Equilibrium and Acid Reactions

Module 5 is worth about 24 marks of the 100-mark HSC paper. The marks are spread fairly evenly across the four inquiry questions, with solution equilibria (Ksp) and factors that affect equilibrium the most heavily tested (about 6–7 marks each a year), then calculating $K_{eq}$ (about 5 marks) and static and dynamic equilibrium (about 3–4 marks).

• Static and dynamic equilibrium: open and closed systems, reversible and irreversible reactions, what is happening at the particle level at equilibrium, and how enthalpy and entropy decide whether a reaction goes to completion.
• Factors that affect equilibrium: Le Chatelier's principle for changes in concentration, pressure (volume) and temperature; why catalysts don't shift equilibrium; explaining shifts with collision theory and rate graphs.
• Calculating the equilibrium constant: writing $K_{eq}$ expressions, ICE tables, the reaction quotient $Q$, what the size of $K_{eq}$ tells you, and how temperature changes $K_{eq}$.
• Solution equilibria: dissolution of ionic compounds, $K_{sp}$ expressions, calculating solubility from $K_{sp}$ (and back), predicting precipitates, the common ion effect, and removing toxins from cycad seeds.

A closed system can exchange energy with its surroundings but not matter. An open system exchanges both. Equilibrium can only be reached in a closed system: if a product (say a gas) escapes, the reverse reaction can never catch up.

A reversible reaction can go in both directions under the same conditions. Dynamic equilibrium is reached when the forward and reverse reactions occur at equal rates, so the concentrations of all species stay constant. Reactions haven't stopped: particles are still reacting in both directions. Concentrations at equilibrium are constant but not usually equal.

Static equilibrium means nothing is changing at the particle level because the reaction has stopped in both directions. Graphite and diamond at room temperature are often quoted: diamond is thermodynamically unstable, but the conversion is so slow that nothing happens.

Evidence for dynamic equilibrium

A classic piece of evidence is isotope labelling. Add a little solid containing radioactive iodide to a saturated solution of a sparingly soluble iodide in contact with its solid: radioactivity soon appears in the solution even though the total amount dissolved doesn't change. Dissolving and crystallising are both still happening at equal rates.

On a concentration–time graph equilibrium is where every line goes flat. On a rate–time graph it is where the forward and reverse rate lines meet and stay together.

Enthalpy, entropy and non-equilibrium systems

Whether a reaction is spontaneous depends on enthalpy change $\Delta H$ (heat released or absorbed) and entropy change $\Delta S$ (the spread of energy and matter; increases when gases form or particles become more disordered). They combine in the Gibbs free energy equation from the data sheet:

Use kJ for $\Delta H$, convert $\Delta S$ from ${\text{J K}}^{-1}{\text{mol}}^{-1}$ to kJ, and use $T$ in kelvin. A reaction with $\Delta G<0$ is spontaneous.

Combustion is the standard example of a non-equilibrium system: it is strongly exothermic, increases entropy (gases form), and the products escape, so it goes essentially to completion. Photosynthesis is non-spontaneous by itself ($\Delta H>0$, $\Delta S<0$) and only proceeds because light energy is continually supplied.

Example. A gas-phase reaction has $\Delta H=-120{\text{kJ mol}}^{-1}$ and $\Delta S=-150{\text{J K}}^{-1}{\text{mol}}^{-1}$. At 298 K:

so it is spontaneous. At 900 K, $\Delta G=+15.0{\text{kJ mol}}^{-1}$, so it is not. The sign changes at $T=120/0.150=800$ K. Exothermic reactions that decrease entropy favour low temperatures.

Le Chatelier's principle: if a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance. "Partially" matters: the system never fully undoes the change.

Explaining shifts with collision theory

Markers increasingly want the "why", not just "Le Chatelier says so". For example, adding reactant increases its concentration, so collisions between reactant particles become more frequent and the forward rate rises above the reverse rate. Products build up until the reverse rate rises to match, and a new equilibrium is reached with higher concentrations of products.

For a temperature rise, both rates increase, but the endothermic direction has the larger activation energy, so a greater fraction of its collisions now succeed and its rate increases more.

A catalyst provides an alternative pathway with lower activation energy for both directions equally. Equilibrium is reached sooner but the position doesn't change.

Reading graphs

On a concentration–time graph:

• A concentration change shows as a sudden vertical jump in one species only, then gradual adjustment of all species.
• A volume change makes every gas concentration jump at once (all up if volume decreases), then the system adjusts.
• A temperature change shows no jump: concentrations simply start drifting to new flat values.

Always check the mole ratios: in the adjustment period, species change in amounts proportional to their coefficients.

