Chemical Equilibrium – Chapter 15 Comprehensive Bullet-Point Notes

Learning Objectives

• Define equilibrium; describe its dynamic nature (forward and reverse rates equal)
• Define and use: equilibrium constant $K$, reaction quotient $Q$, Le Chatelier’s Principle
• Predict and quantify effects of concentration, pressure/volume, temperature changes on equilibrium
• Convert between $K<em>C$ and $K</em>P$; distinguish homogeneous vs. heterogeneous equilibria
• Apply ICE tables to find unknown equilibrium concentrations or constants

Fundamental Concept: Chemical Equilibrium

• Most reactions are reversible; denoted with $\rightleftharpoons$
• Dynamic balance: forward rate = reverse rate → macroscopically constant concentrations
• Example notation: $\text{N}<em>2\text{O}</em>4(g) \rightleftharpoons 2\,\text{NO}_2(g)$
– Blue arrow thicknesses in figures illustrate simultaneous forward & backward flux
• Visual panel exercise (A + B ⇌ C): only the middle panel set shows equilibrium (flat composition lines)

Equilibrium Constant ($K$)

• For generic reaction $aA + bB \rightleftharpoons cC + dD$
$K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
– Unitless (activities)
• Magnitude interpretation:
– $K \gg 1$ (≈ $>10^{1}$): product-favoured, “goes to completion”
– $K \ll 1$ (≈ $<10^{-1}$): reactant-favoured
• Temperature dependent; ↑T can ↑ or ↓ $K$ depending on ΔH

Numerical Example (from slides)

• $K = \frac{[\text{NO}<em>2]^2}{[\text{N}</em>2\text{O}_4]} = 4.63\times10^{-3}$ at given T

Homogeneous vs. Heterogeneous Equilibria

• Homogeneous: all species same phase (all gases or all aqueous)
– Gas equilibria can use concentration $K<em>C$ or partial pressure $K</em>P$ (different numbers!)
• Heterogeneous: ≥2 phases; activities of pure solids/liquids = 1 → they drop from expression
– Example: $\text{CaCO}<em>3(s) \rightleftharpoons \text{CaO}(s)+\text{CO}</em>2(g)$
$K<em>C= [\text{CO}</em>2]$ only

Water in Aqueous Equilibria

• $\text{CH}<em>3\text{COOH}(aq)+\text{H}</em>2O(l) \rightleftharpoons \text{CH}<em>3\text{COO}^-(aq)+\text{H}</em>3O^+(aq)$
– $[\text{H}<em>2O]\approx55.5\,\text{M}$ constant ⇒ absorb into constant: $K</em>C = \frac{[\text{CH}<em>3\text{COO}^-][\text{H}</em>3O^+]}{[\text{CH}_3\text{COOH}]}$

Multiple (Coupled) Equilibria

• Sequential steps: $A+B \rightleftharpoons C+D\;(K'<em>C)$, $C+D \rightleftharpoons E+F\;(K''</em>C)$
– Overall $A+B \rightleftharpoons E+F$ with $K = K'<em>C\times K''</em>C$ (intermediate terms cancel)
• Principle: when combining equations, multiply constants; reversing a reaction inverts $K$; scaling by n powers $K^n$

Converting $K<em>C \leftrightarrow K</em>P$

• Relation: $K<em>P = K</em>C(RT)^{\Delta n}$ where $\Delta n = (\text{mol gas products}) - (\text{mol gas reactants})$
• Constants: $R = 0.08206\,\text{L·atm·mol}^{-1}\text{K}^{-1}$, T in K
• Example: $\text{N}<em>2(g)+3\,\text{H}</em>2(g) \rightleftharpoons 2\,\text{NH}<em>3(g)$ at 745 K – $K</em>C=0.118$, $\Delta n = 2-4 = -2$
– $K<em>P = 0.118(0.08206\times745)^{-2} = 3.16\times10^{5}$ • Another exercise: for $2\,\text{SO}</em>2 +\text{O}<em>2 \rightleftharpoons 2\,\text{SO}</em>3$ at 527 °C, $K<em>P=1.2\times10^{3} ⇒ K</em>C=7.9\times10^{4}$

Reaction Quotient ($Q$)

• Same algebraic form as $K$ but using instantaneous concentrations $[X]<em>0$ • Comparison rules: – $Q < K$: too few products → shifts right – $Q = K$: at equilibrium – $Q > K$: too many products → shifts left • Example (steam reforming) at 900 K: $K</em>C = 2.4\times10^{-4}$ for $\text{CH}<em>4+\text{H}</em>2O \rightleftharpoons \text{CO}+3\,\text{H}<em>2$ – Initial moles in 2 L reactor: $[\text{CH}</em>4]=6.0\times10^{-3}\,\text{M}$, $[\text{H}<em>2O]=4.0\times10^{-3}\,\text{M}$, $[\text{CO}]=8.0\times10^{-3}\,\text{M}$, $[\text{H}</em>2]=3.0\times10^{-3}\,\text{M}$
– $Q = \frac{(8.0\times10^{-3})(3.0\times10^{-3})^3}{(6.0\times10^{-3})(4.0\times10^{-3})} = 1.4\times10^{-7} < K$ ⇒ shift right
• Water–gas shift example: $Q=0.96$ vs $K=0.64$ → shift left

