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Under standard conditions — ∆G o sys = ∆H o sys - T∆S o sys free energy = total energy change for system - energy change in disordering the system

If reaction is exothermic (negative ∆ H o ) (energy dispersed) and entropy increases (positive ∆S o ) (matter dispersed) then ∆G o must be NEGATIVE reaction is spontaneous (and product-favored).

If reaction is endothermic (positive ∆H o ) and entropy decreases (negative ∆S o ) then ∆G o must be POSITIVE reaction is NOT spontaneous (and is reactant-favored).

ENTHALPY and ENTROPY

use Gibb’s equation ∆G o = ∆H o - T∆S o Determine ∆H o rxn and ∆S o rxn

Combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate ∆H o rxn = kJ Use standard molar entropies to calculate ∆S o rxn = J/K or kJ/K ∆G o rxn = kJ - (298 K)( J/K) = Kj Reaction is product-favored in spite of negative ∆S o rxn. Reaction is “enthalpy driven”

Is the dissolution of ammonium nitrate product-favored? If so, is it enthalpy- or entropy-driven? NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

From tables of thermodynamic data we find ∆H o rxn = kJ ∆S o rxn = J/K or kJ/K ∆G o rxn = kJ - (298 K)( kJ/K) = -6.7 kJ Reaction is product-favored in spite of positive ∆H o rxn. Reaction is “entropy driven” NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq)

Use tabulated values of ∆G f o, free energies of formation. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

Note that ∆G˚ f for an element = 0

free energy of a standard state element is 0. ∆G o rxn =∆G f o (CO 2 ) - [∆G f o (graph) + ∆G f o (O 2 )] ∆G o rxn = kJ - [ 0 + 0] ∆G o rxn = kJ Reaction is product-favored as expected. ∆G o rxn =  ∆G f o (products) -  ∆G f o (reactants)

More thermo? You betcha!

∆G o rxn is the change in free energy when pure reactants convert COMPLETELY to pure products. Product-favored systems have K eq > 1. Therefore, both ∆G˚ rxn and K eq are related to reaction favorability.

K eq is related to reaction favorability and so to ∆G o rxn. ∆G o rxn = - RT lnK where R = 8.31 J/Kmol The larger the value of K the more negative the value of ∆G o rxn