Gibbs Free Energy and Bond Dissociation Energy Notes

Gibbs Free Energy

  • Gibbs free energy is the amount of "free" or "available" energy a system possesses to do work.
  • It is a thermodynamic property used to predict whether a process will occur spontaneously at constant pressure and temperature.

Gibbs Equation

  • The Gibbs equation is a thermodynamic equation used to calculate the change in the Gibbs energy of a system as a function of temperature.
  • It is named after Josiah Willard Gibbs and Hermann von Helmholtz.
  • The Gibbs equation is: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
    • ΔG\Delta G: Change in Gibbs free energy
    • ΔH\Delta H: Change in enthalpy
    • TT: Kelvin temperature
    • ΔS\Delta S: Change in entropy
  • Entropy and enthalpy can help determine the spontaneity of a reaction.

Interpreting the Gibbs Equation

  • In terms of enthalpy and entropy:
    • If heat is released in a chemical reaction (ΔH\Delta H is negative), some of this heat may be converted into useful work, while some may increase the order of the system (if ΔS\Delta S is negative).
    • If the system becomes more disordered (\Delta S > 0; positive), more useful energy becomes available to do work.
  • In terms of Gibbs Energy:
    • If ΔG\Delta G is negative, the reaction is spontaneous in the forward direction.
    • If ΔG\Delta G is positive, the reaction is non-spontaneous in the forward direction but spontaneous in the reverse direction.
    • If ΔG\Delta G is zero, the reaction is at equilibrium (reactants and products are equally favored).
  • In terms of spontaneity:
    • For spontaneous reactions, a negative ΔH\Delta H and a positive ΔS\Delta S are preferred.
    • For non-spontaneous reactions, a positive ΔH\Delta H and a negative ΔS\Delta S are preferred.
    • If both have the same signs, temperature determines what happens.

Four Scenarios for Enthalpy and Entropy Changes

  • \Delta S > 0 (increase in entropy):
    • \Delta H > 0 (endothermic): \Delta G < 0 at high temperature. Process is spontaneous at high temperature.
    • \Delta H < 0 (exothermic): \Delta G < 0 at any temperature. Process is spontaneous at any temperature.
  • \Delta S < 0 (decrease in entropy):
    • \Delta H > 0 (endothermic): \Delta G > 0 at any temperature. Process is non-spontaneous at any temperature.
    • ΔH<0\Delta H < 0 (exothermic): ΔG<0\Delta G < 0 at low temperature. Process is spontaneous at low temperature. ΔG>0\Delta G > 0 at high temperature.

Standard Free Energy

  • ΔGΘ<em>reaction=∑nΔG°</em>f(products)−∑nΔGf°(reactants)\Delta G^{\Theta}<em>{reaction} = \sum n \Delta G^{\degree}</em>f (products) - \sum n \Delta G^{\degree}_f (reactants)
  • ΔGfΘ\Delta G^{\Theta}_f (elements in standard state) = 0

Equilibrium Constant (KeqmK_{eqm})

  • It is the ratio of the concentration of products to the concentration of reactants in a reversible chemical process.
  • It tells us if products or reactants are favored.
  • Keqm=[products][reactants]K_{eqm} = \frac{[products]}{[reactants]}

Relationship between ΔG\Delta G and KeqmK_{eqm}

  • ΔGΘ=−RTln⁡K\Delta G^{\Theta} = -RT\ln K
    • R = 8.314 J/K.mol
  • If K > 1, ln K is positive, and ΔG\Delta G is negative (products favored).
  • If K = 1, ln K is 0, products and reactants are equally favored.
  • If K < 1, ln K is negative, and ΔG\Delta G is positive (reactants favored).

Bond Dissociation Energy

  • Bond dissociation energy is the amount of energy required to break one mole of bonds in a gaseous covalent substance to form products in the gaseous state at constant temperature and pressure.
  • It is the enthalpy change required to break a particular bond in one mole of gaseous molecules.
  • Example: H2(g) → H(g) + H(g) ΔHΘ<em>rxn\Delta H^{\Theta}<em>{rxn} or ΔHΘ</em>H−H\Delta H^{\Theta}</em>{H-H} = 435 kJ/mol

Bond Dissociation Energy of Polyatomic Molecules

  • In polyatomic molecules, bond energy varies according to its environment.
  • Example: CH<em>4(g)→CH</em>3(g)+H(g)ΔHΘ=427 kJ/mol\text{CH}<em>4(g) \rightarrow \text{CH}</em>3(g) + \text{H}(g) \quad \Delta H^{\Theta} = 427 \text{ kJ/mol}
    CH<em>3(g)→CH</em>2(g)+H(g)ΔHΘ=439 kJ/mol\text{CH}<em>3(g) \rightarrow \text{CH}</em>2(g) + \text{H}(g) \quad \Delta H^{\Theta} = 439 \text{ kJ/mol}
    CH<em>2(g)→CH(g)+H(g)ΔHΘ=452 kJ/mol\text{CH}<em>2(g) \rightarrow \text{CH}(g) + \text{H}(g) \quad \Delta H^{\Theta} = 452 \text{ kJ/mol}CH(g)→C(g)+H(g)ΔHΘ=347 kJ/mol\text{CH}(g) \rightarrow \text{C}(g) + \text{H}(g) \quad \Delta H^{\Theta} = 347 \text{ kJ/mol}CH</em>4(g)→C(g)+4H(g)ΔHΘ=1665 kJ/mol\text{CH}</em>4(g) \rightarrow \text{C}(g) + 4\text{H}(g) \quad \Delta H^{\Theta} = 1665 \text{ kJ/mol}
  • Average bond energy for C-H = 416 kJ/mol. Hence bond energies are actually average bond enthalpies.

Bond Dissociation Energy of Multiple Bonds

  • Bond energies for double & triple bonds are NOT MULTIPLES of the single bond energies.
  • Triple bonds are STRONGER than double bonds, which in turn are STRONGER than single bonds.
  • Example:
    • C-O ΔHΘ=351 kJ/mol\Delta H^{\Theta} = 351 \text{ kJ/mol}
    • C=O ΔHΘ=741 kJ/mol\Delta H^{\Theta} = 741 \text{ kJ/mol}
    • C≡O ΔHΘ=1070 kJ/mol\Delta H^{\Theta} = 1070 \text{ kJ/mol}

Use of Bond Energies

  • Estimates Enthalpy of Reaction
  • Estimates Enthalpy of Formation
  • For gaseous phase ONLY: ΔHΘ<em>reaction=∑D</em>reactants−∑Dproducts=Energy supplied to break bonds−Energy released to form bonds\Delta H^{\Theta}<em>{reaction} = \sum D</em>{reactants} - \sum D_{products} = \text{Energy supplied to break bonds} - \text{Energy released to form bonds}

Calculating ΔHΘ\Delta H^{\Theta} reaction from Bond Energies

  1. Write balanced equation to show reactants forming products (in the gaseous form).
  2. Draw formula of all species.
  3. Identify the bonds present.
  4. Substitute the values of bond energies.