Reversible change in a closed system

A reversible reaction is a chemical or physical process that can move in both the forward and backward directions under the same conditions. It is shown using the equilibrium sign, $\rightleftharpoons$, rather than a one-way arrow.

A closed system can exchange energy with its surroundings, but not matter. The lid really matters here: if a gas escapes, the backward process may never catch up properly.

Dynamic equilibrium is a state in a closed system where the forward and backward processes continue at equal rates, so there is no overall macroscopic change. Students often miss the force of the word dynamic. Particles are still reacting, evaporating, dissolving or condensing; the two opposing processes simply balance.

For a physical example, picture a volatile liquid in a sealed flask. At the start, evaporation happens faster than condensation. As vapour builds up, condensation happens more often, until the two rates become equal. The amounts of liquid and vapour then stay constant, even though individual particles keep moving between phases.

A physical equilibrium is an equilibrium involving a change of physical state or distribution without changing chemical identity. Examples include liquid $\rightleftharpoons$ vapour, solid $\rightleftharpoons$ dissolved ions in a saturated solution, or $X(g) \rightleftharpoons X(aq)$.

A chemical equilibrium is an equilibrium involving reversible chemical reaction, where reactant and product particles are continually converted into each other. For example, once equilibrium has been reached, a reaction mixture may contain both reactants and products at constant concentrations.

Characteristics of equilibrium

At equilibrium:

• the system is closed;
• the forward and backward rates are equal;
• concentrations of reactants and products remain constant;
• macroscopic properties such as colour, pressure, density or pH remain constant;
• microscopic change continues;
• the same equilibrium composition can often be reached from either direction, provided the temperature is the same.

Do not write that the reaction has stopped. That is the classic giveaway that equilibrium has been treated as static rather than dynamic.

A homogeneous equilibrium is an equilibrium in which all reacting species are in the same phase. A heterogeneous equilibrium is an equilibrium in which reacting species are present in more than one phase.

Writing an equilibrium expression

The equilibrium law describes how, at a fixed temperature, a particular ratio of product and reactant concentrations stays constant for a reaction at equilibrium.

The equilibrium constant, $K$, is a dimensionless number for that ratio, for a specified equilibrium equation at a specified temperature.

For the homogeneous equilibrium

Put the products from the forward reaction on the top. Put the reactants from the forward reaction on the bottom. Use the balancing numbers from the equation as the powers. That’s the whole game.

Balanced homogeneous equilibria matched to K expressions, showing products over reactants and coefficients as powers.

For example, for

$K = [PCl_3][Cl_2] / [PCl_5]$.

For

$K = [SO_3]^2 / ([SO_2]^2[O_2])$.

What to include

In this topic, you deduce expressions for homogeneous reactions, so all assessed reacting species are in the same phase. When an aqueous reaction uses water as the solvent, leave water out because its concentration is effectively constant. Pure liquids and pure solids are left out of concentration expressions for the same reason.

State symbols help in the equation, but don’t put them inside the square brackets. The bracket already means the concentration of that species at equilibrium.

What the size of $K$ tells you

The size of $K$ shows the extent of reaction: how far the forward reaction has gone by the time equilibrium is reached.

When $K$ is very large, the numerator in the equilibrium expression is much larger than the denominator, so products dominate. When $K$ is very small, the mixture is mainly reactants. If $K$ is close to 1, neither side is strongly favoured, though the actual concentrations still depend on the equation stoichiometry.

How the magnitude of K indicates which side is favoured at equilibrium.

A useful verbal scale is:

• $K \ll 1$: very little product at equilibrium; the reactants are strongly favoured.
• $K < 1$: reactants are favoured.
• $K = 1$: neither side is favoured overall by the equilibrium constant.
• $K > 1$: products are favoured.
• $K \gg 1$: the reaction is almost complete in the forward direction.

Reversing an equilibrium equation

The value of $K$ is tied to the equation as written. Reverse the equation at the same temperature, and the new equilibrium constant is the reciprocal of the original.

So if $A \rightleftharpoons B$ has $K = 25$ at a stated temperature, then $B \rightleftharpoons A$ has $K_{\text{rev}} = 1/25 = 0.040$ at that same temperature. Same chemical system, just viewed in the opposite direction.

Temperature dependence

For a given equation, $K$ changes only when the temperature changes. Changing initial concentration, changing pressure, or adding a catalyst may alter the equilibrium composition or the time taken to reach equilibrium, but it does not change $K$ at that temperature.

That’s why published equilibrium constants always need a temperature attached or clearly implied. Giving $K$ without temperature is like giving a solubility value without temperature: the information is incomplete.

