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Chemical equilibrium

1. What is "chemical equilibrium"?

Take a reaction

\[ \nu_A A + \nu_B B \rightleftharpoons \nu_X X + \nu_Y Y \]

This just means:

  • \(\nu_A\) molecules of A react with \(\nu_B\) molecules of B
  • to make \(\nu_X\) molecules of X and \(\nu_Y\) molecules of Y.

In a closed container at fixed temperature and pressure, the system naturally moves to the state where its Gibbs free energy (G) is as small as possible.

  • As the reaction goes forward (A,B → X,Y), the numbers of each type of molecule change.
  • Because G depends on how many molecules of each type you have, G also changes as the reaction proceeds.
  • Equilibrium is reached when you can't lower G any further by letting more reaction happen one way or the other.

So: equilibrium = minimum Gibbs free energy for that T and P.

2. Chemical potentials and the equilibrium condition

The book defines the chemical potential \(\mu\) of a species as

\[ \mu_A = \left( \frac{\partial G}{\partial N_A} \right)_{T,P} \]

That's just "how much G changes if I add one more molecule of A (at fixed T and P)."

If you let the reaction go forward a tiny amount \(\Delta N\), the change in G is

\[ \Delta G = (-\nu_A \mu_A - \nu_B \mu_B + \nu_X \mu_X + \nu_Y \mu_Y)\Delta N \]

At equilibrium, changing the extent of reaction \(\Delta N\) shouldn't change G (we're at the bottom of the valley), so \(\Delta G\) must be zero for that change → the bracket must be zero:

\[ \nu_A \mu_A + \nu_B \mu_B = \nu_X \mu_X + \nu_Y \mu_Y \]

That's the basic equilibrium condition:

A weighted sum of the reactant chemical potentials equals the weighted sum of the product chemical potentials.

If you add a catalyst that appears on both sides of the reaction, its \(\mu\) cancels out of this equation — that's why catalysts don't change the equilibrium, they only change how fast you get there.

3. Connecting μ to concentrations: ideal gas / dilute solution

For an ideal gas (or a very dilute solution), the free energy is basically a sum over all species, and you can show that the chemical potential of species A is

\[ \mu_A = \varepsilon_A + kT \ln(n_A \lambda_A^3) - kT \ln j_A(T) \]

You don't need every symbol here; the important bit is:

μ_A depends on the logarithm of the number density (\(n_A\)) (how many A molecules per volume) plus terms that depend only on T.

So each \(\mu\) looks like

\[ \mu_A = \text{(stuff that only depends on T)} + kT \ln n_A \]

Now plug these \(\mu\)'s into the equilibrium condition.

All the "only temperature" parts combine into a single temperature-dependent constant, called \(\Delta \mu^{(0)}(T)\). The parts with ln of densities give:

\[ kT \ln \left( \frac{[X]^{\nu_X}[Y]^{\nu_Y}}{[A]^{\nu_A}[B]^{\nu_B}} \right) + \Delta\mu^{(0)}(T) = 0 \]

Rearrange:

\[ \frac{[X]^{\nu_X}[Y]^{\nu_Y}}{[A]^{\nu_A}[B]^{\nu_B}} = K(T) \]

where

\[ K(T) = \exp\big(-\beta \Delta\mu^{(0)}(T)\big) \]

and \([A]\) etc. are the (dimensionless) concentrations/densities of A, B, X, Y, scaled by some "standard" density (\(n_0\)).

This is the law of mass action:

At equilibrium, the product of concentrations of products (each raised to its stoichiometric power) divided by the corresponding product for reactants equals a constant (\(K(T)\)) that depends only on temperature.

4. What does K(T) mean?

  • \(K(T)\) tells you how much the reaction prefers products vs reactants at that temperature.
  • If \(K(T)\) is huge → the equilibrium heavily favors products.
  • If \(K(T)\) is tiny → equilibrium heavily favors reactants.
  • It comes from the difference in standard chemical potentials, i.e. the standard Gibbs free energy change of the reaction.

Temperature changes \(K(T)\) (sometimes dramatically).

5. Example: CO vs CO₂ in combustion

They then apply this to combustion inside an engine.

You can combine the reactions to get an effective reaction for carbon monoxide and carbon dioxide:

\[ 2\text{CO} + \text{O}_2 \rightleftharpoons 2\text{CO}_2 \]

Law of mass action gives (their equation (11)):

\[ \frac{[\text{CO}]}{[\text{CO}_2]} = \frac{1}{\sqrt{K(T)[\text{O}_2]}} \]

So:

  • Large \(K(T)\) → the right side is tiny → little CO compared to CO₂ at equilibrium.
  • Increasing \([\text{O}_2]\) (more oxygen) makes the right side smaller → less CO, more CO₂.

They note:

  • At high T (~1500 K), \(K \sim 10^{10}\), so equilibrium CO concentration is already very small (ppm).
  • But inside a real engine the reaction may not have time to reach equilibrium as gases rush out and cool — so you can still have more CO than the equilibrium value.
  • A catalytic converter with platinum/palladium speeds up the conversion CO + ½O₂ → CO₂ so the exhaust gets closer to the equilibrium state at the (lower) exhaust-pipe temperature, reducing CO.
  • Running the engine with slightly extra oxygen (a bit "lean") also lowers \([\text{CO}]/[\text{CO}_2]\) because of that \(1/\sqrt{[\text{O}_2]}\) factor.

6. Big picture

  1. Gibbs free energy wants to be minimal at fixed T, P.

  2. This leads to an equilibrium condition in terms of chemical potentials.

  3. For ideal gases/dilute solutions, chemical potentials depend on ln(concentration).

  4. That gives the law of mass action:

\[ \frac{\text{(products)}^{\nu}}{\text{(reactants)}^{\nu}} = K(T) \]

  1. \(K(T)\) depends only on T (through the reaction's standard free energy change).

  2. Catalysts do not change \(K(T)\); they just help the system reach the equilibrium composition faster.

  3. The combustion/CO example shows how this matters in real engines and why catalytic converters and extra oxygen reduce CO.