� 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 isN2(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:
Express your answer to four significant figures and include the appropriate units.
SubmitHintsMy AnswersGive UpReview Part 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.
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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
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