Equilibrium

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 6.

Why does a cold drink fizz when you open it?

Inside a sealed bottle, CO₂ is squeezed into the liquid under high pressure. The gas going into the drink and the gas leaving it happen at the same rate, so nothing seems to change. Twist the cap and the pressure above the liquid drops. The balance is broken, CO₂ leaves faster than it enters, and you see the bubbles. That balance, and the way it shifts when you disturb it, is the whole chapter. First we meet it for physical changes, then for chemical reactions, and finally for the ions in water that decide pH, buffers and why some salts barely dissolve.

Physical equilibrium: two opposite changes, one steady state

Equilibrium is dynamic, not static. Both the forward and the reverse processes keep going, but at equal rates, so nothing measurable changes. A closed system is essential, because a vapour that escapes can never be balanced by condensation.

A liquid with a higher vapour pressure at a given temperature is more volatile and has a lower boiling point. Henry's law describes the last row: the mass of a gas dissolved in a fixed mass of solvent at a given temperature is proportional to the pressure of the gas above it. That is exactly why the cap matters.

Chemical equilibrium: reactions that never really finish

In a closed vessel, N₂ + 3H₂ ⇌ 2NH₃ does not run until the reactants are used up. As ammonia builds up, the reverse reaction speeds up, and the forward reaction slows down as N₂ and H₂ are consumed. At some stage the two rates become equal and the composition freezes. Two facts show that the reaction is still alive underneath. If you mix an equilibrium mixture of N₂, H₂ and NH₃ with one of N₂, D₂ and ND₃ at the same temperature, partly deuterated ammonia such as NH₂D and NHD₂ soon appears, which can happen only if NH₃ keeps breaking up and re-forming. And if you start with H₂ and I₂, or start with pure HI at the same temperature, you finish with the same equilibrium mixture of H₂, I₂ and HI. The state is set by the conditions, not by the starting side.

The equilibrium constant

For any reversible reaction aA + bB ⇌ cC + dD, the law of chemical equilibrium (Guldberg and Waage, 1864) says that at a given temperature the ratio below is constant:

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

where the brackets are equilibrium concentrations in mol L⁻¹.

Products on top, reactants below, each raised to its stoichiometric coefficient. Pure solids and pure liquids do not appear, because their concentration (density divided by molar mass) is fixed and is absorbed into K.

Reactions with all species in one phase are homogeneous (N₂ + 3H₂ ⇌ 2NH₃). Those with more than one phase are heterogeneous (CaCO₃ decomposition). If you rewrite a reaction, K changes in a predictable way.

Kp for gases. For gases you can use partial pressures, and since $p = (n/V)RT$, the two constants are linked by

$$K_p = K_c\,(RT)^{\Delta n}$$

where Δn = moles of gaseous products − moles of gaseous reactants

Pressure is taken in bar, so R = 0.0831 bar L mol⁻¹ K⁻¹. Count only gases when finding Δn. Strictly, K has no units because it is defined against standard states of 1 mol L⁻¹ and 1 bar, so quote units only when a question does.

Worked example: Kp from Kc. For 2NO(g) + Cl₂(g) ⇌ 2NOCl(g) at 1069 K, Kc = 3.75 × 10⁻⁶. Find Kp.

The common wrong step: using $(RT)^{+1}$ because the number of moles is "changing", or counting solids and liquids in Δn.

What K tells you: how far, and which way

K is a verdict on extent. It says nothing about speed.

Thermodynamics gives the same message in a different language, since ΔG° = −RT ln K: a negative ΔG° means K greater than 1. To tell where a mixture that is not yet at equilibrium will go, calculate the reaction quotient Q, which has the same form as K but uses whatever concentrations are present now.

Worked example: direction and composition. For H₂ + I₂ ⇌ 2HI, Kc = 57.0 at 700 K. (a) A mixture has [H₂] = 0.10 M, [I₂] = 0.20 M and [HI] = 0.40 M. Which way does it go? (b) 1.00 mol each of H₂ and I₂ are placed in a 1 L vessel. Find the equilibrium concentrations.

The common wrong step: writing the change in HI as +x instead of +2x. The coefficient decides the change in the ICE table. Check at the end by putting your numbers back in Kc.

Le Chatelier's principle: how a system answers a push

If a system at equilibrium is disturbed, it shifts in the direction that reduces the disturbance. Use it as a quick predictor.

Only a temperature change alters K. K rises with temperature for an endothermic reaction and falls for an exothermic one. A pressure change affects only reactions in which the number of gas moles changes, and solids or liquids are ignored when counting. A catalyst lowers the activation energy of both directions equally, so it affects how fast equilibrium is reached and not where it settles. Inert gas at constant pressure is different: the volume grows, so partial pressures fall and the equilibrium moves to the side with more gas moles.

