5.3.3Additional lessons
The Equilibrium Constant — IB Chemistry SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 75 of 98
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.
Section 1 of 5What the Size of K Tells Us
K is the ratio [products]/[reactants] at equilibrium, so a large value means a product-rich mixture.
What the Size of K Tells Us
K much less than 1
- K much less than 1 Very few products form, so the equilibrium lies far to the left and the mixture contains mostly reactants.
K less than 1
- K less than 1 Reactants are favoured and the position of equilibrium lies to the left.
K around 1
- K around 1 Significant amounts of both reactants and products are present at equilibrium.
K greater than 1
- K greater than 1 Products are favoured, so the position of equilibrium lies to the right.
K much greater than 1
- K much greater than 1 The reaction goes almost to completion and the mixture is mostly products.
| Value of K | Position of equilibrium | Mixture contains |
|---|---|---|
| ≪ 1 | Far left | Mostly reactants |
| ≈ 1 | Balanced | Both in significant amounts |
| ≫ 1 | Far right | Mostly products |
Section 2 of 5Changing the Equation
Reversing the equation
- Reversing the equation The new constant is the reciprocal of the old one, K′ = 1/K, because products and reactants swap.
Multiplying coefficients
- Multiplying coefficients If every coefficient is multiplied by n, the new constant is Kⁿ, since each power is multiplied by n.
Same temperature
- Same temperature These relationships only apply at one fixed temperature, because K itself depends on temperature.
Rate and K
- Rate and K K says nothing about how fast equilibrium is reached, only about the composition once it is.
Section 3 of 5Temperature and the Value of K
1
K is constant at a given temperature and does not change when concentrations or pressure change.
2
For an endothermic forward reaction, raising the temperature moves the position to the right, so K increases.
3
For an exothermic forward reaction, raising the temperature moves the position to the left, so K decreases.
4
A catalyst speeds up both directions equally, so it does not change K or the position of equilibrium.
Section 4 of 5Worked Example
Question
- For H₂(g) + I₂(g) ⇌ 2HI(g) the equilibrium concentrations are [H₂] = 0.0200 mol dm⁻³, [I₂] = 0.0200 mol dm⁻³ and [HI] = 0.150 mol dm⁻³.
- Calculate K, then K for 2HI(g) ⇌ H₂(g) + I₂(g).
- K = [HI]² / ([H₂][I₂]) = (0.150)² / (0.0200 × 0.0200).
- Numerator = 0.0225 and denominator = 4.00 × 10⁻⁴, so K = 56.25, which is 56 to two significant figures.
- The reverse equation has K′ = 1/K = 1/56.25 = 0.0178.
Answer K = 56 (no units, since the powers cancel) and K′ = 0.018, so the reverse reaction favours the reactant side.
Section 5 of 5Exam Checklist & Tips
5.3.3 – Key Points for the Exam
- Position of equilibrium The relative proportions of reactants and products in the equilibrium mixture.
- Reciprocal relationship K for a reversed reaction equals 1/K of the forward reaction at the same temperature.
- Endothermic reaction
- A reaction that absorbs heat;
- K increases as temperature rises.
- A reaction that absorbs heat;
- Exothermic reaction
- A reaction that releases heat;
- K decreases as temperature rises.
- A reaction that releases heat;
Common Mistakes
- Believing that a large K means the reaction is fast; K only describes the extent of reaction.
- Thinking that changing concentration or pressure changes K; only temperature does.
- Forgetting to invert K when the equation is written in the reverse direction.
- Claiming that a catalyst increases K or the yield at equilibrium.
Last updated:
3 & IGCSE —
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