Chemical equilibrium

A reversible reaction in a closed system can reach dynamic equilibrium. When reactants are first mixed, the forward reaction is fast (high reactant concentration) and the reverse reaction is slow. As reactants are used up the forward rate falls, and as products build up the reverse rate rises.

Eventually the two rates become equal. On a rate-vs-time graph this looks like two curves approaching each other: the forward-rate curve falls, the reverse-rate curve rises, and they meet and then run flat (constant) together. From that moment the concentrations of all species stay constant because each species is being formed and consumed at the same rate.

Key points to remember:

• It is dynamic, not static: both reactions are still happening at the molecular level.
• Concentrations are constant, not equal to each other.
• All macroscopic properties (colour, pressure, density) stay unchanged.
• The system must be closed so nothing escapes.

Equilibrium applies to physical changes (e.g. $\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{O(g)}$ in a sealed flask) as well as chemical changes (e.g. $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$). The half-arrows $\rightleftharpoons$ signal that both directions occur.

For a balanced homogeneous equilibrium (all species in the same phase)

$$a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$$

the equilibrium constant is

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

The rules for writing it:

• Products on top, reactants on the bottom.
• Each concentration is raised to the power of its stoichiometric coefficient from the balanced equation.
• Concentrations are equilibrium values in $\text{mol dm}^{-3}$.

Important: this expression is NOT printed in the data booklet - you must construct it yourself from the balanced equation, so always write the equation first and balance it carefully.

For example, for $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$:

$$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$

The units of $K_c$ depend on the stoichiometry and are usually omitted at IB. Reversing the equation gives the reciprocal value, and doubling all coefficients squares $K_c$, so $K_c$ is only meaningful when tied to a specific written equation.

The size of the equilibrium constant tells you how far a reaction proceeds before equilibrium is reached - its extent.

• $K \gg 1$ (large): the top of the expression dominates, so products are favoured. The reaction goes nearly to completion.
• $K \ll 1$ (small): the bottom dominates, so reactants are favoured. Very little product forms.
• $K \approx 1$: significant amounts of both reactants and products are present at equilibrium.

A useful mental picture: $K = 10^{10}$ means essentially all product; $K = 10^{-10}$ means essentially no reaction; $K$ near $1$ means a genuine mixture.

Crucially, $K$ does not tell you anything about the rate at which equilibrium is reached - a reaction with a huge $K$ can still be extremely slow.

The single most important fact: $K$ depends only on temperature. Changing concentrations or pressure, or adding a catalyst, may shift the *position* of equilibrium but does not change the *value* of $K$. Only a change in temperature gives a new value of $K$ for a given reaction.

Le Chatelier's principle: if a change is imposed on a system at equilibrium, the position of equilibrium shifts to partly oppose that change. Use it to make qualitative predictions.

Concentration: adding more of a species shifts equilibrium away from it (to use it up); removing a species shifts equilibrium towards it. $K$ is unchanged.

Pressure (gases only): increasing pressure shifts equilibrium towards the side with fewer moles of gas; decreasing pressure shifts towards more moles. If both sides have equal gas moles, pressure has no effect. $K$ is unchanged.

Temperature: this is the only change that alters $K$.

• For an exothermic forward reaction, raising temperature shifts equilibrium back (towards reactants) and decreases $K$.
• For an endothermic forward reaction, raising temperature shifts equilibrium forward and increases $K$.

Catalyst: speeds up the forward and reverse reactions equally, so equilibrium is reached faster but the position does not shift and $K$ is unchanged.

A handy check: only temperature changes $K$; everything else only moves the position while $K$ stays fixed.

The reaction quotient $Q$ has exactly the same algebraic form as $K_c$, but it is evaluated using the concentrations present at any instant, not only at equilibrium:

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

Like $K_c$, this expression is not in the data booklet - you build it from the balanced equation.

Comparing $Q$ with $K$ tells you which way a non-equilibrium mixture must change to reach equilibrium:

• $Q < K$: too few products, so the reaction goes forward (to the right) to make more product.
• $Q > K$: too many products, so the reaction goes reverse (to the left) to remake reactants.
• $Q = K$: the system is at equilibrium - no net change.

Think of it as the system always moving so that $Q$ heads towards $K$. This is a quantitative version of Le Chatelier reasoning. $Q$ is especially useful for deciding the direction after a sudden disturbance, or for confirming whether a stated set of concentrations represents a true equilibrium mixture.

An ICE table organises an equilibrium calculation. The rows are Initial, Change, and Equilibrium concentrations (in $\text{mol dm}^{-3}$); changes follow the stoichiometric ratios.

Typical method:

• Write the balanced equation and the $K_c$ expression.
• Fill in initial concentrations; reactants change by $-x$ (times their coefficient), products by $+x$.
• Write equilibrium concentrations as (initial $\pm$ change).
• Substitute into $K_c$ and solve for the unknown.

Small-K approximation: when $K$ is very small, only a tiny amount of reactant converts, so $x$ is negligible compared with the initial reactant concentration. You may then approximate

$$[\text{reactant}]_{\text{eqm}} \approx [\text{reactant}]_{\text{initial}}$$

This avoids having to solve a quadratic. At IB you are not required to solve quadratics, so questions are set so that either a square root or the small-$K$ approximation gives the answer cleanly. Always check that approximations are reasonable (the change should be less than about 5% of the initial value).

Gibbs energy links thermodynamics to the equilibrium constant. The data booklet gives two related equations:

$$\Delta G = \Delta G^{\ominus} + RT\ln Q$$

$$\Delta G^{\ominus} = -RT\ln K$$

The first is the Gibbs energy change at any instant, written in terms of the standard value and the reaction quotient $Q$. As a reaction proceeds, $Q$ changes and $\Delta G$ moves towards zero.

At equilibrium there is no driving force in either direction, so $\Delta G = 0$ and the reaction quotient has reached the equilibrium constant ($Q = K$). Substituting these conditions into the first equation gives the second: $0 = \Delta G^{\ominus} + RT\ln K$, so $\Delta G^{\ominus} = -RT\ln K$.

Interpretation:

• $\Delta G^{\ominus} < 0$ (negative) $\Rightarrow \ln K > 0 \Rightarrow K > 1$: products favoured.
• $\Delta G^{\ominus} > 0$ (positive) $\Rightarrow \ln K < 0 \Rightarrow K < 1$: reactants favoured.
• $\Delta G^{\ominus} = 0 \Rightarrow K = 1$.

Units warning: $R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}$, so $RT\ln K$ comes out in $\text{J mol}^{-1}$, whereas $\Delta G^{\ominus}$ is usually quoted in $\text{kJ mol}^{-1}$. Convert before equating them, and use $T$ in kelvin.