Gibbs Energy Change and Equilibrium

✨ Key Ideas

Gibbs free energy (G) helps us predict two important things about chemical reactions:

  • 🎯 Whether a reaction happens spontaneously
  • ⚡ How much useful work we can get from it

🧪 Equilibrium Condition

At equilibrium, the system’s free energy is at its minimum, and:

\[ \Delta_r G = 0 \]

This means for a reaction like \( A + B \rightleftharpoons C + D \), the forward and reverse reactions perfectly balance each other ⚖️.

🔗 The Magic Equation

For reactions where all substances are in their standard states, we use:

\[ \Delta_r G^\circ = -RT \ln K \quad \text{or} \quad \Delta_r G^\circ = -2.303 RT \log K \]

Where:
– \( \Delta_r G^\circ \) = standard Gibbs energy change
– \( R \) = gas constant (8.314 J/mol·K)
– \( T \) = temperature in Kelvin
– \( K \) = equilibrium constant

🔥🌡️ How Heat (ΔH°) and Chaos (ΔS°) Affect Reactions

ΔH°ΔS°Spontaneity
– (exothermic)+ (more disorder)✅ Spontaneous at all temperatures
– (exothermic)– (more order)✅ Spontaneous at low temperatures
❌ Not spontaneous at high temperatures
+ (endothermic)+ (more disorder)❌ Not spontaneous at low temperatures
✅ Spontaneous at high temperatures
+ (endothermic)– (more order)❌ Never spontaneous

Note: “Low” and “high” temperatures depend on the specific reaction!

📊 Equilibrium Constant Tips

  • If \( K \gg 1 \) (large K): Reaction favors products 🟢
  • If \( K \ll 1 \) (small K): Reaction favors reactants 🔴
  • Exothermic reactions (ΔH° negative) usually have large K
  • Endothermic reactions (ΔH° positive) usually have small K

💡 Problem Solving Toolkit

Example 1: Finding ΔG° from K
For \( \frac{3}{2}O_2(g) \rightarrow O_3(g) \) at 298 K with \( K_p = 2.47 \times 10^{-29} \):

\[ \Delta G^\circ = -2.303 \times 8.314 \times 298 \times \log(2.47 \times 10^{-29}) = 163\ \text{kJ/mol} \]
Interpretation: Large positive ΔG° means ozone formation isn’t spontaneous.

Example 2: Finding K from ΔG°
For \( 2NH_3(g) + CO_2(g) \leftrightharpoons NH_2CONH_2(aq) + H_2O(l) \) with ΔG° = -13.6 kJ/mol at 298 K:

\[ \log K = \frac{13.6 \times 10^3}{2.303 \times 8.314 \times 298} = 2.38 \quad \Rightarrow \quad K = 2.4 \times 10^2 \]
Interpretation: Negative ΔG° gives K > 1, meaning products are favored 👍.

Example 3: Finding K from dissociation
For \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \) at 60°C with 50% dissociation:
– Mole fractions: \( x_{N_2O_4} = \frac{0.5}{1.5} \), \( x_{NO_2} = \frac{1}{1.5} \)
– Partial pressures: \( p_{N_2O_4} = \frac{0.5}{1.5} \ \text{atm} \), \( p_{NO_2} = \frac{1}{1.5} \ \text{atm} \)

\[ K_p = \frac{(p_{NO_2})^2}{p_{N_2O_4}} = 1.33\ \text{atm} \]

🚀 NEET Must-Knows

  1. Equilibrium condition: \( \Delta_r G = 0 \) ⚖️
  2. Master the equation: \( \Delta G^\circ = -RT \ln K \) 🔑
  3. Predict spontaneity using ΔH° and ΔS° signs (Table 5.4) 🔥❄️
  4. Calculate K from ΔG° (and vice versa) 🧮
  5. Find K for dissociation reactions (like N₂O₄ example) 🧪

💎 Key Takeaway

Gibbs energy is your reaction “thermometer” 🌡️ – it tells you if a reaction will run (ΔG < 0), stay stuck (ΔG > 0), or chill at equilibrium (ΔG = 0)! Combine it with K to predict reaction outcomes.