Practical investigations

Know at least two colour-change systems and what you would see:

• Cobalt(II) chloride: pink ${\text{Co(H}}_2{\text{O)}}_6^{2+}$ and blue ${\text{CoCl}}_4^{2-}$; the forward reaction to blue is endothermic, so heating turns it blue.
• Iron(III) thiocyanate: pale yellow ${\text{Fe}}^{3+}$ and deep red ${\text{FeSCN}}^{2+}$; changing either ion's concentration changes the colour.
• Nitrogen dioxide/dinitrogen tetroxide: brown ${\text{NO}}_2$ and colourless ${\text{N}}_2{\text{O}}_4$; responds to temperature and pressure.

A full answer states the change, the direction of shift, the reason, and the observation (colour change, not "the equilibrium moved").

For the general reaction $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$:

Square brackets mean equilibrium concentration in ${\text{mol L}}^{-1}$. Pure solids and pure liquids (including water as solvent) are left out, because their concentration doesn't change. HSC answers usually give $K_{eq}$ without units.

What the value means:

• $K_{eq}\gg 1$: products are favoured; the reaction goes nearly to completion.
• $K_{eq}\ll 1$: reactants are favoured; very little product forms.
• $K_{eq}$ depends only on temperature. For an exothermic forward reaction, raising $T$ decreases $K_{eq}$; for an endothermic one it increases $K_{eq}$.
• Reversing the equation gives $1/K_{eq}$; doubling all coefficients squares $K_{eq}$.

The reaction quotient Q

$Q$ uses the same expression as $K_{eq}$ but with the concentrations at any moment. Compare it with $K_{eq}$:

• $Q<K_{eq}$: too few products, so the reaction proceeds forward.
• $Q>K_{eq}$: too many products, so the reverse reaction is favoured.
• $Q=K_{eq}$: the system is at equilibrium.

Worked example: ICE table

0.80 mol of ${\text{PCl}}_5$ is sealed in a 4.0 L flask and heated. At equilibrium the flask contains 0.30 mol of ${\text{Cl}}_2$. Calculate $K_{eq}$.

Step 1: convert to concentrations. Initial $[{\text{PCl}}_5]=0.80/4.0=0.20{\text{mol L}}^{-1}$; equilibrium $[{\text{Cl}}_2]=0.30/4.0=0.075{\text{mol L}}^{-1}$.

Step 2: ICE table (all in ${\text{mol L}}^{-1}$). The 1 : 1 : 1 ratio means 0.075 of ${\text{PCl}}_5$ reacted.

Step 3: substitute.

Step 4: use Q. At the same temperature another flask contains $[{\text{PCl}}_5]=0.10$, $[{\text{PCl}}_3]=0.15$ and $[{\text{Cl}}_2]=0.20{\text{mol L}}^{-1}$:

$Q>K_{eq}$, so the reaction shifts in the reverse direction: ${\text{PCl}}_5$ forms until $Q$ falls to 0.045.

Practical: determining Keq

In the iron(III) thiocyanate investigation, ${\text{Fe}}^{3+}$ and ${\text{SCN}}^{-}$ solutions are mixed and the equilibrium $[{\text{FeSCN}}^{2+}]$ is found with a colorimeter, using a calibration curve. ICE tables then give $[{\text{Fe}}^{3+}]$ and $[{\text{SCN}}^{-}]$. Several mixtures should give similar $K_{eq}$ values at the same temperature; this consistency is the evidence that $K_{eq}$ is constant.

Keq for acids and other systems

The same idea applies to the dissociation of weak acids and bases ($K_a$, Module 6), the ionisation of water ($K_w$) and saturated solutions ($K_{sp}$). They are all equilibrium constants with solids and water left out.

When an ionic solid dissolves, water molecules (polar) surround ions at the crystal surface, ion–dipole attractions overcome the lattice attractions, and ions move into solution as hydrated ions. Dissolving needs energy to break the lattice and releases energy as ions are hydrated; the balance between these, together with the entropy increase, decides solubility.

A saturated solution in contact with undissolved solid is at dynamic equilibrium. For a salt ${\text{M}}_x{\text{A}}_y$:

The solid is left out. $K_{sp}$ values at 25 °C are given on the data sheet.

Solubility from Ksp

Let $s$ be the molar solubility (mol of solid that dissolves per litre). Write each ion concentration in terms of $s$ using the coefficients, then substitute.

Worked example. Calculate the solubility of magnesium hydroxide at 25 °C in mol per litre and grams per litre, given $K_{sp}=5.61\times 10^{-12}$ and $M=58.33{\text{g mol}}^{-1}$.

$[{\text{Mg}}^{2+}]=s$ and $[{\text{OH}}^{-}]=2s$, so

In grams: $1.12\times 10^{-4}\times 58.33=6.53\times 10^{-3}{\text{g L}}^{-1}$.

