Chemical Equilibrium – Comprehensive Bullet-Point Notes

Types of Chemical Reactions

  • Irreversible reaction
    • One-way; proceeds only from reactants (A) to products (B).
    • Representation: A→BA \rightarrow B
    • Examples: CH<em>4+2O</em>2→CO<em>2+2H</em>2O\text{CH}<em>4 + 2\text{O}</em>2 \rightarrow \text{CO}<em>2 + 2\text{H}</em>2\text{O}, NaOH+HCl→NaCl+H2O\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}
  • Reversible reaction
    • Two-way; products can reconvert to reactants.
    • Representation: A⇌BA \rightleftharpoons B
    • Examples: PCl<em>5(g)⇌PCl</em>3(g)+Cl<em>2(g)\text{PCl}<em>5(g) \rightleftharpoons \text{PCl}</em>3(g)+\text{Cl}<em>2(g), SO</em>3(g)⇌SO<em>2(g)+12O</em>2(g)\text{SO}</em>3(g) \rightleftharpoons \text{SO}<em>2(g)+\tfrac12\text{O}</em>2(g)

State of Equilibrium

  • Achieved when rate of forward process equals rate of backward process: r<em>f=r</em>br<em>f = r</em>b.
  • At equilibrium, measurable properties (concentration, pressure, density, colour) become time-invariant for given conditions.
  • Requires a closed system.

Law of Mass Action & Rate Expressions

  • For an elementary reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD:
    • Forward rate: r<em>f=k</em>f[A]a[B]br<em>f = k</em>f [A]^a [B]^b
    • Backward rate: r<em>b=k</em>b[C]c[D]dr<em>b = k</em>b [C]^c [D]^d.
  • At equilibrium: k<em>f[A]a</em>e[B]b<em>e=k</em>b[C]c<em>e[D]d</em>ek<em>f [A]^{a}</em>e [B]^{b}<em>e = k</em>b [C]^{c}<em>e [D]^{d}</em>e.

Equilibrium Constants

  • Concentration form (Kc): Kc=[C]c[D]d[A]a[B]bK_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b}.
  • Pressure form (Kp) (only gases): K<em>p=P</em>CcP<em>DdP</em>AaPBbK<em>p = \dfrac{P</em>C^{c} P<em>D^{d}}{P</em>A^{a} P_B^{b}}.
  • Relation: K<em>p=K</em>c(RT)Δn<em>gK<em>p = K</em>c (RT)^{\Delta n<em>g} with Δn</em>g=(c+d)−(a+b)\Delta n</em>g = (c+d)-(a+b) (difference in gaseous stoichiometric coefficients).
  • Independence: KeqK_{eq} is unaffected by initial concentrations, catalysts, inert gases (at constant V), or whether the system starts from reactant or product side.

Worked Set-Up Examples

  1. PCl<em>5(g)⇌PCl</em>3(g)+Cl<em>2(g)\text{PCl}<em>5(g) \rightleftharpoons \text{PCl}</em>3(g)+\text{Cl}<em>2(g) • Initial: n</em>0=PCl<em>5=nn</em>0=PCl<em>5=n, none of products. • Equilibrium: n</em>PCl<em>5=n−xn</em>{PCl<em>5}=n-x, n</em>PCl<em>3=n</em>Cl<em>2=xn</em>{PCl<em>3}=n</em>{Cl<em>2}=x. • K</em>c=[PCl<em>3][Cl</em>2][PCl5]=x2(n−x)K</em>c = \dfrac{[PCl<em>3][Cl</em>2]}{[PCl_5]} = \dfrac{x^2}{(n-x)}.
  2. H<em>2(g)+I</em>2(g)⇌2HI(g)\text{H}<em>2(g)+\text{I}</em>2(g) \rightleftharpoons 2\text{HI}(g)
    • K<em>c=[HI]2[H</em>2][I2]=4x2(a−x)(b−x)K<em>c = \dfrac{[HI]^2}{[H</em>2][I_2]} = \dfrac{4x^2}{(a-x)(b-x)}.
  3. N<em>2(g)+3H</em>2(g)⇌2NH<em>3(g)\text{N}<em>2(g)+3\text{H}</em>2(g) \rightleftharpoons 2\text{NH}<em>3(g) • K</em>c=[NH<em>3]2[N</em>2][H2]3=4x2(a−x)(b−3x)3K</em>c = \dfrac{[NH<em>3]^2}{[N</em>2][H_2]^3} = \dfrac{4x^2}{(a-x)(b-3x)^3}.

