([CH_3COOH]))`

`4 = -log(1.8xx10^(-5)) + log (x/(1 - x))`

`4 = 4.74 + log (x/(1-x)) => -0.74 = log (x/(1-x))`

`10^-0.74 = 0.182 = x/(1-x)`

`0.182 - 0.182x = x => 0.182 = 1.182x`

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

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

Since the ratio in the Henderson-Hasselbalch equation is:

`[[CH_3COO^(-)]]/[[CH_3COOH]] = 0.154/0.846`

The equilibrium is skewed `"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. `"10 to 1"` in favor of the products side is approximately where it becomes significant.

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Even a reaction like this:

`H_2SO_4(aq) rightleftharpoons HSO_4^(-)(aq) + H^(+)(aq)`

is simply extremely heavily skewed towards the products side.

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

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_"sp"`, etc), the equilibrium is skewed to a certain extent towards either the products or reactants.

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For a typical reaction with equilibrium constant `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.