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4.3Unit 4. Rates, Equilibria and Further Organic Chemistry

Chemical Equilibria — IAL Edexcel Modular (Chemistry) Revision Notes & Topical Questions

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4.3 Chemical Equilibria — IAL EDEXCEL MODULAR (CHEMISTRY) notes: Write Kc and Kp expressions, calculate K and its units, explain the effect of conditions on K and the position of equilibrium, and relate K to total entropy change.For aA + bB ⇌ cC + dD, Kc = [C]c[D]d ÷ [A]a[B]bWhat…

  • What the symbols mean
  • Significant figures
  • Heterogeneous equilibria
  • Partial pressure
  • Concentration or pressure
  • Exothermic forward reaction

Section 1 of 11Equilibrium constant expression K_c

Equilibrium constant expression Kc
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
  • Square brackets are equilibrium concentrations in mol dm-3.
  • Products go on top, reactants below, each raised to the power of its coefficient.
Solids are left out of the expression
  • For example, Ag+(aq) + Fe2+(aq) ⇌ Ag(s) + Fe3+(aq) has Kc = [Fe3+] ÷ [Ag+][Fe2+].
Constant
  • Kc is specific to a reaction and changes only with temperature

Section 2 of 11Calculating K_c

1
Write the balanced equation and the Kc expression.
2
Find the equilibrium moles of each species; if only initial moles are given, use an initial, change and equilibrium table with the mole ratio.
3
Convert to concentration by dividing the moles by the volume in dm3.
4
Substitute the equilibrium concentrations into the expression.
5
Deduce the units by cancelling the mol dm-3 terms.
Significant figures
  • Significant figures give the answer to the smallest number of significant figures in the data
Units
  • Kc has no units when the powers on top and bottom are equal

Section 3 of 11Equilibrium constant expression K_p

Kp
  • Kp uses the partial pressures of gases, with p in Pa
  • 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
  • The sum of the partial pressures equals the total pressure.
Heterogeneous equilibria
  • Heterogeneous equilibria solids and liquids are left out
  • CaCO3(s) ⇌ CaO(s) + CO2(g) has Kp = p(CO2).
  • N2O4 ⇌ 2NO2 has Kp = p(NO2)2 ÷ p(N2O4).
Equilibrium constant expression Kp
Find the total moles of gas at equilibrium.
Calculate the mole fraction of each gas.
Partial pressure
Partial pressure mole fraction × total pressure.
Substitute into the Kp expression and deduce the units from the pressure terms.
  • Units
    • the units depend on the expression, such as Pa-1 for 2SO2 + O2 ⇌ 2SO3, and none when they cancel

Section 4 of 11Effect of conditions on K and the position

Effect of conditions on K and the position
Concentration or pressure
  • Concentration or pressure changes shift the position of equilibrium temporarily until the quotient equals K again
  • K is unchanged.
Catalyst
  • Catalyst speeds up forward and reverse reactions equally
  • The position and K are unchanged, and equilibrium is reached sooner.
Only temperature changes the value of K

Section 5 of 11Effect of temperature on K

Effect of temperature on K
Exothermic forward reaction
  • Exothermic forward reaction raising the temperature shifts the equilibrium left, so K decreases
  • 2SO2 + O2 ⇌ 2SO3: [SO3] falls and [SO2] and [O2] rise.
Endothermic forward reaction
  • Endothermic forward reaction raising the temperature shifts the equilibrium right, so K increases
  • 2HI ⇌ H2 + I2: [H2] and [I2] rise and [HI] falls.
Shift direction
  • Shift direction the shift opposes the change, as in Le Chatelier's principle

Section 6 of 11Equilibrium constant and entropy

Total entropy
  • ΔStotal = ΔSsystem + ΔSsurroundings, with ΔSsurroundings = −ΔH ÷ T
  • ΔStotal is positive for all spontaneous changes.
  • ΔSsystem changes little with temperature unless there is a change of state.
  • ΔSsurroundings changes significantly with temperature.
Reversible reactions
  • Reversible reactions equilibrium can be approached from either side
  • Both directions are spontaneous towards the equilibrium mixture, so neither goes to completion.
  • At equilibrium the total entropy change is zero.
ΔStotal
R ln K, so K = eΔStotal/R
  • A large positive ΔStotal gives a large K.
Exothermic reaction
  • Exothermic reaction a rise in T makes ΔSsurroundings less positive, so ΔStotal and K decrease

Section 7 of 11Size of K and position of equilibrium

K much greater than 1
  • K much greater than 1 the position lies towards the products
K much less than 1
  • K much less than 1 the position lies towards the reactants
Extremes
  • K above 1010 is effectively complete and below 10-10 means no significant reaction
Rule
  • Rule there is no hard rule linking K and position, so this is a general guide only

Section 8 of 11More from the specification

More from the specification
Esterification Equilibrium Constant (Kc)
  • For the reversible reaction between ethanoic acid and ethanol yielding ethyl ethanoate and water, the equilibrium expression is Kc = / ([CH3COOH][CH3CH2OH]).
    • ([CH3COOCH2CH3][H2O])
Units Cancellation in Ester Systems
  • Because the sum of stoichiometric coefficients is identical on both sides of the equation, concentration dimensions cancel completely, leaving Kc as a dimensionless value with no units.
Equilibrium Mole Calculations
  • Using the initial moles of reactants and the final equilibrium amount of water, molar quantities of ethyl ethanoate, unreacted acid,
  • and alcohol are calculated by 1:1 stoichiometry.
Thermodynamic Relationship to Total Entropy
  • The equilibrium position links directly to the total entropy change of the universe via delta_Stotal = R ln(Kc);
    • a value of Kc > 1 corresponds to a positive delta_Stotal.
Ethanoate
  • The salt of ethanoic acid.
Ethyl ethanoate
  • The ester CH₃COOC₂H₅.

Section 9 of 11More on this topic

  • Industrial compromise
    • the Contact process, 2SO2 + O2 ⇌ 2SO3, uses a V2O5 catalyst, about 450 °C and 1–2 atm
TermMeaning
Dynamic equilibriumForward and reverse rates are equal and concentrations stay constant.
Closed systemA system in which no substances enter or leave.
Le Chatelier's principleEquilibrium shifts to oppose a change.
KcThe ratio of equilibrium concentrations of products to reactants, each raised to its coefficient.
KpThe equilibrium constant using partial pressures.
Brønsted–Lowry acidA proton donor.
Brønsted–Lowry baseA proton acceptor.
Conjugate pairAn acid and base differing by one H⁺.

Section 10 of 11Key terms

TermMeaning
Equilibrium constant, KcThe ratio of product concentrations to reactant concentrations, each raised to its coefficient, at equilibrium.
Equilibrium constant, KpThe equivalent ratio using partial pressures of gases.
Partial pressureThe pressure a gas would exert alone: mole fraction × total pressure.
Mole fractionThe moles of one gas divided by the total moles of gas.
Total entropy changeΔSsystem + ΔS_surroundings, related to K by ΔStotal = R ln K.
Heterogeneous equilibriumAn equilibrium involving substances in more than one phase.

Section 11 of 11Exam technique

1
Including solids in a Kc or Kp expression
2
Using moles instead of concentrations in Kc, or mole fractions instead of partial pressures in Kp
3
Saying a catalyst changes K — only temperature does
4
Saying a change in pressure or concentration changes K — it only shifts the position temporarily
5
Writing the expression the wrong way up — products go on top

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