Chemical Thermodynamics

Why it matters

Chemical thermodynamics is essential for understanding how energy is transferred and transformed in chemical reactions and processes. This knowledge is crucial for designing efficient engines, reactors, and other systems in mechanical engineering.

Key ideas

• Chemical Potential: The partial molar Gibbs energy μ_i = (∂G/∂n_i) at constant T, P and amounts of all other species.
• Gibbs Free Energy (G): Its decrease gives maximum reversible non-expansion work under suitable constant-temperature/pressure constraints. It helps predict the direction of chemical reactions.
• Enthalpy (H): The state property H = U + PV; its change equals heat in suitable constant-pressure processes with only pressure–volume work.
• Entropy (S): A measure of disorder or randomness in a system, crucial for determining the feasibility of a reaction.
• Equilibrium Constant (K): Relates to the concentrations of reactants and products at equilibrium, providing insight into reaction extents.

Formulas

• ΔG = ΔH - TΔS

  • ΔG: Change in Gibbs Free Energy (J)
  • ΔH: Change in Enthalpy (J)
  • T: Temperature (K)
  • ΔS: Change in Entropy (J/K)

• K = e^(-ΔG°/RT)

  • K: Equilibrium Constant (dimensionless)
  • ΔG°: Standard molar Gibbs energy of reaction (J/mol of reaction)
  • R: Universal Gas Constant (8.314 J/(mol·K))
  • T: Temperature (K)

For actual mixture composition, reaction Gibbs energy is Δ_rG = Δ_rG° + RT ln Q_r, where Q_r is the dimensionless activity-based reaction quotient. At equilibrium Δ_rG = 0 and Q_r = K; Δ_rG° is generally not zero. Temperature-dependent standard values should be used when conditions change. The relation ΔG = ΔH-TΔS assumes the compared states have the same T.

Worked example

Problem: Calculate the equilibrium constant for a reaction at 298 K, given standard reaction values ΔH° = -100 kJ/mol and ΔS° = -200 J/(mol·K) at 298 K. Use these to compute ΔG°; the equilibrium relation requires standard-state values.

1. Convert ΔH to J/mol: ΔH = -100 kJ/mol × 1000 J/kJ = -100000 J/mol
2. Calculate ΔG using ΔG = ΔH - TΔS
   • ΔG = -100000 J/mol - 298 K × (-200 J/(mol·K))
   • ΔG = -100000 J/mol + 59600 J/mol
   • ΔG = -40400 J/mol
3. Calculate K using K = e^(-ΔG°/RT)
   • K = e^(-(-40400 J/mol) / (8.314 J/(mol·K) × 298 K))
   • K = e^(16.3063)
   • K ≈ 1.207 × 10^7

Answer: 1.207 × 10^7 (dimensionless)

Common mistakes

• Confusing units, especially when converting kJ to J.
• Misapplying the sign of ΔS when calculating ΔG.
• Forgetting to convert temperature to Kelvin.

For GATE ME

Questions often involve calculating Gibbs Free Energy, equilibrium constants, or predicting reaction spontaneity. Practice problems involving unit conversions and understanding the physical significance of thermodynamic properties.

Quick check

1. What does Gibbs Free Energy indicate about a reaction?
2. How is the equilibrium constant related to Gibbs Free Energy?
3. Why is entropy important in chemical thermodynamics?

Answers: 1. Reaction Gibbs energy indicates the favored infinitesimal reaction direction under fixed T and P; 2. Through the formula K = e^(-ΔG°/RT); 3. It determines reaction feasibility.