Chemical Equilibrium: Reversible Reactions, Equilibrium Constant, Le Chatelier’s Principle

TAKEAWAYS:

• Chemical equilibrium is a dynamic balance where forward and reverse reactions occur at equal rates, and concentrations of reactants and products remain constant.
• The equilibrium constant (Kc or Kp) quantifies the ratio of product to reactant concentrations at equilibrium, and its value depends only on temperature.
• Le Chatelier’s Principle predicts how a system at equilibrium responds to changes in concentration, pressure, or temperature.
• Equilibrium calculations require setting up an ICE table (Initial, Change, Equilibrium) and solving for unknowns using stoichiometry and Kc/Kp.
• Real-world applications include industrial processes (e.g., Haber process for ammonia) and biological systems (e.g., oxygen transport in blood).
• NEB exam focus: Problems often test ICE tables, equilibrium expressions, and predicting shifts—practice these step-by-step.

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### **1. What is Chemical Equilibrium?**
Chemical equilibrium occurs when a **reversible reaction** proceeds in both forward and reverse directions at the **same rate**. At this point:
- The concentrations of reactants and products **stop changing** (they become constant).
- The system is **not static**—reactants are still forming products, and products are still forming reactants, but the **net change is zero**.

#### **Key Observations:**
1. **Dynamic Equilibrium**: Reactants → Products **and** Products → Reactants happen simultaneously.
2. **Closed System**: Equilibrium can only be reached in a closed system (no reactants/products can escape).
3. **Constant Macroscopic Properties**: Color, pressure, concentration, etc., appear unchanged over time.