Link to acid strength

For an acid dissociation equilibrium, the acid dissociation constant, $K_a$, is the equilibrium constant for ionization of an acid in water (dimensionless in IB treatment). A larger $K_a$ means a greater extent of ionization, so the acid is stronger. A smaller $K_a$ means the equilibrium lies more to the undissociated acid side, so the acid is weaker.

The principle

Le Châtelier's principle states that when a system at dynamic equilibrium is disturbed, the equilibrium shifts in the direction that tends to oppose the disturbance.

A shift in equilibrium position means the equilibrium composition changes because either the forward or backward reaction is favoured until a new equilibrium is reached. A shift to the right increases the amount of products. A shift to the left increases the amount of reactants.

Changing concentration

Add a reactant, and the system tends to use up some of what was added, so the equilibrium shifts to the product side. Remove a reactant, and the system tends to replace it, so the equilibrium shifts to the reactant side.

Products follow the same pattern: adding product shifts left; removing product shifts right. The equilibrium composition changes, but $K$ is unchanged because temperature is unchanged.

For a coloured equilibrium, you can often see the shift happening. If adding acid increases $[H^+]$ and $H^+$ appears on the left of the equilibrium equation, the system shifts right to consume some $H^+$. The colour then changes toward the species on the right. The colour is not magic; it is a concentration change you can see.

Changing pressure in gaseous equilibria

Pressure changes mainly matter for gases. Increasing pressure shifts equilibrium toward the side with fewer moles of gas particles. Decreasing pressure shifts equilibrium toward the side with more moles of gas particles. If both sides have the same number of moles of gas, changing pressure has no effect on the equilibrium position.

Decreasing volume is the same as increasing pressure. Increasing volume is the same as decreasing pressure. Pressure and volume changes do not change $K$ if temperature is constant.

For heterogeneous equilibria, count only gaseous species when deciding the pressure effect. Solids, liquids and aqueous species do not contribute significantly to pressure changes in this context.

The guide example $X(g) \rightleftharpoons X(aq)$ is useful. Increasing pressure favours removal of gas particles from the gas phase, so more $X$ dissolves and the equilibrium shifts toward $X(aq)$. Decreasing pressure favours $X(g)$.

Changing temperature

Temperature behaves differently: it changes both the equilibrium composition and the value of $K$.

The standard enthalpy change of reaction, $\Delta H_r^\circ$, is the enthalpy change when the molar amounts in the balanced equation react under standard conditions (usually $\mathrm{kJ\ mol^{-1}}$; SI unit $\mathrm{J\ mol^{-1}}$). If $\Delta H_r^\circ < 0$, the forward reaction is exothermic. If $\Delta H_r^\circ > 0$, the forward reaction is endothermic.

For an exothermic forward reaction, heat can be treated as a product. Increasing temperature shifts the equilibrium left and decreases $K$. Decreasing temperature shifts the equilibrium right and increases $K$.

For an endothermic forward reaction, heat can be treated as a reactant. Increasing temperature shifts the equilibrium right and increases $K$. Decreasing temperature shifts the equilibrium left and decreases $K$.

Catalysts and equilibrium

A catalyst is a substance that increases reaction rate by providing an alternative pathway with lower activation energy and is regenerated by the end of the reaction. In a reversible reaction, the catalyst speeds up both forward and backward reactions. It helps the system reach equilibrium faster, but it does not change $K$ and does not change the equilibrium composition.

So the link with rates is simple: catalysts affect how fast equilibrium is reached, not how far the reaction has gone at equilibrium.

Summary of how common disturbances affect equilibrium position and K.

Using $Q$ before equilibrium

The reaction quotient, $Q$, is a dimensionless ratio found from the equilibrium expression, using the concentrations present at one particular moment rather than necessarily at equilibrium.

For the same general reaction

$aA + bB \rightleftharpoons cC + dD$

It looks just like the expression for $K$; what changes is whether the concentrations are equilibrium concentrations.

Predicting direction

Compare $Q$ with $K$ at the same temperature:

• If $Q < K$, there is too little product compared with the equilibrium mixture, so the forward reaction is favoured.
• If $Q > K$, there is too much product compared with the equilibrium mixture, so the reverse reaction is favoured.
• If $Q = K$, the mixture is at equilibrium.

I like to think of $Q$ as the mixture asking, “where am I now?”, while $K$ says, “where must I end up at this temperature?”. The reaction goes in the direction that moves $Q$ toward $K$.

