Chapter 1: Chemical Equilibrium

Overview

Chemical equilibrium is a fundamental concept in chemistry that describes the state where the forward and reverse reactions occur at the same rate, resulting in no net change in the concentrations of reactants and products. This chapter explores the dynamic nature of equilibrium, the factors that influence equilibrium position, and Le Chatelier's principle, which provides a powerful tool for predicting how systems respond to changes. Understanding equilibrium is crucial for industrial processes, environmental chemistry, and biological systems.

Learning Objectives

After studying this chapter, you should be able to:

Define chemical equilibrium and explain its dynamic nature
Write equilibrium expressions and calculate equilibrium constants (Kc and Kp)
Apply Le Chatelier's principle to predict how systems respond to changes
Understand the factors affecting equilibrium position (concentration, temperature, pressure)
Analyze industrial applications of chemical equilibrium
Perform calculations involving equilibrium constants and concentrations

1.1 Introduction to Chemical Equilibrium

What is Chemical Equilibrium?

Chemical equilibrium is a dynamic state in a reversible reaction where:

The forward reaction rate equals the reverse reaction rate
Concentrations of reactants and products remain constant over time
The reaction appears to have stopped, but molecular collisions continue

Dynamic Nature of Equilibrium

Unlike static equilibrium, chemical equilibrium is dynamic - reactions continue to occur at the molecular level, but there is no net change in macroscopic properties.

For a general reversible reaction:

aA + bB \rightleftharpoons cC + dD

At equilibrium:

Rate of forward reaction = Rate of reverse reaction
[A], [B], [C], [D] remain constant
The system is closed (no matter enters or leaves)

Types of Equilibrium

Physical Equilibrium

Phase equilibrium: Liquid ⇌ Gas (evaporation-condensation)
Solution equilibrium: Solute ⇌ Dissolved solute (dissolution-precipitation)

Chemical Equilibrium

Involves chemical reactions where reactants ⇌ products
Example: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ (Haber process)

1.2 Equilibrium Constant and Expression

Writing Equilibrium Expressions

For a general reaction:

aA + bB \rightleftharpoons cC + dD

The equilibrium constant expression (Kc) is:

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

Where:

= molar concentration (mol/d $m^3$ )
Kc = equilibrium constant in terms of concentration
Exponents = stoichiometric coefficients

Characteristics of Equilibrium Constants

Constant at constant temperature: Kc is constant for a given reaction at a specific temperature
Temperature dependent: Kc changes with temperature
No units: Kc is dimensionless for most reactions
Large Kc (>1): Products favored at equilibrium
Small Kc (<1): Reactants favored at equilibrium

Example Equilibrium Calculations

Example 1: For the reaction $\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ , the equilibrium constant expression is:

K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 \times [\text{O}_2]}

Example 2: For the reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ , the equilibrium constant expression is:

K_c = \frac{[\text{PCl}_3] \times [\text{Cl}_2]}{[\text{PCl}_5]}

1.3 Factors Affecting Equilibrium

Le Chatelier's Principle

Le Chatelier's principle states: "If a system at equilibrium is disturbed by a change in concentration, temperature, pressure, or volume, the system will shift its equilibrium position to counteract the disturbance."

1. Concentration Changes

Increase [reactants]: Equilibrium shifts to the right (toward products)
Increase [products]: Equilibrium shifts to the left (toward reactants)
Decrease [reactants]: Equilibrium shifts to the left
Decrease [products]: Equilibrium shifts to the right

Practical Example

For the reaction $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$ (red color):

Adding more $\text{Fe}^{3+}$ or $\text{SCN}^-$ intensifies the red color
Adding water dilutes all concentrations, shifting equilibrium to reduce total particles

2. Temperature Changes

3. Pressure and Volume Changes

Only affects reactions involving gases. The system shifts to reduce the number of gas molecules when pressure increases.

General rule: System shifts toward the side with fewer moles of gas to reduce pressure.

Examples:

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ (4 moles ⇌ 2 moles)
- Increase pressure: Shifts right (fewer gas molecules)
- Decrease pressure: Shifts left (more gas molecules)
$\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ (1 mole ⇌ 2 moles)
- Increase pressure: Shifts left (fewer gas molecules)
- Decrease pressure: Shifts right (more gas molecules)

4. Catalyst Effects

Catalysts speed up both forward and reverse reactions equally
No change in equilibrium position or equilibrium constant
Only helps system reach equilibrium faster

1.4 Industrial Applications of Equilibrium

Haber Process for Ammonia Synthesis

\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92 \text{ kJ/mol}

Optimization conditions:

High pressure (200-300 atm): Shifts equilibrium to right
Moderate temperature (450°C): Balances rate and yield
Catalyst: Iron (Fe) to speed up reaction
Recycling unreacted gases: Increases overall yield

Contact Process for Sulfuric Acid

2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \quad \Delta H = -197 \text{ kJ/mol}

Optimization conditions:

High pressure (1-2 atm)
High temperature (400-450°C)
Catalyst: Vanadium(V) oxide ( $V_2O_5$ )