The commonest error is forgetting to double $s$ before squaring it.

Comparing solubilities: you can only compare $K_{sp}$ values directly when the salts have the same ion ratio (for example AgCl and AgBr). For different ratios, calculate $s$ for each.

Will a precipitate form?

Calculate $Q$ (the ionic product) using concentrations after mixing, then compare with $K_{sp}$:

• $Q>K_{sp}$: the solution is supersaturated, so a precipitate forms.
• $Q\le K_{sp}$: no precipitate.

Worked example. 20.0 mL of $1.0\times 10^{-3}{\text{mol L}}^{-1}$ ${\text{AgNO}}_3$ is mixed with 30.0 mL of $2.0\times 10^{-3}{\text{mol L}}^{-1}$ ${\text{Na}}_2{\text{CO}}_3$. Does silver carbonate precipitate? ($K_{sp}=8.46\times 10^{-12}$)

The total volume is 50.0 mL, so each solution is diluted:

$Q>K_{sp}$, so a precipitate of ${\text{Ag}}_2{\text{CO}}_3$ forms.

The common ion effect

Adding an ion already in the equilibrium pushes it to the left (Le Chatelier), so the salt is less soluble. For magnesium hydroxide in $0.10{\text{mol L}}^{-1}$ NaOH, $[{\text{OH}}^{-}]\approx 0.10$, so $s=5.61\times 10^{-12}/0.10^2=5.6\times 10^{-10}{\text{mol L}}^{-1}$, about 200 000 times less than in pure water. Making $s$ small is how gravimetric analysis (Module 8) gets nearly all of an ion out of solution.

Solubility rules

Know the general patterns: all nitrates and all sodium, potassium and ammonium salts are soluble; most chlorides are soluble except those of silver and lead (lead chloride is slightly soluble); most sulfates are soluble except those of barium, lead and calcium (slightly); most carbonates and phosphates are insoluble except those of group 1 and ammonium; most hydroxides are insoluble except group 1 (barium hydroxide is soluble and calcium hydroxide slightly soluble). These rules drive the ion tests in Module 8.

Cycad seeds

Some Aboriginal and Torres Strait Islander Peoples prepare cycad seeds, which contain the water-soluble toxin cycasin, by crushing them and leaching them in running water for days to weeks. Crushing increases surface area, and running water keeps carrying away dissolved toxin, so the solution never reaches saturation and the dissolving equilibrium keeps shifting forward until the toxin is removed. Questions usually want you to link this to an open system and Le Chatelier's principle.

• Saying concentrations of reactants and products are equal at equilibrium. Rates are equal; concentrations are constant.
• Including solids or water in $K_{eq}$ or $K_{sp}$ expressions.
• Using moles instead of concentrations in $K_{eq}$ when the volume isn't 1 L.
• Forgetting that only temperature changes $K_{eq}$; a concentration or pressure change shifts the position but leaves $K_{eq}$ unchanged.
• Saying a catalyst "shifts equilibrium to the right" or "increases yield".
• In $K_{sp}$ problems, writing $(2s{)}^2$ as $2s^2$, or using concentrations from before mixing.
• Writing "the equilibrium shifts" as an observation. Observations are colour, temperature, bubbles or precipitate.
• Mixing kJ and J in $\Delta G=\Delta H-T\Delta S$, or using °C instead of K.

• Explain a shift: state the change, use Le Chatelier's principle to give the direction, justify it with collision theory or rates, and give the observable result. Three-mark answers usually reward all three of direction, reason and observation.
• Calculate $K_{eq}$: write the balanced equation and the expression first, show the ICE table, then substitute. Even with a wrong final number, the expression and table earn marks.
• Predict direction: calculate $Q$, compare it with $K_{eq}$ in words ("$Q>K_{eq}$ so…"), and state which way the reaction goes.
• For graph questions, annotate the graph itself if allowed and refer to specific times.
• Assess/evaluate questions on industrial conditions (such as ammonia production) need the trade-off between yield, rate and cost, ending in a judgement.
• Give answers to the precision of the data, usually 2 or 3 significant figures, and keep unrounded values in the calculator between steps.

• Module 6: $K_a$, $K_w$ and pH are equilibrium constants; buffers work by Le Chatelier's principle.
• Module 7: esterification is an equilibrium; excess reactant and removing water push it forward.
• Module 8: precipitation tests, gravimetric analysis and the colorimetry used in the $K_{eq}$ practical; industrial synthesis trade-offs between equilibrium yield and rate.
• Year 11 Module 4: enthalpy, entropy, Gibbs free energy and collision theory are assumed knowledge.