4.3Unit 4. Rates, Equilibria and Further Organic Chemistry
Chemical Equilibria — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 12 parts Lesson 3 of 56 1 quiz
Curriculum Equation Key معادلة تفاعل تكوين الإستر (الأسترة) - Esterification Reaction EquationFast Exam Solution
R-COOH + R2-OH ⇌ (conc. H₂SO₄) R-COO-R2 + H₂O
Word Equation & Conditions: حمض كربوكسيلي + كحول (في وجود حمض الكبريتيك المركز كعامل حفاز ونازع للماء) ⇌ إستر + ماء
Location & Specifications: الكيمياء العضوية للثانوية العامة والتحصيلي و Edexcel Chemistry.
Lesson summary
4.3 Chemical Equilibria — Kc, Kp, effect of conditions, temperature and entropy — IAL EDEXCEL LINEAR (CHEMISTRY) notes: Write Kc and Kp expressions, calculate K and its units, explain which changes affect K, and relate K to total entropy change.For aA + bB ⇌ cC + dD, Kc = [C]c[D]…
- What the symbols mean
- Partial pressure
- Equilibrium constant, Kc
- Equilibrium constant, Kp
- Mole fraction
- Heterogeneous equilibrium
Part 1 of 10Equilibrium constant expression
What the symbols mean:
For aA + bB ⇌ cC + dD, Kc = [C]c[D]d ÷ [A]a[B]b.
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
The concentrations are equilibrium concentrations in mol dm−3, raised to the powers of their coefficients.
Solids (and pure liquids in heterogeneous systems) are omitted, e.g. Ag+ + Fe2+ ⇌ Ag(s) + Fe3+ has Kc = [Fe3+] ÷ [Fe2+][Ag+].
K is specific to a reaction and changes only if the temperature changes.
Examples: N2 + 3H2 ⇌ 2NH3 has Kc = [NH3]2 ÷ [N2][H2]3; 2SO2 + O2 ⇌ 2SO3 has Kc = [SO3]2 ÷ [SO2]2[O2].
Part 2 of 10Calculating Kc
1
Write the balanced equation and the Kc expression.
2
If initial and change data are given, use an initial / change / equilibrium table with the mole ratio.
3
Divide equilibrium moles by the volume (dm3) to find concentrations.
4
Substitute into the expression.
5
Work out the units by cancelling the mol dm−3 terms.
Calculating Kc
When the powers cancel, Kc has no units, as in the esterification CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O.
Quote the answer to the smallest number of significant figures in the data.
Part 3 of 10Kp expressions
What the symbols mean:
For gaseous equilibria, Kp uses partial pressures: Kp = p(C)cp(D)d ÷ p(A)ap(B)b.
What the symbols mean: Kp = equilibrium constant in terms of partial pressures · p = partial pressure of each gas at equilibrium (Pa or kPa) · powers come from the balanced equation
Solids and liquids are omitted, e.g. CaCO3(s) ⇌ CaO(s) + CO2(g) has Kp = p(CO2).
The units of Kp depend on the expression, e.g. Pa−1 for 2SO2 + O2 ⇌ 2SO3, or none for H2 + I2 ⇌ 2HI.
Kp is constant at a given temperature.
Part 4 of 10Partial pressures and mole fractions
- Partial pressure
- the pressure a gas would exert if it alone occupied the container.
1
Add up the moles of all gases to find the total.
2
Mole fraction of a gas = its moles ÷ total moles.
3
Partial pressure = mole fraction × total pressure.
4
Substitute the partial pressures into the Kp expression.
- The sum of the partial pressures equals the total pressure.
Part 5 of 10Effect of conditions on K and position
Effect of conditions on K and position
Changing concentration or pressure changes the reaction quotient temporarily; the system shifts until the quotient equals K again, so K is unchanged.
A catalyst speeds both directions equally, so K and the position are unchanged and equilibrium is reached faster.
Only a change in temperature changes the value of K.
| Change | Effect on K | Effect on position |
|---|---|---|
| Concentration | None | Shifts to restore K |
| Pressure | None | Shifts toward fewer (or more) gas moles |
| Catalyst | None | None; equilibrium reached faster |
| Temperature (exothermic forward) | K decreases as T rises | Shifts left |
| Temperature (endothermic forward) | K increases as T rises | Shifts right |
Part 6 of 10Temperature and K
Temperature and K
Endothermic reaction (2HI ⇌ H2 + I2)
- Raising T increases [H2] and [I2] and decreases [HI], so K increases.