Characteristics of Equilibrium State

  • Dynamic; forward and reverse reactions continue at equal rates.
  • Can be attained from any initial composition (reactants, products, or mixture).
  • Catalyst lowers time to attain equilibrium but does not change KeqK_{eq}.
  • Activity (active mass) of pure solids/liquids is taken as constant (unity).
  • Only temperature changes KeqK_{eq}.

Factors Affecting KeqK_{eq}

  • Mode of writing reaction: reversing reaction inverts K<em>eqK<em>{eq}; multiplying stoichiometry by n raises K</em>eqK</em>{eq} to nthn^{th} power.
  • Temperature dependence – van’t Hoff equation:
    dln⁡K<em>eqdT=ΔH∘RT2\dfrac{d\ln K<em>{eq}}{dT}=\dfrac{\Delta H^\circ}{RT^2} Integrated form: ln⁡K</em>2K<em>1=ΔH∘R(1T</em>1−1T<em>2)\ln\dfrac{K</em>2}{K<em>1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T</em>1}-\dfrac{1}{T<em>2}\right) • Endothermic ( ΔH∘>0\Delta H^\circ>0 ): K</em>eqK</em>{eq} increases with temperature.
    • Exothermic ( ΔH∘<0\Delta H^\circ<0 ): K<em>eqK<em>{eq} decreases with temperature. • ΔH∘=0\Delta H^\circ=0 : K</em>eqK</em>{eq} independent of T.

Reaction Quotient (Q)

  • Same algebraic form as KcK_c, but concentrations at any instant.
  • Comparison:
    • Q<K<em>eqQ<K<em>{eq} → reaction proceeds forward. • Q>K</em>eqQ>K</em>{eq} → reaction proceeds backward.
    • Q=KeqQ=K_{eq} → system at equilibrium.

Gibbs Free Energy & Equilibrium

  • ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q.
  • At equilibrium ΔG=0\Delta G = 0, Q=K<em>eqQ=K<em>{eq}, hence ΔG∘=−RTln⁡K</em>eq\Delta G^\circ = -RT \ln K</em>{eq}.
  • Significance:
    • K<em>eq>1K<em>{eq}>1 → ΔG∘<0\Delta G^\circ<0 (product-favoured). • K</em>eq<1K</em>{eq}<1 → ΔG∘>0\Delta G^\circ>0 (reactant-favoured).

Activation Energy (Eₐ) & Energy Profile

  • Minimum energy needed for reactants to form activated complex.
  • Energy diagram parameters:
    • E<em>aE<em>a (forward), E</em>a′E</em>a' (reverse).
    • Enthalpy change ΔH=E<em>p−E</em>r\Delta H = E<em>p - E</em>r.
    • Endothermic: ΔH>0\Delta H>0, peak closer to products; Exothermic: ΔH<0\Delta H<0.

Degree of Dissociation (α) & Molecular Mass

  • Definition: α=moles dissociatedinitial moles\alpha = \dfrac{\text{moles dissociated}}{\text{initial moles}}.
  • For PCl<em>5⇌PCl</em>3+Cl2\text{PCl}<em>5 \rightleftharpoons \text{PCl}</em>3+\text{Cl}_2, if one mole starts:
    • Equilibrium moles = 1−α+2α=1+α1-\alpha + 2\alpha = 1+\alpha.
  • Relation with molar mass or vapour density:
    α=M<em>t−M</em>oM<em>o(n−1)=d</em>t−d<em>od</em>o(n−1)\alpha = \dfrac{M<em>t - M</em>o}{M<em>o (n-1)} = \dfrac{d</em>t - d<em>o}{d</em>o (n-1)}
    where nn = total moles after complete dissociation per mole of reactant (here n=2n=2), M<em>tM<em>t theoretical molar mass, M</em>oM</em>o observed.