Pause and predict: argon is pumped into a sealed rigid vessel holding N₂, H₂ and NH₃ in equilibrium, at constant temperature. Does the ammonia yield change?

Check your answer: no. The volume is fixed, so the concentrations and the partial pressures of the three gases are exactly what they were. The reaction quotient still equals K, so nothing shifts.

Colour makes it visible. N₂O₄(g) (colourless) ⇌ 2NO₂(g) (brown) is endothermic in the forward direction. Heat the mixture and it darkens as NO₂ is favoured. Cool it and it fades. Compress it and, after the first flash of darker colour from the higher concentration, it fades as the equilibrium moves to the side with fewer gas moles, N₂O₄.

Beyond the textbook. For the exam: for N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92.4 kJ mol⁻¹), high pressure and low temperature favour ammonia, and a catalyst speeds things up. In reality: at low temperature the reaction is far too slow to be useful, so the Haber process runs at about 700 K and roughly 200 atm with an iron catalyst. Ammonia is removed as it forms. Why it matters: industry always trades yield against speed, and K and rate are two separate questions.

Ionic equilibrium: electrolytes and acid–base ideas

Strong electrolytes (NaCl, HCl, NaOH) ionise completely in water. Weak electrolytes (CH₃COOH, NH₃) ionise only partly, so an equilibrium exists between the molecules and their ions. Three definitions of acids and bases are in use, each wider than the one before.

In the Brønsted–Lowry picture every acid has a conjugate base (one H⁺ less) and every base a conjugate acid (one H⁺ more).

Species such as H₂O, HCO₃⁻ and HSO₄⁻ can act as either acid or base and are called amphoteric. The golden rule: the stronger the acid, the weaker its conjugate base. Cl⁻ is so weak a base that it hardly picks up H⁺, while the conjugate base of a very weak acid like HCN is a fairly strong base. Strong acids are HClO₄, HCl, HBr, HI, HNO₃ and H₂SO₄ (first step). BF₃ is a Lewis acid because boron has an empty orbital, and in BF₃ + :NH₃ → F₃B←NH₃ the lone pair of nitrogen is donated to it.

Water, Kw and the pH scale

Pure water ionises very slightly: H₂O + H₂O ⇌ H₃O⁺ + OH⁻. The product of the ion concentrations is the ionic product of water:

$$K_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14}\ \text{(at 298 K)}$$

In pure water [H⁺] = [OH⁻] = 10⁻⁷ mol L⁻¹. Writing such tiny numbers is awkward, so we use logarithms.

$$\text{pH} = -\log[\text{H}^+], \qquad \text{pOH} = -\log[\text{OH}^-], \qquad \text{pH} + \text{pOH} = 14\ \text{(298 K)}$$

At 298 K, pH below 7 is acidic, 7 is neutral and above 7 is basic. Because the scale is logarithmic, each pH unit means a tenfold change in [H⁺].

Pause and predict: what is the pH of a 1.0 × 10⁻⁸ M HCl solution?

Check your answer: not 8. An acid can never give a pH above 7. At this dilution the H⁺ from the water itself (1 × 10⁻⁷ M) is no longer negligible. Solving $[\text{H}^+] = 10^{-8} + K_w/[\text{H}^+]$ gives [H⁺] ≈ 1.05 × 10⁻⁷ M and pH ≈ 6.98.

Beyond the textbook. For the exam: neutral means pH = 7 and Kw = 10⁻¹⁴. In reality: Kw rises with temperature because ionisation of water is endothermic. At 100 °C, Kw is about 5 × 10⁻¹³ and the pH of pure water is about 6.1. Why it matters: the water is still neutral there, since [H⁺] = [OH⁻]. Neutral means equal ions, and 7 is just where this happens at 298 K.

Weak acids and bases: Ka, Kb and the degree of ionisation

For a weak acid HA ⇌ H⁺ + A⁻, the acid ionisation constant is

$$K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$$

If the initial concentration is c and a fraction α ionises, then $K_a = c\alpha^2/(1-\alpha)$, which for a small α becomes $K_a = c\alpha^2$. So

$$\alpha = \sqrt{\dfrac{K_a}{c}}, \qquad [\text{H}^+] = c\alpha = \sqrt{K_a c}$$

The same applies to a weak base with Kb and [OH⁻]. A larger Ka (or smaller pKa) means a stronger acid. Dilution increases α.

Worked example: pH of a weak acid. Find the pH and the degree of ionisation of 0.10 M CH₃COOH (Ka = 1.8 × 10⁻⁵).

