Why a Sealed Soda Bottle Stays Fizzy Until You Open It

Many reactions do not run to completion. They slow down and seem to stop, leaving a mixture of reactants and products that no longer changes. Nothing has stopped, though — the forward and backward reactions carry on at equal rates.

Understanding that balance lets chemists predict how much product forms and how to push a reaction to make more.

This part covers dynamic equilibrium, equilibrium constants, predicting extent and direction, and Le Chatelier's principle. Take $R=0.0831$ L bar/mol K.

At equilibrium, two opposite processes continue at the same rate, so measurable properties stay constant even though particles keep changing sides.

Physical equilibria:

- Solid–liquid: ice and water at $0$ °C and $1$ atm — melting and freezing happen at equal rates
- Liquid–vapour: water in a closed flask — evaporation and condensation balance, giving a constant vapour pressure
- Dissolution: a saturated sugar solution with extra sugar at the bottom — dissolving and crystallising balance

General characteristics:

- Reached only in a closed system at constant temperature
- Measurable properties — concentration, pressure, colour — stay constant
- It is dynamic, not static
- It can be reached from either direction
- A catalyst helps reach equilibrium faster but does not change it

An everyday example. A busy local-train platform can hold a steady crowd while hundreds board and alight every minute, as long as the two flows match.

The substance. Equilibrium does not mean equal amounts of reactants and products — only unchanging amounts.

**For $aA+bB\rightleftharpoons cC+dD$, $K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$ in molar concentrations and $K_p$ uses partial pressures; pure solids and liquids are left out, and $K_p=K_c(RT{)}^{\Delta n}$, where $\Delta n$ is gaseous product moles minus gaseous reactant moles.

Worked example 1 — finding $K_c$.** For H${}_2$(g) + I${}_2$(g) $\rightleftharpoons$ 2HI(g), equilibrium concentrations are $[{\text{H}}_2]=[{\text{I}}_2]=0.10$ M and $[\text{HI}]=0.70$ M:

$$K_c=\frac{(0.70{)}^2}{(0.10)(0.10)}=49$$

**Worked example 2 — converting to $K_p$.** For PCl${}_5$(g) $\rightleftharpoons$ PCl${}_3$(g) + Cl${}_2$(g), $K_c=0.040$ at $500$ K and $\Delta n=2-1=+1$:

$$K_p=0.040\times (0.0831\times 500{)}^1\approx 1.66$$

Worked example 3 — heterogeneous. For CaCO${}_3$(s) $\rightleftharpoons$ CaO(s) + CO${}_2$(g), the solids are omitted:

$$K_p=p_{{\text{CO}}_2}$$

An everyday example. A sealed soda bottle holds carbon dioxide gas and dissolved carbon dioxide in equilibrium, CO${}_2$(g) $\rightleftharpoons$ CO${}_2$(aq).

The substance. **$K$ depends on how the equation is written** — reversing it gives $\frac{1}{K}$, and doubling all coefficients gives $K^2$.

**A very large $K$ means products dominate at equilibrium, a very small $K$ means reactants dominate, and comparing the reaction quotient $Q$ — the same expression with the current concentrations — with $K$ shows which way the reaction will move.

- $K$ greater than about $10^3$ — reaction goes nearly to completion
- $K$ less than about $10^{-3}$ — very little product forms
- In between — appreciable amounts of both

Direction:

- $Q<K$ — net forward reaction
- $Q>K$ — net backward reaction
- $Q=K$ — already at equilibrium

Worked example 1 — direction.** For the HI equilibrium with $K_c=49$, a mixture has $[{\text{H}}_2]=[{\text{I}}_2]=0.20$ M and $[\text{HI}]=0.50$ M:

$$Q=\frac{(0.50{)}^2}{(0.20)(0.20)}=6.25<49\Rightarrow \text{more HI forms}$$

Worked example 2 — equilibrium amounts. Starting with $0.50$ M each of H${}_2$ and I${}_2$, let $x$ M of each react:

$$\frac{(2x{)}^2}{(0.50-x{)}^2}=49\Rightarrow \frac{2x}{0.50-x}=7\Rightarrow x\approx 0.389$$

So $[\text{HI}]\approx 0.778$ M and $[{\text{H}}_2]=[{\text{I}}_2]\approx 0.111$ M.

An everyday example. A water tank with a float valve lets water in while the level is below the mark and stops when it reaches it — like a reaction moving until $Q=K$.

The substance. **$K$ is fixed at a given temperature, while $Q$ changes** as the reaction proceeds.

Le Chatelier's principle states that when an equilibrium is disturbed, it shifts in the direction that tends to undo the disturbance.

- Adding a reactant or removing a product — shifts forward
- Increasing pressure by reducing volume — shifts towards fewer moles of gas; no shift if $\Delta n=0$
- Raising temperature — favours the endothermic direction and changes $K$
- Adding a catalyst — equilibrium is reached faster, but its position does not change
- Adding an inert gas at constant volume — no change

Worked example — ammonia manufacture (Haber process).

$${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)\Delta H=-92.4\text{ kJ}$$

- $\Delta n=2-4=-2$, so high pressure favours ammonia
- The forward reaction is exothermic, so lower temperature favours ammonia — but makes the reaction too slow, so a moderately high temperature is used as a compromise
- An iron catalyst speeds up attainment of equilibrium

Contact process. 2SO${}_2$(g) + O${}_2$(g) $\rightleftharpoons$ 2SO${}_3$(g) is also exothermic with fewer gas moles on the right; a vanadium pentoxide catalyst and moderate temperature give a good yield at a useful rate.

**For H${}_2$ + I${}_2$ $\rightleftharpoons$ 2HI**, $\Delta n=0$, so changing pressure has no effect.

An everyday example. Opening a soda bottle lowers the pressure, so dissolved carbon dioxide shifts to the gas side and bubbles out.

The substance. **Only a temperature change alters the value of $K$**; concentration and pressure changes shift the position but leave $K$ unchanged.