Calculational Caveats

  • In α problems, initial moles may be taken as 1, but use actual initial concentration or pressure values (do NOT set to 1) when writing KeqK_{eq} expressions.

Le Châtelier’s Principle

  • If a system at equilibrium experiences a change in concentration, pressure/volume, or temperature, it shifts in the direction that counteracts the change.

Concentration Changes

  • Adding reactant / removing product → shift forward.
  • Removing reactant / adding product → shift backward.

Pressure / Volume Changes (gaseous systems)

  • Decrease V (increase P): equilibrium shifts to side with fewer moles of gas.
  • Increase V (decrease P): shifts to side with more moles of gas.
  • Cases:
    • Δn<em>g=0\Delta n<em>g=0 (e.g., H</em>2+I<em>2⇌2HI\text{H}</em>2+I<em>2 \rightleftharpoons 2\text{HI}) → no effect. • Δn</em>g<0\Delta n</em>g<0 (e.g., 2NO<em>2⇌N</em>2O<em>42\text{NO}<em>2 \rightleftharpoons \text{N}</em>2\text{O}<em>4) → increasing P shifts right. • Δn</em>g>0\Delta n</em>g>0 (e.g., PCl<em>5⇌PCl</em>3+Cl2\text{PCl}<em>5 \rightleftharpoons \text{PCl}</em>3+\text{Cl}_2) → increasing P shifts left.
  • Addition of inert gas:
    • At constant V: no effect on KeqK_{eq} or position.
    • At constant P: behaves like volume increase, apply mole-based rule.

Temperature Changes

  • Endothermic: raise T → shift forward; lower T → shift backward.
  • Exothermic: opposite trend.
  • ΔH=0\Delta H = 0: no shift.

Applications to Physical Equilibria

  • Liquid ⇌ Vapour (boiling): endothermic; raising P raises boiling point.
  • Solid ⇌ Liquid (melting):
    • For water/ice (volume decreases on melting), pressure lowers melting point.
    • For metals (volume increases on melting), pressure raises melting point.
  • Gas ⇌ Solution (Henry’s law): increased external P increases gas solubility.
  • Endothermic dissolution (e.g., NH4Cl\text{NH}_4\text{Cl} in water): solubility rises with T.
  • Exothermic dissolution (e.g., NaOH\text{NaOH}): solubility decreases with T.

Simultaneous & Sequential Equilibria

  • Simultaneous: two reactions sharing common species occur together; each has its own K<em>eqK<em>{eq}. Example: A⇌B+CA \rightleftharpoons B + C ( K</em>1K</em>1 ) and B⇌D+EB \rightleftharpoons D + E ( K2K_2 ).
    Overall relationships can be derived via algebraic combination of individual equilibria.
  • Sequential: product of first becomes reactant for second; treat stepwise, using intermediate concentrations.

Summary of Key Formulae

  • r=k[active mass]coeffr = k [\text{active mass}]^{\text{coeff}} (Law of mass action).
  • K<em>c,  K</em>p,  K<em>p=K</em>c(RT)ΔngK<em>c,\;K</em>p,\;K<em>p=K</em>c(RT)^{\Delta n_g}.
  • QQ vs KeqK_{eq} for predicting direction.
  • ΔG∘=−RTln⁡Keq\Delta G^\circ = -RT \ln K_{eq}.
  • van’t Hoff: ln⁡K<em>2K</em>1=ΔH∘R(1T<em>1−1T</em>2)\ln\dfrac{K<em>2}{K</em>1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T<em>1}-\dfrac{1}{T</em>2}\right).
  • Degree of dissociation–molar mass relation: α=M<em>t−M</em>oMo(n−1)\alpha = \dfrac{M<em>t-M</em>o}{M_o(n-1)}.

Ethical & Practical Implications

  • Industrial optimisation (Haber, Contact processes) relies on manipulating T, P, and catalysts per Le Châtelier to maximise yield while balancing economic cost.
  • Understanding equilibrium is foundational for biochemical homeostasis, environmental chemistry (ozone balance, CO₂ solubility in oceans), and safety (pressure vessels, reactors).