Dynamic equilibrium and the equilibrium constant
Many reactions never go fully to completion. They reach a balance where forward and reverse reactions continue at equal rates.
Reversible reactions
Many reactions can run in both directions: \(\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3\). At first the forward reaction dominates, but as products build up, the reverse reaction speeds up. Eventually both rates are equal and concentrations stop changing.
Dynamic equilibrium
At equilibrium, reactions have not stopped; forward and reverse rates are equal, so there is no net change. It is like a crowded escalator system where as many people go up as come down: the numbers on each floor stay constant while individuals keep moving.
The equilibrium constant
For \(a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}\), the equilibrium constant is \(K = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\), using equilibrium concentrations. Pure solids and liquids are left out. \(K\) depends only on temperature, not on starting amounts.
K = \frac{[C]^c[D]^d}{[A]^a[B]^b}Interpreting K
If \(K \gg 1\), products dominate at equilibrium; if \(K \ll 1\), reactants dominate; if \(K \approx 1\), significant amounts of both exist. The ionisation of a weak acid has a small \(K\), while the combustion of methane has an enormous one, effectively going to completion.
The reaction quotient
The reaction quotient \(Q\) has the same form as \(K\) but uses current concentrations. Comparing them predicts direction: if \(Q < K\), the reaction proceeds forward; if \(Q > K\), it proceeds in reverse; if \(Q = K\), the system is at equilibrium.
Worked example
For \(\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}\), equilibrium concentrations are \([\text{H}_2] = 0.10\), \([\text{I}_2] = 0.10\), \([\text{HI}] = 0.70\) M. \(K = \frac{0.70^2}{0.10 \times 0.10} = 49\). If a mixture instead has \(Q = 10\), it will shift forward, producing more HI until \(Q\) reaches 49.
Calculating equilibrium concentrations
ICE tables organise the work: Initial concentrations, Change (in terms of an unknown \(x\) using the coefficients), and Equilibrium concentrations. Substituting the equilibrium expressions into \(K\) gives an equation for \(x\), often simplified when \(K\) is small.
Gases and Kp
For gas reactions, equilibrium can be expressed with partial pressures, \(K_p\). It relates to the concentration form by \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the change in moles of gas. When \(\Delta n = 0\), the two constants are equal.
Key takeaways
Equilibrium is dynamic: forward and reverse rates are equal. \(K\) is products over reactants raised to their coefficients, excluding solids and liquids, and depends only on temperature. Large \(K\) favours products. Comparing \(Q\) with \(K\) predicts the direction of change; ICE tables find equilibrium concentrations.
Flashcard
Can proceed in both directions
Check your understanding
For a reaction with K = 4, a mixture has Q = 10. What happens?