ICE Tables, Quadratics & "Small-x" Approximation

• To predict equilibrium concentrations when $K$ and initials given, or vice-versa
• Structure:
− Row 1: Initial (I)
− Row 2: Change (C) using stoichiometric ratios (let unknown be $x$)
− Row 3: Equilibrium (E) = I ± C
• Solve for $x$ via substitution into $K$ expression
• If $K$ extremely small/large, assume $x \ll$ initial, drop $x$ in denominator → validate afterward (error <5 %)
• Example: Decomposition of $\text{NOCl}$ at 227 °C
– $2\,\text{NOCl} \rightleftharpoons 2\,\text{NO}+\text{Cl}<em>2$ – In 2 L: $[\text{NOCl}]</em>0 = 0.50\,\text{M}$; measured $[\text{Cl}<em>2]</em>e = 0.028\,\text{M}$
– Via ICE, $x = 0.028\,\text{M}$ ⇒ $K_C = \frac{(0.028)^1(0.056)^2}{(0.50-0.056)^2} ≈$ (students compute)

Sample Forward Calculations

• For $\text{H}<em>2 + \text{CO}</em>2 \rightleftharpoons \text{H}_2O+\text{CO}$, $K=0.106$ at 700 K, start with $0.0150$ M each reactant, 0 product
– ICE: let $x$ be amount reacted → $K = \frac{x^2}{(0.0150-x)^2}$ → solve $x \approx 4.0\times10^{-3}$
– Final: reactants 0.011, products 0.0040 M

Le Chatelier’s Principle

• "External stress" (Δconc, ΔP/ΔV, ΔT) causes system to shift to reduce the stress

Concentration Changes

• Adding reactant → shift right; removing → shift left
• Example iron–thiocyanate equilibrium: $\text{Fe}^{3+}+\text{SCN}^- \rightleftharpoons \text{Fe(SCN)}^{2+}$
– Adding NaSCN (common ion) shifts right (more red complex)
– Adding Fe(NO₃)₃ also shifts right (additional $\text{Fe}^{3+}$)
– Adding complexing agent $\text{H}<em>2\text{C}</em>2\text{O}_4$ or NaCl can bind $\text{Fe}^{3+}$, effectively decreasing its concentration, shifting left

Pressure/Volume (Gas-Phase)

• $P \propto \frac{n}{V}$
• ↓Volume (↑P) shifts toward side with fewer gas moles; ↑Volume shifts toward more moles
• Example: $\text{N}<em>2\text{O}</em>4 \rightleftharpoons 2\,\text{NO}_2$ (Δn=+1): compress → shift left (dimer formation)
• Adding inert gas at constant V: no shift (no concentration change)

Temperature

• Treat heat as reagent: endothermic (ΔH>0) behaves like reactant; exothermic (ΔH<0) like product
• Only temperature changes actual $K$ value
• Example: $\text{N}<em>2\text{O}</em>4 + \text{heat} \rightleftharpoons 2\,\text{NO}<em>2$ (ΔH=+58 kJ) – ↑T adds heat → shift right, $K$ increases • Practice Q: $2\,\text{O}</em>3 \rightleftharpoons 3\,\text{O}_2$, ΔH° = –284 kJ (exothermic)
– Volume ↓ (↑P, Δn=+1) → shift left (more O₃)
– Add O₂ → shift left
– ↓T (remove heat) → shift right (favor exothermic)
– Catalyst: no position change
– Correct choice in MC list: option (c)

Catalyst Effect

• Lowers activation energy equally for forward & reverse → faster attainment of equilibrium but no change in $K$ or position

Propane Combustion Multiple-Choice Example

• Balanced propane combustion: $\text{C}<em>3\text{H}</em>8 + 5\,\text{O}<em>2 \rightleftharpoons 3\,\text{CO}</em>2 + 4\,\text{H}<em>2\text{O}$ • Equilibrium expression: $K = \frac{[\text{CO}</em>2]^3[\text{H}<em>2\text{O}]^4}{[\text{C}</em>3\text{H}<em>8][\text{O}</em>2]^5}$ → matches choice (d)

Solubility & Complex-Ion Equilibria

• Slightly soluble salts dissolve more with complexation
– Steps: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-\;(K<em>{sp}=2.0\times10^{-10})$; $\text{Ag}^+ + 2\,\text{NH}</em>3 \rightleftharpoons \text{Ag(NH}<em>3)</em>2^+\;(K<em>f=1.5\times10^{7})$ – Overall: $\text{AgCl}+2\,\text{NH}</em>3 \rightleftharpoons \text{Ag(NH}<em>3)</em>2^+ + \text{Cl}^-$
– $K = K<em>{sp}\times K</em>f = (2.0\times10^{-10})(1.5\times10^{7}) = 3.0\times10^{-3}$
– Expression: $K = \frac{[\text{Cl}^-][\text{Ag(NH}<em>3)</em>2^+]}{[\text{NH}_3]^2}$ → choice (b)

Graphical Depictions (Rate vs. Time)

• Provided time-course plots show [N₂O₄] decreasing, [NO₂] rising until plateau (equilibrium) while rates of forward & reverse converge

Page-1 Classroom Example (mixed H₂ & I₂)

• Initial: $0.100\,\text{M H}<em>2$ + $0.100\,\text{M I}</em>2$ → equilibrium of $\text{H}<em>2 + \text{I}</em>2 \rightleftharpoons 2\,\text{HI}$
– Students determine which direction proceeds and equilibrium $K$ sign (positive by definition)

Practical & Industrial Relevance

• Steam reforming + water–gas shift → bulk H₂ for ammonia (Haber), fuel cells, etc.
• Propane stoves, catalytic converters, atmospheric ozone balance; understanding equilibrium vital for design & environmental predictions

Ethical & Philosophical Considerations

• Industrial manipulation of equilibria (e.g., removing NH₃ to drive Haber process) balances societal demand for fertilizer vs. energy use & emissions
• Catalyst development affects sustainability by lowering required temperatures/pressures → reduced carbon footprint

Remember: Always specify phase labels, balance equations, check Δn before using $K_P$ conversions, and verify approximations.