Chemical Equilibrium

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Chemical Equilibrium — Key Facts for WAEC WASSCE

• Definition: A reaction is at equilibrium when the forward and reverse reaction rates are equal, and the concentrations of reactants and products remain constant (but not necessarily equal).
• The Equilibrium Constant $K_c$: For $aA + bB \rightleftharpoons cC + dD$: $$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$ $K_c > 1$: equilibrium lies to the right (products favoured). $K_c < 1$: equilibrium lies to the left (reactants favoured).
• Le Chatelier’s Principle: If a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system adjusts to partially oppose the change.
• Factors Affecting Equilibrium:
  • Concentration: Adding reactant shifts equilibrium to the right (towards products); removing product shifts right.
  • Temperature: Increasing temperature shifts equilibrium towards the endothermic direction; decreasing shifts towards the exothermic direction.
  • Pressure: Increasing pressure shifts equilibrium towards the side with fewer moles of gas; decreasing pressure shifts towards more moles.
  • Catalyst: Does not shift equilibrium position — it only speeds up the attainment of equilibrium.

⚡ WAEC Exam Tip: A catalyst has NO effect on the equilibrium constant $K_c$ and no effect on the position of equilibrium. It only increases both forward and reverse reaction rates equally. This is a common trick question in WAEC!

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Chemical Equilibrium — WAEC WASSCE Chemistry Study Guide

Dynamic Nature of Equilibrium

Chemical equilibrium is dynamic, not static. At equilibrium, the forward and reverse reactions continue to occur, but their rates are equal so there is no net change in concentrations. This can be demonstrated isotopically — for example, in the Haber process equilibrium: $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$, if radioactive nitrogen is introduced as $\mathrm{N_2}$, it will be found in both reactants and products at equilibrium, proving the reaction is ongoing in both directions.

The Equilibrium Constant in Terms of Concentration ($K_c$)

For gaseous reactions, $K_c$ is expressed in terms of molar concentrations (mol dm⁻³). For example:

$$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$$

$$K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$$

The units of $K_c$ depend on the stoichiometry. For the Haber process, the units are $\text{mol}^{-2}\text{dm}^6$.

The Equilibrium Constant in Terms of Partial Pressure ($K_p$)

For gaseous equilibria, it is often more convenient to use partial pressures:

$$K_p = \frac{(P_{\text{NH}3})^2}{(P{\text{N}2})(P{\text{H}_2})^3}$$

The relationship between $K_p$ and $K_c$ is: $K_p = K_c(RT)^{\Delta n}$, where $\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$.

Homogeneous and Heterogeneous Equilibrium

• Homogeneous: All reactants and products are in the same phase (e.g., gaseous or aqueous). Example: $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$
• Heterogeneous: Reactants and products are in different phases. Example: $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$. For heterogeneous equilibria, the activities of pure solids and liquids are taken as 1 (they have constant concentration), so only gases appear in the $K_c$ expression.

Application of Le Chatelier’s Principle

Industrial Example — The Haber Process for Ammonia Synthesis: $$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)} \quad \Delta H = -92 \text{ kJ mol}^{-1}$$

This reaction is exothermic (heat is given out). To maximise ammonia yield:

• Temperature: Lower temperature favours the forward reaction (exothermic, right), but the rate is slower. In practice, a compromise temperature of about 450–500°C is used with a catalyst (iron) for reasonable yield and acceptable rate.
• Pressure: Higher pressure favours the right (fewer moles of gas on product side: 2 moles vs 4 moles). Pressures of 200–300 atm are used industrially.
• Removal of ammonia: As ammonia is formed, it is removed by liquefaction, pulling the equilibrium further to the right.

The Effect of Catalysts on Equilibrium

A catalyst lowers the activation energy for both forward and reverse reactions by the same amount. This increases both rates equally, so the equilibrium position is unchanged. The catalyst allows equilibrium to be reached more quickly but does not change the composition of the equilibrium mixture.

Relationship Between $K_c$ and Reaction Quotient $Q$

The reaction quotient $Q$ has the same expression as $K_c$ but uses current (non-equilibrium) concentrations:

• If $Q < K_c$: the forward reaction is favoured (products increase)
• If $Q = K_c$: the system is at equilibrium
• If $Q > K_c$: the reverse reaction is favoured (reactants increase)

This is useful for predicting the direction of shift when conditions change.

Comparison Table — Factors Affecting Equilibrium

Factor	Change	Effect on Equilibrium Position	Effect on $K$?
Concentration of reactant ↑	Added reactant	Shifts right (towards products)	No change
Concentration of product ↓	Product removed	Shifts right	No change
Temperature ↑	Heat added	Shifts towards endothermic direction	Changes
Temperature ↓	Heat removed	Shifts towards exothermic direction	Changes
Pressure ↑	Volume decreased	Shifts towards fewer gas moles	No change
Pressure ↓	Volume increased	Shifts towards more gas moles	No change
Catalyst added	—	No shift	No change

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🔴 Extended — Deep Study (3mo+)

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Chemical Equilibrium — Comprehensive WAEC Chemistry Notes

Derivation of the Equilibrium Constant Expression

For a general reversible reaction: $$aA + bB \rightleftharpoons cC + dD$$

At equilibrium, the rates of the forward and reverse reactions are equal. If the rate laws follow elementary steps (which is assumed for the purpose of $K_c$ derivation at this level):

Rate forward = $k_f[A]^a[B]^b$ Rate reverse = $k_r[C]^c[D]^d$

At equilibrium: $k_f[A]^a[B]^b = k_r[C]^c[D]^d$

$$\frac{k_f}{k_r} = \frac{[C]^c[D]^d}{[A]^a[B]^b} = K_c$$

Thus $K_c$ is the ratio of the forward and reverse rate constants — it is constant at a given temperature. This also explains why $K_c$ changes with temperature: $k_f$ and $k_r$ change differently with temperature, so their ratio changes.

