Lectures 3-4 Gibbs energy and bond energies
Page 1: Reaction Information
Reaction: H2O(l) ↔ H2O(g)
Standard Enthalpy of Formation (ΔHӨ):
H2O(l): -286 kJ/mol
H2O(g): -242 kJ/mol
Standard Entropy of Formation (Sө):
H2O(l): 70 J/K·mol
H2O(g): 189 J/K·mol
Question: Is this reaction spontaneous at room temperature?
Page 2: Objectives
Objective 1: Define Gibbs free energy.
Objective 2: Relate Gibbs free energy (ΔG°) to reaction spontaneity, incorporating enthalpy (ΔH°) and entropy (ΔS°).
Objective 3: Show the relationship between ΔG° and equilibrium constant (Keqm).
Page 3: Definition of Gibbs Free Energy
Gibbs Free Energy: The amount of "free" or "available" energy a system possesses to do work.
It predicts whether a process will occur spontaneously at constant pressure and temperature.
Page 4: Gibbs Equation
Thermodynamic equation used to calculate the change in Gibbs energy of a system as a function of temperature.
Named after: Josiah Willard Gibbs and Hermann von Helmholtz.
Page 5: Gibbs Equation and Spontaneity
Gibbs Equation: ΔG = ΔH – TΔS
ΔH: Change in enthalpy
T: Temperature in Kelvin
ΔS: Change in entropy
Relationship: Entropy (ΔS) and enthalpy (ΔH) help determine reaction spontaneity.
Page 6: Interpreting the Gibbs Equation
If heat is released (ΔH is negative), some may convert into useful work.
Some heat increase order of the system (ΔS is negative).
More disorder (ΔS>0) leads to more available energy for work.
Page 7: Interpreting Gibbs Energy
If ΔG < 0: Reaction is spontaneous forward/right direction.
If ΔG > 0: Non-spontaneous forward direction, spontaneous reverse/left direction.
If ΔG = 0: Reaction is at equilibrium (reactants and products equally favored).
Page 8: Interpreting Gibbs for Spontaneity
Preferred conditions:
Spontaneous: Negative ΔH and positive ΔS.
Non-spontaneous: Positive ΔH and negative ΔS.
Same signs: Temperature determines spontaneity.
Page 9: Diagram Summary of Gibbs Equation
Summary of ΔH, ΔS, ΔG:
+ΔH, +ΔS: Reaction spontaneous at high temperature, reverse at low.
-ΔH, +ΔS: Spontaneous at all temperatures.
+ΔH, -ΔS: Spontaneous reverse at all temperatures.
-ΔH, -ΔS: Spontaneous at low temperature, reverse at high temperature.
Page 10: Standard Free Energy
Standard Gibbs Free Energy for reaction (ΔGⓇreaction):
ΔGⓇreaction = [ΣΔG°f (products) - ΣΔG°f (reactants)]
Elements in standard state: ΔG°f = 0.
Page 11: Equilibrium Constant (Keqm)
Definition: Ratio of product concentrations to reactant concentrations in a reversible process.
Example Reaction: A + B ↔ C + D
K = ([C][D])/([A][B])
Indicates if products or reactants are favored.
Page 12: ΔG and Keqm Relationship
Relationship: ΔGӨ = -RTlnK
Interpretation:
K > 1: Products favored (ΔG < 0).
K = 1: Products and reactants equally favored (ΔG = 0).
K < 1: Reactants favored (ΔG > 0).
Page 13: Specific Reaction Evaluation
Reaction: H2O(l) ↔ H2O(g)
Given:
ΔG = -RTlnKc
ΔGӨreaction = +8.5 kJ/mol
R = 8.314 J/K·mol
At 25°C, evaluation:
lnK = - [ΔG/(RT)] = - [8500/(8.314*298)] = -3.43
K = 0.03 (non-spontaneous).
Page 14: Second Set of Objectives
Objective 1: Define bond dissociation energy.
Objective 2: Illustrate bond dissociation energy for determining standard enthalpies of reaction and formation.
Page 15: The Chemical Reaction
Represented:
1 Methane (CH4) molecule + 2 Oxygen (O2) molecules → 1 Carbon Dioxide (CO2) + 2 Water (H2O).
Page 16: Chemical Reaction Description
Chemical reactions involve breaking and forming chemical bonds.
Energy needed to break bonds (supplied as heat).
Energy released when bonds are formed.
Page 17: Bond Dissociation Energy Definition
Bond Dissociation Energy (D): Energy needed to break one mole of bonds in a gaseous covalent substance.
Example: H2(g) → H(g) + H(g) with ΔHӨ = 435 kJ/mol.
Page 18: Bond Dissociation Energy in Polyatomic Molecules
Bond energy varies; examples given:
CH4(g) → CH3(g) + H(g) at ΔHӨ = 427 kJ/mol
Average bond enthalpy for C-H = 416 kJ/mol.
Page 19: Multiple Bonds Energy
Bond energies for double & triple bonds are not multiples of single bond energies.
Strength order: Triple > Double > Single Bonds.
Example energies are provided for C-O and C=O bonds.
Page 20: Uses of Bond Energies
Applications:
Estimate Enthalpy of Reaction
Estimate Enthalpy of Formation (for gaseous phase only).
Formula: ΔHӨreaction = ΣDreactants - ΣDproducts.
Page 21: ΔHӨreaction Calculation Steps
Write balanced equation for reactants forming products (gaseous).
Draw structural formula of all species.
Identify bonds present.
Input bond energy values.
Page 22: Example Problem Setup
Given:
DC-I = 343 kJ/mol
DH-H = 435 kJ/mol
DH-Cl = 432 kJ/mol
Task: Calculate ΔHⓇformation of HCl(g).
Page 23: Example Calculation Steps
Balanced Equation: ½ H2(g) + ½Cl2(g) → HCl(g)
Bonds: ½ H-H + ½ Cl-Cl → H-Cl.
Energy Calculation: ½ DH-H + ½ DCl-Cl – DH-Cl = [½(435) + ½(343)] – [432] = -43 kJ.
Page 24: Another Calculation Setup
Given:
DC-H = 416 kJ/mol
DH-H = 435 kJ/mol
Enthalpy of atomization of graphite = 717 kJ/mol
Task: Calculate ΔHӨformation of CH4(g).