Equilibrium isn't limited to chemical reactions — many physical processes also reach a state of balance. In a closed container, when a liquid is left to evaporate, molecules leave the liquid surface (evaporation) while some vapour molecules return to the liquid (condensation). Initially evaporation dominates, but as vapour builds up, condensation speeds up until both rates become equal — this is physical equilibrium.
Solid–Liquid: $\text{H}_2\text{O(s)} \rightleftharpoons \text{H}_2\text{O(l)}$ at the melting point, where the rate of melting equals the rate of freezing. Liquid–Vapour: $\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{O(g)}$ in a closed vessel, where the rate of evaporation equals the rate of condensation, giving a constant vapour pressure at a given temperature. Solid–Vapour (sublimation): e.g., $\text{I}_2\text{(s)} \rightleftharpoons \text{I}_2\text{(g)}$. Dissolution equilibrium: a saturated solution in contact with undissolved solute, where the rate of dissolution equals the rate of crystallisation.
Many chemical reactions don't go to completion — they proceed only partway before reactants and products reach a state where their concentrations no longer change, even though the reaction hasn't stopped. Such reactions are called reversible reactions, shown with a double arrow: $\rightleftharpoons$.
Chemical equilibrium is reached when the rate of the forward reaction becomes exactly equal to the rate of the reverse reaction, so the concentrations of all reactants and products remain constant over time (at constant temperature). Like physical equilibrium, this is a dynamic state — both reactions keep occurring, just at matched rates.
Example: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ — starting with only N₂ and H₂, the forward reaction is fast initially; as NH₃ builds up, the reverse reaction (decomposition of NH₃) speeds up, until eventually both rates match and concentrations stop changing.
Equilibrium does not mean equal concentrations of reactants and products — it means the concentrations have stopped changing. The equilibrium mixture can be almost all products, almost all reactants, or anywhere in between, depending on the reaction.
For a general reversible reaction $aA + bB \rightleftharpoons cC + dD$ at a given temperature, the ratio of the product of equilibrium concentrations of products (raised to their stoichiometric powers) to the product of equilibrium concentrations of reactants (raised to their stoichiometric powers) is a constant, called the equilibrium constant, $K_c$.
Concentrations here refer to equilibrium (molar) concentrations, usually in mol/L, and each is raised to the power of its coefficient in the balanced equation.
Write the equilibrium constant expression for $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$, and calculate Kc if at equilibrium [SO₂] = 0.2 M, [O₂] = 0.1 M, [SO₃] = 0.4 M.
$K_c = \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \dfrac{(0.4)^2}{(0.2)^2(0.1)} = \dfrac{0.16}{0.004} = 40$
For the reaction $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$, at equilibrium [N₂O₄] = 0.05 M and [NO₂] = 0.02 M. Find Kc.
$K_c = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \dfrac{(0.02)^2}{0.05} = \dfrac{0.0004}{0.05} = 0.008$
A homogeneous equilibrium has all reactants and products in the same physical phase (e.g., all gases, as in the $\text{N}_2/\text{H}_2/\text{NH}_3$ system). A heterogeneous equilibrium involves reactants/products in more than one phase (e.g., a solid and a gas).
The concentration (or activity) of a pure solid or pure liquid is taken as constant and is omitted from the equilibrium constant expression, since its concentration doesn't change no matter how much of it is present.
Write the equilibrium constant expression for the decomposition of solid calcium carbonate: $\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$.
Since CaCO₃ and CaO are pure solids, they are omitted: $K_c = [\text{CO}_2]$, or in terms of pressure, $K_p = p_{\text{CO}_2}$.
For reactions involving gases, the equilibrium constant can be expressed either in terms of molar concentrations ($K_c$) or partial pressures ($K_p$).
where $\Delta n_g$ = (moles of gaseous products) − (moles of gaseous reactants), $R$ = gas constant (0.0821 L atm/mol K when pressure is in atm), and $T$ = absolute temperature.
If $\Delta n_g = 0$, then $K_p = K_c$.
For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ at 500 K, $K_c = 6.0 \times 10^{-2}$. Calculate $K_p$.
$\Delta n_g = 2 - (1+3) = -2$
$K_p = K_c(RT)^{-2} = (6.0\times10^{-2})(0.0821 \times 500)^{-2} = (6.0\times10^{-2})(41.05)^{-2} \approx 3.56 \times 10^{-5}$
If $K_c$ for $\text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO}$ is 4, find $K_c$ for $2\text{NO} \rightleftharpoons \text{N}_2 + \text{O}_2$ and for $\tfrac{1}{2}\text{N}_2 + \tfrac{1}{2}\text{O}_2 \rightleftharpoons \text{NO}$.
