The thermodynamic properties for a reaction are related by the equation that def

Question

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:

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

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)

Explanation / Answer

Part – A: The given thermochemical reaction is

N2(g) + O2(g) ------à2 NO(g),

DeltaH(rxn) = 180.5 kJ/mol = 180.5kJx(1000J /1kJ)/mol = 180500J/mol

Since we need stadrad enthalpy change(DeltaH0) to use the formulae

DeltaG0 = DeltaH0 – TxDeltaS0

We need to calculate stadrad enthalpy change(DeltaH0). The reaction enthalpy will be said to be standard enthalpy change(DeltaH0) when we consider the formation of 1 mole of NO(g). In this case, we need to divide the abve thermochemical reaction by 2. Hence

1/2N2(g) +1/2 O2(g) ------à NO(g), DeltaH0 = (180500J/mol)x1/2 = 90250 J/mol

DeltaS0 = 24.80J/molxK

For the reaction to be spontaneous, DeltaG0 should be negative. Hence the reaction will be spontaneous above the minimum temperature at which DeltaG0 will be 0.

Let the minium temperature be T.

Now applying the formulae for DeltaG0

DeltaG0 = DeltaH0 – TxDeltaS0

=> 0 = 90250J/mol – Tx(24.80J/molxK )

=> Tx(24.80J/molxK ) = 180500J/mol

=> T =   90250J/mol /(24.80J/molxK ) = 3639 K

Part-B: Given temperature, T = 25 DegC = 25 + 273 = 298K

DeltaH0 = (180500J/mol)x1/2 = 90250 J/mol

DeltaS0 = 24.80J/molxK

Now DeltaG0 can be calculated as

DeltaG0 = DeltaH0 – TxDeltaS0

= 90250 J/mol – 298Kx24.80J/molxK

= 82860J/mol   = (82860J/mol)x(1kJ/1000J) = 82.86kJ/mol

The standard free energy change DeltaG0 and equilibrium constant K for a reaction are related by the following equation

DeltaG0 = - RTlnK = -2.303RTlogK

=> logK = DeltaG0/ (-2.303RT) = - (82860J/mol) /2.303x(8.314 J/molxK)x298K

=> logK = - 14.52

=> K = antilog(- 14.52) = (answer)

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