Gibbs Free Energy & Equilibrium Constants

19.7-8

K: Equilibrium Constant

• For a general reaction:

aA + bB --> cC + dD

• K = [C]c[D]d

[A]a[B]b

Equilibrium Molarities orPressures of R/P

K >1 product favored

K <1 reactant favored

Gibbs Free E and Equilibrium Constant

Gº = − RT lnKeq

- R: ideal gas constant, 8.314 J/mole·K

- T: Kelvin

- K: value of equilibrium constant

∆G° and K• At equilibrium: Gº = − RT lnKeq

• If G < 0, then K > 1; product favored• If G = 0, then K = 1; the rx is at

equilibrium.• If G > 0, then K < 1; reactant favored

Practice Problems

1. For a reaction, the G = -16.37 kJ/mol. Calculate the equilibrium constant at 25 C. P or R favored?

2. For a reaction, the K = 1.8 x10-10. Calculate Gº. P or R favored?

Bond Dissociation Energy

[Ch 9: pp.355-357]

• The enthalpy change for breaking a specific type of bond in a molecule

• Breaking bonds is endothermic; forming bonds is exothermic

• ∆H°rxn = Energy to break bonds - Energy to form Bonds

Bond Dissociation Energy

• Estimate the enthalpy for the combustion of propane

• C3H8 + O2 --> CO2 + H2O

Bond: Ave bond diss. energy in kJ mol-1C-H +413C-C +347O=O +498C=O +805H-O +464

All endothermic

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Bond: Ave bond diss. energy in kJ mol-1C-H +413C-C +347O=O +498C=O +805H-O +464