5.3.2Additional lessons
The Equilibrium Law — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 110 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 5Writing the Expression
The equilibrium law states that the ratio of product to reactant concentrations, each raised to the power of its coefficient, is constant at a given temperature.
General case
- General case For aA + bB ⇌ cC + dD the expression is Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ), using equilibrium concentrations in mol dm⁻³.
Products on top
- Products on top Products of the forward reaction go in the numerator and reactants in the denominator.
Powers
- Powers Each concentration is raised to the power of its coefficient in the balanced equation.
Fixed temperature
- Fixed temperature Kc has one value at a given temperature and changes only when the temperature changes.
Part 2 of 5Special Cases
Special Cases
Solids and liquids
- Solids and liquids Pure solids and pure liquids are left out of the expression because their concentrations do not change.
Aqueous and gases
- Aqueous and gases Only species that are gases or dissolved in solution (aq) appear in Kc.
Units
- Units The units depend on the powers, for example N₂O₄(g) ⇌ 2NO₂(g) has Kc in mol dm⁻³, while some equilibria have no units.
Pressure form
- Pressure form For gases an expression using partial pressures, Kp, can also be written, but only Kc is needed here.
Part 3 of 5Changing the Equation
1
If the equation is reversed, the new constant is the reciprocal of the original: K′ = 1/K.
2
If all the coefficients are multiplied by n, the constant is raised to the power n: K′ = Kⁿ.
3
Halving the coefficients therefore gives the square root of the original constant.
4
If two equations are added together, their constants are multiplied.
5
Always state the balanced equation with the constant, since the value of K depends on how the equation is written.
Part 4 of 5Worked Example
Question
- At a given temperature the equilibrium concentrations for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) are [SO₂] = 0.050, [O₂] = 0.020 and [SO₃] = 0.10 mol dm⁻³.
- Calculate Kc and Kc for the reverse reaction.
- Kc = [SO₃]² / ([SO₂]²[O₂]) = (0.10)² / ((0.050)² × 0.020).What the symbols mean: Kc = equilibrium constant in terms of concentration · [ ] = concentration at equilibrium (mol dm−3) · each power is the number in front of that substance in the balanced equation
- Numerator = 0.010; denominator = 0.0025 × 0.020 = 5.0 × 10⁻⁵, so Kc = 200.
- Units
- mol⁻¹ dm³.
- For the reverse reaction K′ = 1/200 = 5.0 × 10⁻³.
Answer
- Kc = 200 mol⁻¹ dm³;
- K′ = 5.0 × 10⁻³ mol dm⁻³
Part 5 of 5Exam Checklist & Tips
5.3.2 – Key Points for the Exam
- Equilibrium law At constant temperature, the ratio of product to reactant concentrations raised to their coefficients is constant.
- Kc The equilibrium constant expressed using concentrations at equilibrium.
- Homogeneous equilibrium An equilibrium in which all species are in the same phase.
- Heterogeneous equilibrium An equilibrium involving species in more than one phase.
Common Mistakes
- Putting pure solids and liquids into the Kc expression.
- Writing the reactant concentrations on the top of the fraction.
- Using initial concentrations instead of equilibrium concentrations.
- Forgetting to change K when the equation is reversed or multiplied.
Last updated:
3 & IGCSE —
Kc
- : Kc! (ΔH < 0) Kc. (ΔH > 0) Kc.
- : ( ).
- : Kc .