Reverse reaction: $K = 1/4 = 0.25$
Halved equation: $K = \sqrt{4} = 2$
The reaction quotient, Q, has the same mathematical form as K, but uses concentrations at any point in the reaction, not necessarily at equilibrium. Comparing Q to K tells us which direction a reaction must proceed to reach equilibrium.
| Comparison | What It Means |
|---|---|
| Q < K | Too few products relative to equilibrium — reaction proceeds forward |
| Q > K | Too many products relative to equilibrium — reaction proceeds in reverse |
| Q = K | System is already at equilibrium — no net change |
For $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$, $K_c = 50$ at a given temperature. If a mixture has [H₂] = 0.1 M, [I₂] = 0.1 M, [HI] = 0.5 M, predict the direction of reaction.
$Q = \dfrac{(0.5)^2}{(0.1)(0.1)} = \dfrac{0.25}{0.01} = 25$
Since $Q (25) < K (50)$, the reaction will proceed in the forward direction to produce more HI.
If a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the equilibrium shifts in the direction that partially counteracts that change, establishing a new equilibrium.
Adding more reactant shifts equilibrium forward (towards products) to partly consume the added reactant. Removing a product (or adding more reactant) has the same forward-shifting effect; removing a reactant or adding more product shifts equilibrium in reverse.
For gaseous equilibria, increasing pressure (decreasing volume) shifts equilibrium towards the side with fewer moles of gas, since that side occupies less volume and partly relieves the pressure increase. If both sides have equal moles of gas, pressure changes have no effect on the equilibrium position.
Increasing temperature shifts equilibrium in the endothermic direction (absorbing the extra heat); decreasing temperature shifts it in the exothermic direction. So for an exothermic forward reaction, raising temperature shifts equilibrium backward (decreasing K), and vice versa for endothermic reactions.
A catalyst speeds up the approach to equilibrium but does not shift its position. Adding an inert gas at constant volume doesn't change partial pressures of reactants/products, so it has no effect on equilibrium; but adding inert gas at constant pressure increases total volume, effectively diluting reactants and products, which shifts equilibrium towards the side with more moles of gas.
For the exothermic Haber process, $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, $\Delta H < 0$, predict the effect of (a) increasing pressure and (b) increasing temperature on NH₃ yield.
(a) Reactant side has 4 moles of gas, product side has 2 — increasing pressure shifts equilibrium towards the side with fewer moles, i.e., forward, increasing NH₃ yield.
(b) The forward reaction is exothermic; increasing temperature shifts equilibrium in the endothermic (reverse) direction, decreasing NH₃ yield — this is why industrial ammonia synthesis uses a moderate, not extremely high, temperature.
| Concept | Acid | Base |
|---|---|---|
| Arrhenius | Increases [H⁺] in water | Increases [OH⁻] in water |
| Brønsted-Lowry | Proton (H⁺) donor | Proton (H⁺) acceptor |
| Lewis | Electron-pair acceptor | Electron-pair donor |
In the Brønsted-Lowry concept, when an acid donates a proton, it becomes its conjugate base; when a base accepts a proton, it becomes its conjugate acid. E.g., $\text{HCl} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{Cl}^-$ — HCl/Cl⁻ and H₂O/H₃O⁺ are conjugate acid-base pairs. A strong acid always has a very weak conjugate base, and vice versa.
The Lewis concept extends beyond species with H⁺ — e.g., $\text{BF}_3$ is a Lewis acid (accepts an electron pair, incomplete octet) even though it has no H to donate, while $\text{NH}_3$ is a Lewis base (donates its lone pair).
Strong acids/bases (HCl, HNO₃, NaOH) ionize almost completely in water. Weak acids/bases (CH₃COOH, NH₄OH) ionize only partially, establishing an ionic equilibrium characterised by an ionization constant, $K_a$ (for acids) or $K_b$ (for bases).
A larger $K_a$ means a stronger (more ionized) acid. Similarly, $K_b$ for a weak base measures its strength.
The fraction of the total initial concentration that has ionized at equilibrium. If initial concentration is $C$ and degree of dissociation is $\alpha$, then at equilibrium: $[\text{H}^+] = C\alpha$, $[\text{A}^-] = C\alpha$, $[\text{HA}] = C(1-\alpha)$.
This shows that as a weak electrolyte is diluted (C decreases), its degree of dissociation α increases.
Calculate the degree of dissociation and [H⁺] for 0.1 M acetic acid ($K_a = 1.8 \times 10^{-5}$).
$\alpha \approx \sqrt{K_a/C} = \sqrt{(1.8\times10^{-5})/0.1} = \sqrt{1.8\times10^{-4}} = 0.0134$
$[\text{H}^+] = C\alpha = 0.1 \times 0.0134 = 1.34\times10^{-3}\ \text{M}$
A 0.01 M solution of a weak acid has a degree of dissociation of 0.042. Calculate its Ka.