The common wrong step: treating a weak acid as strong and writing [H⁺] = 0.10 M, which gives pH = 1.

Conjugate pair link. For an acid and its conjugate base,

$$K_a \times K_b = K_w, \qquad pK_a + pK_b = 14$$

This is why a very strong acid has a negligible conjugate base.

What makes a stronger acid. Down a group, the H–A bond weakens as A gets bigger, so acidity runs HF < HCl < HBr < HI. Across a period, increasing electronegativity of A stabilises the anion, so acidity runs CH₄ < NH₃ < H₂O < HF. Polyprotic acids lose protons one at a time, and each step is harder because you pull H⁺ off an ion that is already negative: for H₃PO₄, Ka₁ = 7.5 × 10⁻³, Ka₂ = 6.2 × 10⁻⁸ and Ka₃ = 4.2 × 10⁻¹³.

Common ion effect

Add a substance that supplies an ion already present in a weak electrolyte's equilibrium, and Le Chatelier tells you the equilibrium shifts back, suppressing ionisation. Add sodium acetate to acetic acid and the acetate ion pushes CH₃COOH ⇌ H⁺ + CH₃COO⁻ to the left. In 0.10 M acetic acid, $K_a = [\text{H}^+][\text{A}^-]/[\text{HA}]$ gives $[\text{H}^+] = 1.8\times10^{-5}\times 0.10/0.10 = 1.8\times10^{-5}$ once 0.10 M sodium acetate is added, so the pH jumps from 2.87 to 4.74. Likewise, adding NH₄Cl to ammonia solution lowers [OH⁻] and the pH.

Pause and predict: NH₄Cl is dissolved in a solution of NH₃. Does the pH go up or down?

Check your answer: down. NH₄⁺ is a common ion in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, so the equilibrium shifts left and [OH⁻] falls.

Salt hydrolysis: when salts are not neutral

Ions of a salt may react with water, and the result decides the pH. The ion of a strong acid or strong base does nothing. The ion of a weak acid or weak base grabs from or gives to the water.

The winner is "the stronger parent". A salt takes the character of whichever of its parent acid or base is stronger. For the weak acid and weak base case the final pH depends on which of Ka and Kb is larger. For a salt of a weak acid and strong base at concentration c, $\text{pH} = 7 + \tfrac{1}{2}(pK_a + \log c)$.

Buffer solutions

A buffer resists a change in pH when small amounts of acid or base are added, or when it is diluted. It always contains a weak acid and its conjugate base (or a weak base and its conjugate acid), so it has something ready to soak up either H⁺ or OH⁻.

When [salt] = [acid], pH = pKa, and a buffer works best within about one pH unit of the pKa. Blood is buffered at about 7.4.

Worked example: a buffer at work. A litre of buffer contains 0.10 mol CH₃COOH and 0.10 mol CH₃COONa (pKa = 4.74). Find its pH, then the pH after adding 0.01 mol HCl.

The buffer moved by only 0.09 units where water would have moved by five. For a basic buffer, say 0.10 M NH₃ with 0.10 M NH₄Cl (pKb = 4.75), pOH = 4.75 and pH = 9.25.

Solubility product: how little is "insoluble"?

In a saturated solution of a sparingly soluble salt, the solid and its ions are in equilibrium. For a salt MₓXᵧ ⇌ xMᵖ⁺ + yX^q⁻, the solid is left out of the expression:

$$K_{sp} = [\text{M}^{p+}]^x\,[\text{X}^{q-}]^y$$

If the molar solubility is S mol L⁻¹, then $[\text{M}^{p+}] = xS$ and $[\text{X}^{q-}] = yS$, so

$$K_{sp} = x^x\,y^y\,S^{x+y}$$

If the ionic product exceeds Ksp the salt precipitates, and if it is less the solution is unsaturated. The common ion effect works here too: a common ion cuts the solubility.

Worked example: solubility and the common ion. For Ni(OH)₂, Ksp = 2.0 × 10⁻¹⁵. Find the solubility in pure water and in 0.10 M NaOH.

The solubility has fallen by about 40 million times. The common wrong step: writing $K_{sp} = S \times S$ when the formula has two hydroxides. Always put the coefficient into the concentration and then raise it to the power.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Before this: Thermodynamics gives ΔG° = −RT ln K, which explains why a reaction has the K it does.
• Next: Redox Reactions moves from proton transfer to electron transfer, another kind of balance between two species.
• Forward link: Chemical Kinetics explains how fast equilibrium is reached, and Electrochemistry uses equilibrium ideas for cells and the Nernst equation.