Question

� Gibbs Free Energy: Equilibrium Constant

Nitric oxide, NO, also known as nitrogen monoxide, is one of the primary contributors to air pollution, acid rain, and the depletion of the ozone layer. The reaction of oxygen and nitrogen to form nitric oxide in an automobile engine is

N2(g)+O2(g)?2NO(g)

The spontaneity of a reaction can be determined from the free energy change for the reaction, ?G?.

A reaction is spontaneous when the free energy change is less than zero.

A reaction is nonspontaneous when the free energy change is greater than zero.

A reaction is in equilibrium when the free energy change is equal to zero.

Relation among thermodynamic quantities

The thermodynamic properties for a reaction are related by the equation that defines the standard free energy, ?G?, in kJ/mol:

?G?=?H??T?S?

where ?H? is the standard enthalpy change in kJ/mol and ?S? is the standard entropy change in J/(mol?K). A good approximation of the free energy change at other temperatures, ?GT, can also be obtained by utilizing this equation and assuming enthalpy (?H?) and entropy (?S?) change little with temperature.

Part A

For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data:

?H?rxn	180.5kJ/mol
?S?rxn	24.80J/(mol?K)

Calculate the temperature in kelvins above which this reaction is spontaneous.

Express your answer to four significant figures and include the appropriate units.

T =

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Relationship between free energy and the equilibrium constant

The standard free energy change, ?G?, and the equilibrium constant K for a reaction can be related by the following equation:

?G?=?RTlnK

where T is the Kelvin temperature and R is equal to 8.314 J/(mol?K).

Part B

Calculate the equilibrium constant for the following reaction at room temperature, 25 ?C:

N2(g)+O2(g)?2NO(g)

Express your answer numerically to three significant figures.

K =

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Answer 1

A) the given reaction is

N2 + 02 ---> 2 N0

we know that

for a reaction to be spontaneous

dG < 0

also

dG = dH - TdS

so

dH - TdS < 0

given

dH = 180.5 kJ /mol

also

dS = 24.80 J / mol K

so

dH - TdS < 0

so

( 180.5 x 1000) - ( T x 24.80) < 0

180.5 x 1000 < 24.80T

T > 7278.22 K

so

the reaction is spontaneous at 7278 K

B)

we know that

dGo = dHo - TdSo

so

dGo = ( 180.5 x1000) - ( 298 x 24.80)

dGo = 173.1096 x 1000 J / mol

we know that

at equilibrium

dGo = -RT lnKeq

so

173.1096 x 1000 = - 8.314 x 298 x ln Keq

ln Keq = 4.52 x 10-31

so

the equilibrium constant at 25 C is 4.52 x 10-31