$K_a \approx C\alpha^2 = (0.01)(0.042)^2 = (0.01)(0.001764) = 1.764\times10^{-5}$
Water undergoes self-ionization: $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$. At 25°C, $K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}$. In pure water, $[\text{H}^+] = [\text{OH}^-] = 10^{-7}\ \text{M}$.
pH < 7 → acidic; pH = 7 → neutral; pH > 7 → basic (at 25°C).
Calculate the pH of a 0.001 M HCl solution (assume complete ionization).
$[\text{H}^+] = 0.001 = 10^{-3}\ \text{M}$. $\text{pH} = -\log(10^{-3}) = 3$
Calculate the pH of a 0.01 M NaOH solution (assume complete ionization).
$[\text{OH}^-] = 0.01 = 10^{-2}\ \text{M}$. $\text{pOH} = -\log(10^{-2}) = 2$. $\text{pH} = 14 - 2 = 12$
The common ion effect is the suppression of the ionization of a weak electrolyte when a strong electrolyte containing an ion common to it is added to the solution — a direct consequence of Le Chatelier's principle.
Example: adding sodium acetate (which fully dissociates to give CH₃COO⁻) to a solution of acetic acid pushes the equilibrium $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$ backward, since the added CH₃COO⁻ is a common ion — this decreases the degree of ionization of acetic acid and lowers [H⁺].
Explain qualitatively why adding NH₄Cl to a solution of NH₄OH decreases the solution's pH.
NH₄Cl fully dissociates to give NH₄⁺, which is common to the NH₄OH equilibrium ($\text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$). The excess NH₄⁺ shifts this equilibrium backward (Le Chatelier), reducing [OH⁻] — a lower [OH⁻] means a lower pOH is not achieved, so pH decreases (moves closer to neutral/acidic) compared to pure NH₄OH.
A buffer solution resists significant changes in pH when small amounts of acid or base are added to it. An acidic buffer is made from a weak acid and its salt with a strong base (e.g., CH₃COOH + CH₃COONa); a basic buffer is made from a weak base and its salt with a strong acid (e.g., NH₄OH + NH₄Cl).
For a basic buffer: $\text{pOH} = \text{p}K_b + \log\dfrac{[\text{Salt}]}{[\text{Base}]}$
Adding a small amount of acid gets consumed by the buffer's conjugate base component (converting it to the weak acid form); adding a small amount of base gets consumed by the weak acid component (converting it to the conjugate base form). Because the buffer has a large reservoir of both forms, small additions don't drastically shift the ratio, so pH barely changes.
Calculate the pH of a buffer containing 0.1 M acetic acid and 0.1 M sodium acetate ($K_a = 1.8 \times 10^{-5}$, $\text{p}K_a = 4.74$).
$\text{pH} = \text{p}K_a + \log\dfrac{[\text{Salt}]}{[\text{Acid}]} = 4.74 + \log\dfrac{0.1}{0.1} = 4.74 + 0 = 4.74$
Calculate the pH of a buffer with 0.2 M acetic acid and 0.05 M sodium acetate (pKa = 4.74).
$\text{pH} = 4.74 + \log\dfrac{0.05}{0.2} = 4.74 + \log(0.25) = 4.74 - 0.602 = 4.14$
For a sparingly soluble salt in equilibrium with its saturated solution, e.g., $\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^-$, the solubility product is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient: $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$.
| Comparison | Result |
|---|---|
| Ionic product < Ksp | Unsaturated solution — more solid can dissolve |
| Ionic product = Ksp | Saturated solution — at equilibrium |
| Ionic product > Ksp | Precipitation occurs until equilibrium is restored |
The solubility of AgCl is $1.3 \times 10^{-5}$ mol/L. Calculate its Ksp.
Let solubility $= s$. $[\text{Ag}^+] = [\text{Cl}^-] = s = 1.3\times10^{-5}$
$K_{sp} = s^2 = (1.3\times10^{-5})^2 = 1.69\times10^{-10}$
Ksp of $\text{Mg(OH)}_2$ is $8.9 \times 10^{-12}$. Calculate its molar solubility.
$\text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^-$. If solubility = $s$, then $[\text{Mg}^{2+}] = s$, $[\text{OH}^-] = 2s$.
$K_{sp} = s(2s)^2 = 4s^3 \Rightarrow s^3 = \dfrac{8.9\times10^{-12}}{4} = 2.225\times10^{-12}$
$s = (2.225\times10^{-12})^{1/3} \approx 1.3\times10^{-4}\ \text{mol/L}$
Adding a common ion (e.g., adding NaCl to a saturated AgCl solution) decreases AgCl's solubility, since the increased Cl⁻ concentration pushes the dissolution equilibrium backward (Le Chatelier) — this is exploited in qualitative salt analysis to selectively precipitate ions.