Consider the following reaction: 3 H2(g) + N2(g) 2 NH3(g) The enthalpy change at

Question

Consider the following reaction:

3 H2(g) + N2(g) 2 NH3(g)


The enthalpy change at 298 K for this reaction is Horxn = -92.38 kJ/mol. Under the same conditions the entropy change for the reaction is Sorxn = -198.24 J/mol-K. Calculate the value of Gorxn at a temperature of 758 K where the reaction is under standard conditions other than the temperature. Assume that Horxn and Sorxn do not change with temperature.




a) At this temperature Gorxn = 1.50e+05 kJ/mol.
b) At this temperature Gorxn = 5.79e+01 kJ/mol.
c) At this temperature Gorxn = 5.90e+04 kJ/mol.
d) At this temperature Gorxn = -1.51e+02 kJ/mol.
e) At this temperature Gorxn = -1.50e+05 kJ/mol.

Explanation / Answer

ans is c Gorxn=5.9e+04 KJ/mol

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