Equilibrium Constant Applications

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Equilibrium Constant ($K_{\text{eq}}$): The quotient of product to reactant concentrations at equilibrium.

Overview

• $K_{\text{eq}}$ establishes the extent to which a reaction progresses before reaching equilibrium.
• It offers understanding of the ratio of reactants to products.
• Guiding Questions:
  • "What insights does $K_{\text{eq}}$" provide about the equilibrium state of products and reactants?"
  • "How can $K_{\text{eq}}$ be utilised to foresee reaction results?"

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Mathematical Representation

• The expression for the equilibrium constant is:

  $K_{\text{eq}} = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

  • [A], [B]: Reactant concentrations.
  • [C], [D]: Product concentrations.
  • a, b, c, d: Stoichiometric coefficients.

• Example Calculation:

  • Reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
  • Measured Concentrations: $[NH_3] = 2.0$ M, $[N_2] = 1.0$ M, $[H_2] = 3.0$ M
  • Calculation: $K_{\text{eq}} = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(2.0)^2}{(1.0)(3.0)^3} = \frac{4.0}{27.0} = 0.148$

  

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Significance of $K_{\text{eq}}$

• Relation between $K_{\text{eq}}$ and Reaction Favourability:

  • A elevated $K_{\text{eq}}$ indicates preference towards product formation.
  • A diminished $K_{\text{eq}}$ signifies preference towards reactant formation.

• Industrial Applications:

  • Haber Process: Engineers adjust conditions to enhance ammonia yield.
  • Modifying temperature and pressure influences $K_{\text{eq}}$, optimising production outcomes.

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Introduction to Ionisation Process

• Ionisation: The process in which ionic compounds dissolve in water, establishing a dynamic equilibrium between dissolved ions and undissolved solids.

  • Real-world Context: Dissolving table salt results in sodium ions ($Na^+$) and chloride ions ($Cl^-$).
  • Interactive Engagement: Identify ions formed from potassium chloride (KCl) dissolution in water.
  infoNote

  Dissolving KCl in water dissociates it into potassium ions ($K^+$) and chloride ions ($Cl^-$).

• Phase Transition: This involves a change from solid (crystalline structure) to an aqueous phase (ions freely moving in the solution), constituting a reversible process that sustains equilibrium.

  

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Solubility Product Constant ($K_{\text{sp}}$)

• Definition and Explanation:

  • $K_{\text{sp}}$: Represents the product of ion concentrations at equilibrium, each raised to the power of their respective coefficients.
  • It determines solubility limits and helps in predicting precipitate formation.

• Practical Significance:

  • Assesses if a solution will become saturated or if precipitation of a solid occurs.
  infoNote

  Example: For $AB_{(s)} \leftrightarrow A^+(aq) + B^-(aq)$, $K_{\text{sp}}$ is expressed as $K_{\text{sp}} = [A^+][B^-]$.

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Dissociation of Acids and Bases

Acid Dissociation Constant ($K_a$):

• Formula: $K_a = \frac{[H^+][A^-]}{[HA]}$

• A greater $K_a$ reflects stronger acid dissociation.

• Base Dissociation Constant ($K_b$):

  • Formula: $K_b = \frac{[BH^+][OH^-]}{[B]}$
  • A higher $K_b$ reflects stronger base dissociation.

Conjugate Acid-Base Pairs

• Example: HCl/Cl$^-$.
  • Displaying reversible transformations through proton transfer.

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Influence of Temperature and Pressure

Temperature Effects

• Endothermic Reactions: Absorb heat, favouring product formation with higher $K_{\text{eq}}$ as temperature increases.
• Exothermic Reactions: Release heat, favouring reactant formation, resulting in a decrease in $K_{\text{eq}}$ with rising temperature.

Pressure Effects

• In systems involving gases, increasing pressure favours the formation of the side with fewer gaseous molecules.

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Common Misconceptions and Clarifications

• Solubility vs. $K_{\text{sp}}$: These terms are distinct; solubility refers to the amount of solute that can dissolve, whereas $K_{\text{sp}}$ concerns the ion concentration product at equilibrium.
• Dynamic vs. Static Equilibrium: Authentic equilibrium involves continuous reactions; misconceiving it as static is an error.

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Problem-solving Strategies

Structured Problem-Solving

• Employ a balanced chemical equation, ICE tables, and apply stoichiometry accurately.

Equilibrium Calculations

• Example Problem:

  • Consider the reaction: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
  • Initial concentrations: $[N_2]_0 = 0.5$ M, $[H_2]_0 = 1.5$ M, $[NH_3]_0 = 0$ M
  • After reaction reaches equilibrium: $[NH_3] = 0.4$ M

  ICE Table:

  	$N_2$	$H_2$	$NH_3$
  Initial	0.5 M	1.5 M	0 M
  Change	-0.2 M	-0.6 M	+0.4 M
  Equilibrium	0.3 M	0.9 M	0.4 M

  Calculating $K_{\text{eq}}$: $K_{\text{eq}} = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.4)^2}{(0.3)(0.9)^3} = \frac{0.16}{0.3 \times 0.729} = \frac{0.16}{0.2187} = 0.731$

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Common Mistakes to Avoid

• Pitfall: Overlooking the impact of temperature and pressure.
• Preserve unit consistency for accuracy.
• Remember that $K_{\text{eq}}$ values must use equilibrium concentrations, not initial concentrations.