Why Ammonia Factories Run Their Reactors at Enormous Pressure

In a sealed bottle of soda water, carbon dioxide keeps escaping from the liquid and dissolving back at the same rate, so nothing seems to change. Many reactions reach this kind of balance, and knowing how to shift it lets chemists and industries obtain more of the product they want.

This lesson covers dynamic equilibrium, equilibrium constants, and Le Chatelier's principle.

Chemical equilibrium is dynamic because the forward and backward reactions continue at equal rates, so the concentrations of reactants and products stay constant even though molecules keep reacting in both directions.

Features of equilibrium:

- Reached only in a closed system at constant temperature
- The forward and backward rates become equal
- The concentrations of all species stay constant, though not necessarily equal
- It can be reached from either side — starting with reactants or with products
- A catalyst speeds up both reactions equally, so equilibrium is reached sooner but not changed

Physical equilibria:

- Ice and water at 273 K and 1 atm — melting and freezing at equal rates
- Water and its vapour in a closed container — evaporation and condensation at equal rates
- Carbon dioxide in a sealed soft drink — escaping and dissolving at equal rates

Chemical equilibria. In ${\mathrm{H}}_2(\mathrm{g})+{\mathrm{I}}_2(\mathrm{g})\rightleftharpoons 2HI(\mathrm{g})$, hydrogen iodide forms and breaks down at the same rate once equilibrium is reached.

Evidence that it is dynamic. If some hydrogen in an ammonia equilibrium mixture is replaced by deuterium, deuterium soon appears in the ammonia molecules — so reactions keep happening even though the overall amounts stay the same.

An everyday example. A closed bottle of water left in a hot car has water evaporating and vapour condensing at equal rates, so the water level stops falling.

The substance. Equal rates do not mean equal amounts — at equilibrium, reactants and products can be present in very different concentrations.

**For a reaction $a\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}$, the equilibrium constant is $K_c=\frac{[\mathrm{C}{]}^c[\mathrm{D}{]}^d}{[\mathrm{A}{]}^a[\mathrm{B}{]}^b}$, and for gaseous reactions it is linked to the pressure constant by $K_p=K_c(RT{)}^{\Delta n}$.

Law of chemical equilibrium:**

$$K_c=\frac{[\mathrm{C}{]}^c[\mathrm{D}{]}^d}{[\mathrm{A}{]}^a[\mathrm{B}{]}^b}$$

Rules for writing expressions:

- Products go in the numerator and reactants in the denominator, each raised to its coefficient
- Pure solids and pure liquids are left out, because their concentrations are constant
- Reversing a reaction gives $1/K_c$; multiplying the equation by n gives $K_c^n$

Worked example 1. For ${\mathrm{H}}_2(\mathrm{g})+{\mathrm{I}}_2(\mathrm{g})\rightleftharpoons 2HI(\mathrm{g})$, with $[{\mathrm{H}}_2]=[{\mathrm{I}}_2]=0.10$ M and $[\mathrm{HI}]=0.80$ M at equilibrium:

$$K_c=\frac{(0.80{)}^2}{(0.10)(0.10)}=\frac{0.64}{0.010}=64$$

**Relation between $K_p$ and $K_c$:**

$$K_p=K_c(RT{)}^{\Delta n}$$

where $\Delta n$ is the moles of gaseous products minus the moles of gaseous reactants, and $R=0.0831$ bar L mol${}^{-1}$ K${}^{-1}$.

Worked example 2. For $\mathrm{PC}{\mathrm{l}}_5(\mathrm{g})\rightleftharpoons \mathrm{PC}{\mathrm{l}}_3(\mathrm{g})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})$, $K_c=0.040$ at 500 K and $\Delta n=2-1=1$:

$$K_p=0.040\times (0.0831\times 500)=0.040\times 41.6=1.66$$

Worked example 3. For ${\mathrm{H}}_2+{\mathrm{I}}_2\rightleftharpoons 2HI$, $\Delta n=0$, so $K_p=K_c$.

An everyday example. Fertiliser plants that make ammonia use equilibrium constants to calculate how much ammonia a reactor can produce under given conditions.

The substance. A large K means products are favoured, but says nothing about speed — a reaction can have a very large K and still be extremely slow.

Le Chatelier's principle states that if a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the change.

Change in concentration:

- Adding a reactant or removing a product shifts equilibrium forward
- Adding a product or removing a reactant shifts it backward
- The value of K does not change

Change in pressure, for gaseous reactions:

- Raising the pressure shifts equilibrium towards the side with fewer moles of gas
- Lowering the pressure shifts it towards the side with more moles of gas
- If $\Delta n=0$, as in ${\mathrm{H}}_2+{\mathrm{I}}_2\rightleftharpoons 2HI$, pressure has no effect

Change in temperature:

- For an exothermic forward reaction, raising the temperature shifts equilibrium backward and lowers K
- For an endothermic forward reaction, raising the temperature shifts equilibrium forward and raises K
- Temperature is the only one of these factors that changes K

Other factors. A catalyst does not shift equilibrium, and adding an inert gas at constant volume has no effect.

Worked example — the Haber process. ${\mathrm{N}}_2(\mathrm{g})+3{\mathrm{H}}_2(\mathrm{g})\rightleftharpoons 2N{\mathrm{H}}_3(\mathrm{g})$, $\Delta H=-92.4$ kJ:

- 4 moles of gas become 2, so high pressure favours ammonia
- The reaction is exothermic, so lower temperature favours ammonia, but the rate then becomes too slow
- Industry uses high pressure, a moderate temperature of about 700 K and an iron catalyst as a compromise

An everyday example. Opening a bottle of soda water lowers the pressure above the liquid, so dissolved carbon dioxide comes out as bubbles and the drink slowly goes flat.

The substance. Le Chatelier's principle predicts the direction of a shift, not a full return — the new equilibrium only partly offsets the change.