Ostwald Process for Nitric Acid

4\text{NH}_3(g) + 5\text{O}_2(g) \rightleftharpoons 4\text{NO}(g) + 6\text{H}_2\text{O}(g) \quad \Delta H = -906 \text{ kJ/mol}

Optimization conditions:

High temperature (800-900°C)
Catalyst: Platinum-Rhodium gauze

Equilibrium Calculations

Calculating Equilibrium Constants

Step-by-step approach:

Write the balanced equation
Write the equilibrium constant expression
Calculate concentrations at equilibrium
Substitute into the K expression
Calculate K value

Example: For the reaction $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ , if at equilibrium $[\text{H}_2] = 0.22 \text{ M}$ , $[\text{I}_2] = 0.22 \text{ M}$ , and $[\text{HI}] = 1.56 \text{ M}$ :

K_c = \frac{[\text{HI}]^2}{[\text{H}_2] \times [\text{I}_2]}

K_c = \frac{(1.56)^2}{(0.22) \times (0.22)}

K_c = \frac{2.4336}{0.0484} = 50.3

Using K to Find Equilibrium Concentrations

Example: For $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$ , $K_c = 4.00$ . If $[\text{CO}] = 0.20 \text{ M}$ and $[\text{H}_2\text{O}] = 0.30 \text{ M}$ at equilibrium, find $[\text{CO}_2]$ and $[\text{H}_2]$ :

4.00 = \frac{[\text{CO}_2] \times [\text{H}_2]}{(0.20) \times (0.30)}

4.00 = \frac{[\text{CO}_2] \times [\text{H}_2]}{0.06}

[\text{CO}_2] \times [\text{H}_2] = 0.24

Since $[\text{CO}_2] = [\text{H}_2]$ , then:

[\text{CO}_2]^2 = 0.24

[\text{CO}_2] = \sqrt{0.24} = 0.49 \text{ M}

Did You Know?

The Haber process produces approximately 150 million tons of ammonia annually, which is essential for fertilizers that feed about half the world's population. Without this equilibrium-controlled process, global food production would be severely limited.

1.5 Common Equilibrium Systems

Solubility Equilibrium

\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)

K_{sp} = [\text{Ag}^+] \times [\text{Cl}^-] \quad \text{(Solubility product constant)}

Acid-Base Equilibrium

\text{HA}(aq) \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq)

K_a = \frac{[\text{H}^+] \times [\text{A}^-]}{[\text{HA}]} \quad \text{(Acid dissociation constant)}

Complex Ion Equilibrium

Ag^+(aq) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^+(aq)

K_f = \frac{[[Ag(NH_3)_2]^+]}{[Ag^+] \times [NH_3]^2} \quad \text{(Formation constant)}

SPM Exam Tips

Always check if the reaction is reversible (⇌) before applying equilibrium concepts

When writing Kc expressions, remember:

Only include aqueous and gas species

Solids and liquids have concentration = 1 (not included)

Exponents come from balanced equation coefficients

For pressure changes, only consider gaseous species

Exothermic: ΔH < 0; Endothermic: ΔH > 0

1.6 Common Equilibrium Systems

Solubility Equilibrium

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag⁺] × [Cl⁻]  (Solubility product constant)

Acid-Base Equilibrium

HA(aq) ⇌ H⁺(aq) + A⁻(aq)
Ka = [H⁺] × [A⁻] / [HA]  (Acid dissociation constant)

Complex Ion Equilibrium

Ag⁺(aq) + 2N$H_3$(aq) ⇌ [Ag(N$H_3$)₂]⁺(aq)
Kf = [[Ag(N$H_3$)₂]⁺] / [Ag⁺] × [N$H_3$]²  (Formation constant)

Safety Reminder

When working with equilibrium experiments:

Use proper eye protection and lab coats

Handle chemicals in a well-ventilated area

Be careful with concentrated acids and bases

Follow proper waste disposal procedures

Never heat closed containers

Summary

Key Concepts

Chemical equilibrium is a dynamic state where forward and reverse reaction rates are equal
Equilibrium constant (Kc) relates concentrations of reactants and products at equilibrium
Le Chatelier's principle predicts how systems respond to disturbances
Factors affecting equilibrium:
- Concentration: Shifts to oppose changes
- Temperature: Shifts to absorb/release heat
- Pressure: Shifts to reduce gas molecules (gases only)
- Catalysts: Speed up both reactions equally
Industrial applications optimize conditions for maximum yield and efficiency

Equilibrium Problem-Solving Strategy

Identify if the system is at equilibrium
Write the equilibrium constant expression
Determine what changes occur (concentration, temperature, pressure)
Apply Le Chatelier's principle to predict the shift
Calculate new equilibrium concentrations if needed

Practice Questions

For the reaction 2NO(g) + $O_2$(g) ⇌ 2N$O_2$(g), write the Kc expression and explain how equilibrium shifts when: a) Oxygen is added b) Temperature increases (ΔH = -114 kJ/mol) c) Pressure decreases
In the Haber process, explain why high pressure is used but moderate temperature is preferred.
Calculate Kc for $N_2O_4$(g) ⇌ 2N$O_2$(g) if at equilibrium [ $N_2O_4$ ] = 0.042 M and [N $O_2$ ] = 0.122 M.

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