Exothermic reaction (2SO2 + O2 ⇌ 2SO3)
- Raising T decreases [SO3] and increases [SO2] and [O2], so K decreases.
Only temperature changes permanently affect Kc or Kp.
Part 7 of 10Equilibrium constant and entropy
1
ΔStotal = ΔSsystem + ΔSsurroundings, and ΔStotal is positive for spontaneous changes.
2
ΔSsystem changes little with temperature unless a state changes, but ΔSsurroundings = −ΔH/T changes significantly.
- CaCO3 → CaO + CO2 (ΔH = +177.9 kJ mol−1, ΔSsys = +160.4 J K−1 mol−1) is not spontaneous at 293 K but is at 1173 K.
3
At equilibrium the total entropy change is zero in net terms.
- The approach to equilibrium from either side has a positive ΔStotal, so neither forward nor backward reaction goes to completion.
4
ΔStotal = R ln K, so K = eΔStotal/R.
- For an exothermic reaction, a rise in T makes ΔSsurroundings less positive, so ΔStotal and K decrease.
Part 8 of 10Magnitude of K
Magnitude of K
A very large K means the equilibrium lies towards the products (right).
A very small K means it lies towards the reactants (left).
There is no hard rule, but K > 1010 is effectively complete and K < 10−10 means no significant reaction.
Part 9 of 10More from the specification
1Equilibrium constant
- Kc is the ratio of product to reactant concentrations at equilibrium, each raised to its coefficient power.
2Esterification
- Ethanol and ethanoic acid form ethyl ethanoate and water reversibly; Kc = [ester][water]/[acid][ethanol].
3Equilibrium mixture
- Use an ICE table to deduce concentrations from initial amounts and the amount reacted.
4Correct units
- Units for Kc depend on the stoichiometry, and cancel when moles are equal on both sides.
5Le Chatelier
- The equilibrium shifts to oppose changes in concentration, pressure or temperature;
- only temperature changes Kc.
6Mixture
- Two or more substances mixed but not chemically combined.
7Ethanoate
- The salt of ethanoic acid.
8Ethyl ethanoate
- The ester CH₃COOC₂H₅.
Part 10 of 10More on this topic
1Dynamic equilibrium
- Dynamic equilibrium forward and reverse rates are equal and concentrations are constant in a closed system.
2Le Chatelier's principle
- Le Chatelier's principle the system shifts to oppose a change.
3
A catalyst does not move the position of equilibrium; it only reaches it sooner.
4Contact process
- 2SO2 + O2 ⇌ 2SO3, V2O5 catalyst, about 450 °C, 1–2 atm.
5
Writing that a catalyst increases yield — it does not change the yield at equilibrium, only the time taken to reach it.
| Term | Meaning |
|---|---|
| Dynamic equilibrium | A state in a closed system where forward and reverse rates are equal and concentrations are constant. |
| Closed system | A system in which no substances enter or leave. |
| Le Chatelier's principle | A system at equilibrium shifts to oppose a change imposed on it. |
| Kc | The ratio of equilibrium concentrations of products to reactants, each raised to its coefficient. |
| Kp | The equilibrium constant written with partial pressures. |
| Conjugate pair | An acid and a base that differ by one H+. |
Key terms
| Term | Meaning |
|---|---|
| Equilibrium constant, Kc | The ratio of equilibrium concentrations of products to reactants, each raised to its stoichiometric power. |
| Equilibrium constant, Kp | The equilibrium constant written with partial pressures of gases. |
| Partial pressure | The pressure a gas in a mixture would exert if it alone filled the container. |
| Mole fraction | Moles of one gas divided by the total moles of gas. |
| Heterogeneous equilibrium | An equilibrium with species in different phases; solids are omitted from K. |
| Position of equilibrium | The relative amounts of products and reactants at equilibrium. |
Avoid these mistakes
1
Including solids in the K expression — they are omitted.
2
Saying a catalyst changes K — it does not.
3
Saying pressure changes K — only the position shifts, K is unchanged.
4
Forgetting to divide moles by volume before calculating Kc.
5
Writing K as reactants over products — it is products over reactants.
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