Chemical Equilibrium

By the end of this lesson, you will be able to say what dynamic equilibrium actually means, write and interpret an equilibrium constant, and predict which way a system shifts when it is disturbed.

Both Directions at Once

Some reactions do not finish. They reach a point where reactants and products coexist and the amounts stop changing.

$$\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3}$$

The double arrow means both directions are happening.

Dynamic equilibrium is not stopping. Both reactions continue at full speed; they simply proceed at equal rates, so ammonia is being made exactly as fast as it is being destroyed and the concentrations hold steady.

$$\text{rate}{\text{forward}} = \text{rate}{\text{reverse}}$$

A picture that fits: two escalators side by side, one up and one down, with people stepping on at the same rate. The number of people on each floor stops changing while nobody is standing still.

Equal rates, not equal amounts. At equilibrium there may be far more product than reactant, or far more reactant than product. What is equal is the speed of the two conversions, and confusing the two is the central misunderstanding of this topic.

Equilibrium requires a closed system. If a product escapes, the reverse reaction has nothing to work with and the reaction runs to completion instead. That is why an open beaker of decomposing limestone never reaches equilibrium: the $\mathrm{CO_2}$ leaves.

Sort It: True at Equilibrium

The Equilibrium Constant

$$aA + bB \rightleftharpoons cC + dD \qquad K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$

Products on top, reactants on the bottom, each raised to its coefficient.

$$\mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3} \qquad K = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$$

What the value tells you:

$$K \gg 1 \quad \text{products dominate; the reaction went nearly to completion}$$

$$K \approx 1 \quad \text{comparable amounts of both}$$

$$K \ll 1 \quad \text{reactants dominate; barely anything happened}$$

Pure solids and pure liquids are LEFT OUT. Their concentration cannot change: adding more solid gives you more solid, not a more concentrated one, since only the surface participates. So for

$$\mathrm{CaCO_3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO_2}(g)$$

$$K = [\mathrm{CO_2}]$$

which is a genuinely surprising result: the equilibrium depends on nothing but the carbon dioxide pressure.

Not with concentration, not with pressure, not with a catalyst. Those things change WHERE the system sits, and $K$ is the relationship that always holds once it settles.

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