#### **Example of a Reversible Reaction:**
```mermaid
flowchart LR
    A["Reactants (N₂ + 3H₂)"] -->|Forward Reaction| B["Products (2NH₃)"]
    B -->|Reverse Reaction| A
    B & A -->|Equilibrium| C["Dynamic Balance"]

Real-Life Example: The reaction between nitrogen and hydrogen to form ammonia (Haber process): $${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$$

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2. Writing Equilibrium Expressions (Kc and Kp)

For a general reversible reaction: $$aA+bB\rightleftharpoons cC+dD$$ The equilibrium constant expression is written as: $$K_c=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$$

• Square brackets [ ] denote molar concentrations at equilibrium (mol/L).
• Solids and pure liquids are not included in the expression (their concentrations are constant).
• Kc is called the equilibrium constant in terms of concentration.

Example 1: Writing Kc for a Reaction

Reaction: $${\text{PCl}}_5(g)\rightleftharpoons {\text{PCl}}_3(g)+{\text{Cl}}_2(g)$$ Equilibrium Expression: $$K_c=\frac{[{\text{PCl}}_3][{\text{Cl}}_2]}{[{\text{PCl}}_5]}$$ (No solids/liquids are present here.)

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3. The Equilibrium Constant (Kc and Kp)

Properties of Kc:

1. Constant at a Given Temperature: Kc changes only if temperature changes.
2. Unitless or Has Units: Depends on the reaction (e.g., $K_c$ for ${\text{N}}_2+3{\text{H}}_2\rightleftharpoons 2{\text{NH}}_3$ is unitless because the exponents cancel out).
3. Predicts Reaction Direction:
   • If $K_c\gg 1$: Products are favored (reaction goes mostly to the right).
   • If $K_c\ll 1$: Reactants are favored (reaction goes mostly to the left).

Kp: Equilibrium Constant in Terms of Pressure

For gaseous reactions, we can also express equilibrium in terms of partial pressures (Kp): $$K_p=K_c(RT{)}^{\Delta n}$$ where:

• $R$ = gas constant (0.0821 L·atm·K⁻¹·mol⁻¹),
• $T$ = temperature (K),
• $\Delta n$ = moles of gaseous products – moles of gaseous reactants.

Example 2: Calculating Kp from Kc

Reaction: $${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$$ Given:

• $K_c=0.50$ at 400°C,
• $T=400°C=673\text{K}$,
• $\Delta n=2-(1+3)=-2$.

Calculation: $$K_p=K_c(RT{)}^{\Delta n}=0.50\times (0.0821\times 673{)}^{-2}$$ (Note: This is a simplified example; actual calculations may vary.)

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4. ICE Tables: Solving Equilibrium Problems

The ICE table (Initial, Change, Equilibrium) is a step-by-step method to solve equilibrium problems.

Steps to Solve an ICE Table:

1. Write the balanced equation.
2. List initial concentrations (include initial concentrations of all species, even if zero).
3. Define the change (use a variable $x$ for the amount reacted).
4. Write equilibrium concentrations (Initial + Change).
5. Plug into the equilibrium expression and solve for $x$.

Example 3: Solving an Equilibrium Problem

Problem: For the reaction ${\text{H}}_2(g)+{\text{I}}_2(g)\rightleftharpoons 2\text{HI}(g)$, $K_c=50.0$ at 400°C. If initially $[{\text{H}}_2]=[{\text{I}}_2]=0.10\text{M}$ and $[\text{HI}]=0$, what are the equilibrium concentrations?

Solution:

Equilibrium Expression: $$K_c=\frac{[\text{HI}{]}^2}{[{\text{H}}_2][{\text{I}}_2]}=50.0$$ $$\frac{(2x{)}^2}{(0.10-x)(0.10-x)}=50.0$$ $$\frac{4x^2}{(0.10-x{)}^2}=50.0$$ $$4x^2=50.0(0.10-x{)}^2$$ (Solve the quadratic equation to find $x=0.095$.)

Equilibrium Concentrations:

• $[{\text{H}}_2]=[{\text{I}}_2]=0.10-0.095=0.005\text{M}$,
• $[\text{HI}]=2\times 0.095=0.19\text{M}$.

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5. Le Chatelier’s Principle: Predicting Shifts in Equilibrium

Le Chatelier’s Principle states:

"If a system at equilibrium is subjected to a change (stress), the system will shift in a direction to counteract that change."

Factors Affecting Equilibrium:

Example 4: Applying Le Chatelier’s Principle

Reaction: $$\text{CO}(g)+{\text{H}}_2\text{O}(g)\rightleftharpoons {\text{CO}}_2(g)+{\text{H}}_2(g)\Delta H=+41\text{kJ}$$ (Endothermic reaction: absorbs heat.)

Questions:

1. What happens if we increase the pressure?
   • Answer: Shift left (3 mol gas → 2 mol gas).
2. What happens if we increase the temperature?
   • Answer: Shift right (endothermic reaction absorbs heat).
3. What happens if we add more CO?
   • Answer: Shift right (to reduce excess CO).

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6. Real-World Applications of Chemical Equilibrium

1. Haber Process (Ammonia Production):

   • ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$
   • Conditions: High pressure (shift right), moderate temperature (compromise between rate and yield), catalyst (Fe).
   • Why? Ammonia is used in fertilizers and explosives.

2. Contact Process (Sulfuric Acid Production):

   • $2{\text{SO}}_2(g)+{\text{O}}_2(g)\rightleftharpoons 2{\text{SO}}_3(g)$
   • Conditions: High pressure, low temperature, V₂O₅ catalyst.

3. Blood Oxygen Transport:

   • Hemoglobin (Hb) binds oxygen reversibly: $$\text{Hb}+{\text{O}}_2\rightleftharpoons {\text{HbO}}_2$$
   • Le Chatelier’s Principle: When ${\text{O}}_2$ is low (e.g., in tissues), the equilibrium shifts right to release ${\text{O}}_2$.

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7. Common Mistakes to Avoid

1. Forgetting to include all species in the ICE table (e.g., missing initial concentrations of products).
2. Incorrectly setting up the equilibrium expression (e.g., including solids/liquids).
3. Assuming equilibrium is reached instantly (it takes time, especially without a catalyst).
4. Ignoring the effect of temperature on Kc (Kc changes only with temperature, not pressure/concentration).
5. Miscounting moles of gas for pressure effects (always count gaseous species only).

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Exam Tip: How to Score Full Marks in NEB Questions

NEB questions on Chemical Equilibrium typically test:

1. Writing equilibrium expressions (Kc/Kp).
2. Solving ICE tables (always show all steps).
3. Applying Le Chatelier’s Principle (predict shifts clearly).
4. Relating Kc to reaction conditions (temperature dependence).

NEB-Style Questions & Solutions

Question 1 (Short Answer): For the reaction ${\text{PCl}}_5(g)\rightleftharpoons {\text{PCl}}_3(g)+{\text{Cl}}_2(g)$, write the expression for $K_c$. Answer: $$K_c=\frac{[{\text{PCl}}_3][{\text{Cl}}_2]}{[{\text{PCl}}_5]}$$

Question 2 (Calculation): For the reaction ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$, $K_c=0.060$ at 400°C. If initially $[{\text{N}}_2]=1.0\text{M}$, $[{\text{H}}_2]=2.0\text{M}$, and $[{\text{NH}}_3]=0$, find the equilibrium concentration of ${\text{NH}}_3$. Solution:

Equilibrium Expression: $$K_c=\frac{[{\text{NH}}_3{]}^2}{[{\text{N}}_2][{\text{H}}_2{]}^3}=0.060$$ $$\frac{(2x{)}^2}{(1.0-x)(2.0-3x{)}^3}=0.060$$ (Assume $x$ is small: $1.0-x\approx 1.0$, $2.0-3x\approx 2.0$.) $$\frac{4x^2}{1.0\times (2.0{)}^3}=0.060$$ $$4x^2=0.060\times 8$$ $$x^2=0.12$$ $$x=0.346\text{M}$$ Equilibrium $[{\text{NH}}_3]=2x=0.692\text{M}$. (Note: Exact solution requires solving the cubic equation, but approximation is often acceptable in exams.)

Question 3 (Le Chatelier’s Principle): How will the equilibrium $\text{CO}(g)+{\text{H}}_2\text{O}(g)\rightleftharpoons {\text{CO}}_2(g)+{\text{H}}_2(g)$ shift if:

1. Pressure is increased? Answer: Shift left (3 mol → 2 mol).
2. Temperature is decreased? Answer: Shift left (exothermic in reverse).
3. More ${\text{CO}}_2$ is added? Answer: Shift left (to reduce excess ${\text{CO}}_2$).

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Summary Table: Key Concepts at a Glance

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Final Advice for NEB Exams

1. Practice ICE tables until you can solve them quickly.
2. Memorize Le Chatelier’s Principle and apply it systematically.
3. Check units in equilibrium expressions (especially for Kp).
4. Show all steps in calculations—partial credit is given for correct reasoning.
5. Review past NEB questions to identify common patterns.

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