The van’t Hoff Equation (Temperature Dependence of $K_c$)

From the Arrhenius equation and thermodynamic relationships, the effect of temperature on $K_c$ is given by:

$$\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$$

For an exothermic reaction ($\Delta H^\circ < 0$): as $T$ increases, $\ln K$ decreases, so $K_c$ decreases — consistent with Le Chatelier’s principle (heat added favours endothermic direction).

For an endothermic reaction ($\Delta H^\circ > 0$): as $T$ increases, $\ln K$ increases, so $K_c$ increases.

Equilibrium in Ionic Reactions — Solubility Product

For a sparingly soluble salt like silver chloride: $$\mathrm{AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)}$$

The solubility product $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$

When $K_{sp}$ is exceeded, precipitation occurs. This is a heterogeneous equilibrium involving a solid and its ions in solution.

Similarly, for calcium fluoride: $\mathrm{CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)}$ $$K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2$$

The Common Ion Effect

In the equilibrium $\mathrm{AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)}$, adding a source of $\text{Ag}^+$ ions (e.g., $\text{AgNO}_3$) increases $[\text{Ag}^+]$, causing the equilibrium to shift left, decreasing the solubility of AgCl. This is the common ion effect — the suppression of solubility of a salt when a common ion is added.

ICE Tables (Initial-Change-Equilibrium)

ICE tables are systematic ways to solve equilibrium problems:

Example: At 700 K, $K_c = 0.52$ for $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$. If 2.0 mol of N₂ and 6.0 mol of H₂ are placed in a 1.0 dm³ container, find equilibrium concentrations.

	N₂	H₂	NH₃
Initial (I)	2.0	6.0	0
Change (C)	$-x$	$-3x$	$+2x$
Equilibrium (E)	$2.0-x$	$6.0-3x$	$2x$

$$K_c = \frac{(2x)^2}{(2.0-x)(6.0-3x)^3} = 0.52$$

Solve for $x$ (quadratic equation) to find equilibrium concentrations.

Henderson-Hasselbalch Equation (for Acid-Base Equilibrium)

For the weak acid dissociation equilibrium $\mathrm{HA \rightleftharpoons H^+ + A^-}$: $$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$$

Taking logarithms: $$\text{p}K_a = -\log K_a$$

$$\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}$$

This is the Henderson-Hasselbalch equation, useful for buffer solutions. A buffer resists changes in pH because it contains both a weak acid and its conjugate base.

Equilibrium in Contact Process (Sulfuric Acid Production)

$$\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)} \quad \Delta H = -197 \text{ kJ mol}^{-1}$$

This is also exothermic, so lower temperatures favour higher $SO_3$ yield. However, the reaction is very slow at low temperatures. A compromise temperature of about 450°C with a vanadium(V) oxide catalyst is used industrially. High pressure also favours the product (3 moles → 2 moles of gas).

WAEC Past Question Patterns

Typical WAEC questions on chemical equilibrium include:

1. Writing $K_c$ expressions for given equilibrium reactions
2. Applying Le Chatelier’s principle to predict the effect of changes in conditions
3. Calculating equilibrium concentrations using ICE tables
4. Explaining why a catalyst does not affect $K_c$ or equilibrium position
5. Explaining industrial processes using equilibrium principles (Haber process, Contact process)
6. Calculating $K_{sp}$ and predicting precipitation
7. Using the reaction quotient $Q$ to predict the direction of spontaneous change

⚡ WAEC Exam Tip: When writing $K_c$ expressions, make sure you ONLY include gaseous and aqueous species. Pure solids and pure liquids are omitted because their concentrations (activities) are constant. Also watch for reactions where $\Delta n = 0$ — for such reactions, $K_p = K_c$ and pressure changes have no effect on the equilibrium position. Watch out for the trick in equilibrium constant questions: ALWAYS write the balanced equation first before substituting into the $K_c$ expression!

Numerical Worked Example (WAEC-style)

For the equilibrium $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$, $K_c = 0.025$ at 500 K. If the initial concentration of PCl₅ is 0.80 mol dm⁻³, find the equilibrium concentrations of all species.

Solution (ICE table):

	PCl₅	PCl₃	Cl₂
Initial	0.80	0	0
Change	$-x$	$+x$	$+x$
Equilibrium	$0.80-x$	$x$	$x$

$$K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{x^2}{0.80-x} = 0.025$$

$x^2 = 0.025(0.80-x) = 0.020 - 0.025x$

$x^2 + 0.025x - 0.020 = 0$

Using quadratic formula: $x = \frac{-0.025 \pm \sqrt{0.025^2 + 4(0.020)}}{2} = \frac{-0.025 \pm 0.286}{2}$

Taking positive root: $x = \frac{0.261}{2} = 0.1305$ mol dm⁻³

$[\text{PCl}_5] = 0.80 - 0.1305 = 0.670$ mol dm⁻³ $[\text{PCl}_3] = [\text{Cl}_2] = 0.1305$ mol dm⁻³

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