From K to equilibrium concentrations

Since K gives a mathematical link between equilibrium concentrations, you can use it to calculate an unknown equilibrium concentration from known initial or equilibrium data. In assessed questions, the equilibria are homogeneous.

A neat way to do the working is to set up an initial-change-equilibrium table. Many teachers call this an ICE table; the name doesn't matter much, as long as the stoichiometry is right.

ICE table for A + 2B ⇌ C; use the equilibrium row in Kc = [C]/([A][B]²).

For a reaction such as

$A(g) + 2B(g) \rightleftharpoons C(g)$

if the change in $[A]$ is $-x$, then the change in $[B]$ is $-2x$ and the change in $[C]$ is $+x$, where $x$ is the concentration change linked to one stoichiometric unit of reaction ($\text{mol dm}^{-3}$; SI unit $\text{mol m}^{-3}$). Put the equilibrium row into the K expression, then solve.

Working backwards

Sometimes you are given the equilibrium concentrations and asked to find the initial concentrations. Work backwards using the stoichiometric changes. If $0.20\ \text{mol dm}^{-3}$ of C has formed in $A + 2B \rightleftharpoons C$, then $0.20\ \text{mol dm}^{-3}$ of A and $0.40\ \text{mol dm}^{-3}$ of B were used up.

This is not a new equilibrium idea; it is conservation of atoms wearing an equilibrium hat.

The small-K approximation

When K is very small, the equilibrium sits far to the reactant side. Only a small amount of reactant changes into product, so

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

Here $[reactant]_{\text{initial}}$ is the initial reactant concentration ($\text{mol dm}^{-3}$; SI unit $\text{mol m}^{-3}$), and $[reactant]_{\text{eqm}}$ is the equilibrium reactant concentration ($\text{mol dm}^{-3}$; SI unit $\text{mol m}^{-3}$).

This approximation is particularly useful for weak acid and weak base equilibria, where only a small extent of ionization occurs. The syllabus does not expect you to solve quadratic equations here, so use the approximation when the chemistry justifies it.

Link to pH calculations

The equilibrium law can help determine pH because it gives the equilibrium concentration of $H^+$ or $OH^-$ in weak acid and weak base systems. For a weak acid, use the equilibrium expression to find $[H^+]$, then use the pH relationship from Reactivity 3.1 to convert $[H^+]$ into pH. For a weak base, the equilibrium expression gives $[OH^-]$, which can then be related to $[H^+]$ using water equilibrium. In buffer solutions, the same idea applies to the equilibrium between a weak acid and its conjugate base: the ratio of the pair controls $[H^+]$.

Two ways to describe the same position

The Gibbs energy change, $\Delta G$, is the energy change for a reaction that shows whether the forward or reverse direction is thermodynamically favoured under the current conditions (J mol$^{-1}$). If $\Delta G < 0$, the forward reaction is favoured. If $\Delta G > 0$, the reverse reaction is favoured. At equilibrium, $\Delta G = 0$.

The standard Gibbs energy change, $\Delta G^\circ$, is the Gibbs energy change for a reaction when the reactants and products are in their standard states at a stated temperature (J mol$^{-1}$, often reported as kJ mol$^{-1}$).

Here’s the link between $\Delta G^\circ$ and $K$:

, where

The equation is in the data booklet. The unit conversion, though, is on you: kJ mol$^{-1}$ must be converted to J mol$^{-1}$ before using $R$ in J K$^{-1}$ mol$^{-1}$.

Interpreting the signs

The signs line up with the equilibrium position:

• $K$ > 1 gives $\ln K$ > 0, so $\Delta G^\circ < 0$: products are favoured under standard conditions.
• $K$ < 1 gives $\ln K$ < 0, so $\Delta G^\circ > 0$: reactants are favoured under standard conditions.
• $K$ = 1 gives $\ln K$ = 0, so $\Delta G^\circ = 0$: neither side is favoured under standard conditions.

This links back to Reactivity 1.4. Before equilibrium is reached, Gibbs energy tells us which direction is favoured. The system moves in the direction that lowers Gibbs energy until $\Delta G$ becomes zero. At that point, the forward and reverse tendencies balance; that’s the thermodynamic description of equilibrium.

Calculating $K$ or $\Delta G^\circ$

If $K$ is known, substitute it directly into $\Delta G^\circ = -RT \ln K$. If $\Delta G^\circ$ is known and $K$ is required, rearrange:

, where

A negative $\Delta G^\circ$ gives $K$ greater than 1. A positive $\Delta G^\circ$ gives $K$ less than 1. Use that as a quick check on any calculator answer.