6.1 Chemical Equilibrium and the Equilibrium Constant

A sealed bottle of soda water can stand on a shelf for a year and nothing seems to happen. Open it and it fizzes; leave it open and it goes flat. Inside the sealed bottle carbon dioxide molecules are still escaping from the liquid into the gas space above it — and others are dissolving back into the liquid at exactly the same rate, so the amount in each place never changes. The same balance appears in chemical reactions. Heat hydrogen and iodine vapour in a sealed flask and the violet colour of iodine fades, but it never disappears: the reaction stops changing while all three substances, $\mathrm{H_{2}}$, $\mathrm{I_{2}}$ and $\mathrm{HI}$, are still present.

Such a state is called equilibrium. In this section you will meet equilibria in physical processes, see why chemical equilibrium is dynamic, and learn the law that turns every equilibrium into a number — the equilibrium constant. That number, $K_{c}$ for concentrations or $K_{p}$ for partial pressures, is the starting point for every calculation in this chapter.

Equilibrium in Physical Processes

Put crushed ice and water in a perfectly insulated flask at $273\ \mathrm{K}$ ($0\ ^{\circ}\mathrm{C}$) and atmospheric pressure. The masses of ice and of water stay constant, and so does the temperature. This does not mean that nothing happens: molecules leave the surface of the ice and join the liquid, and molecules from the liquid stick to the ice, at the same rate. We write $\mathrm{H_{2}O(s)\rightleftharpoons H_{2}O(l)}$, the double half-arrows showing that both processes go on together.

Liquid ⇌ vapour. Pour a little water into a closed, evacuated flask kept at $25\ ^{\circ}\mathrm{C}$ and connect a manometer. The pressure of water vapour rises and then levels off at about $3.17\ \mathrm{kPa}$: the rate of evaporation is now equal to the rate of condensation, $\mathrm{H_{2}O(l)\rightleftharpoons H_{2}O(g)}$. This constant pressure is the equilibrium vapour pressure of water at that temperature. It depends on the liquid and on the temperature only — not on how much liquid is present — and the temperature at which it equals $1.013\ \mathrm{bar}$ (1 atm) is the normal boiling point ($100\ ^{\circ}\mathrm{C}$ for water). In an open vessel vapour escapes as fast as it forms and equilibrium is never reached: the water evaporates completely. Likewise, the normal melting point is the temperature at which solid and liquid are in equilibrium at $1.013\ \mathrm{bar}$.

Solid ⇌ vapour. A few crystals of iodine in a closed jar fill it with violet vapour whose intensity soon becomes constant: $\mathrm{I_{2}(s)\rightleftharpoons I_{2}(g)}$. Camphor and naphthalene sublime in the same way.

Solids in liquids. Stir sugar into water until no more dissolves: the solution is saturated and the undissolved sugar is in equilibrium with the dissolved sugar, $\text{sugar (solid)}\rightleftharpoons\text{sugar (solution)}$. The concentration of a saturated solution at a given temperature is the solubility. If some radioactively labelled sugar is added to a saturated solution of ordinary sugar, radioactivity is soon found both in the solution and in the solid, although the amount of solid does not change: molecules keep dissolving and crystallising.

Gases in liquids. In a sealed bottle of soda water, $\mathrm{CO_{2}(g)\rightleftharpoons CO_{2}(aq)}$. The amount of gas that can dissolve depends on the pressure of the gas above the liquid.

Soda water is bottled under a high pressure of $\mathrm{CO_{2}}$. When the cap is removed the pressure falls to that of the air, whose partial pressure of $\mathrm{CO_{2}}$ is tiny, so the dissolved gas escapes as bubbles; on a warm day it escapes faster because the gas is less soluble.

Equilibrium in Chemical Processes

Most reactions can go in both directions. Seal hydrogen and iodine vapour in a flask at about $720\ \mathrm{K}$: hydrogen iodide forms, $\mathrm{H_{2}(g)+I_{2}(g)\rightarrow2HI(g)}$, and the violet colour fades. At first the reactants are plentiful and the forward reaction is fast; as they are used up it slows down. Meanwhile hydrogen iodide molecules begin to split up again, $\mathrm{2HI(g)\rightarrow H_{2}(g)+I_{2}(g)}$; this backward reaction speeds up as $\mathrm{HI}$ accumulates. Eventually the two rates become equal, and from then on the concentrations stay constant. If instead you start with pure hydrogen iodide in the same flask, it partly decomposes and the colour of iodine appears — and if the amounts of H and I atoms are the same as before, the flask ends with exactly the same mixture.

How do we know that the reactions continue at equilibrium? Haber's ammonia equilibrium, $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$, gives a neat proof. Two equilibrium mixtures are prepared under identical conditions, one with ordinary hydrogen $\mathrm{H_{2}}$ and one with deuterium $\mathrm{D_{2}}$ (the heavy isotope ${}^{2}\mathrm{H}$), with the same starting amounts. The two mixtures are combined. A few minutes later, mass spectrometry shows $\mathrm{NH_{2}D}$, $\mathrm{NHD_{2}}$ and $\mathrm{ND_{3}}$ as well as $\mathrm{NH_{3}}$, and $\mathrm{HD}$ as well as $\mathrm{H_{2}}$ and $\mathrm{D_{2}}$ — yet the total amount of ammonia is unchanged. Isotopes can only become scrambled if ammonia molecules keep breaking up and re-forming.

The Law of Chemical Equilibrium and $K_{c}$

Different starting mixtures of hydrogen and iodine give different equilibrium mixtures. Is anything the same about them? The table shows three experiments at the same temperature (about $720\ \mathrm{K}$). The concentrations themselves differ, but one combination of them does not.

The ratio $[\mathrm{HI}]^{2}/([\mathrm{H_{2}}][\mathrm{I_{2}}])$ is the same in every experiment; the simpler-looking ratio without the square is not. Each concentration is raised to the power of its coefficient in the balanced equation. In 1864 the Norwegian chemists Cato Guldberg and Peter Waage generalised this observation.

Homogeneous and Heterogeneous Equilibria; $K_{p}$

For reactions between gases it is often easier to measure pressures than concentrations.

For example, for $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons2NH_{3}(g)}$, $K_{p}=\dfrac{(c_{\mathrm{NH_{3}}}RT)^{2}}{(c_{\mathrm{N_{2}}}RT)(c_{\mathrm{H_{2}}}RT)^{3}}$ $=K_{c}(RT)^{2-4}=K_{c}(RT)^{-2}$.

In a heterogeneous equilibrium a pure solid or a pure liquid takes part, but its "concentration" — its amount per unit volume of that solid or liquid, which equals density divided by molar mass — is fixed, however much of it is present. A constant like this is absorbed into the equilibrium constant.

Exercises