Consider the following reaction:
N2(g) + 2 H2O(g) 2 NO(g) + 2
The enthalpy change at 298 K for this reaction is Horxn = 664.38 kJ/mol. Under the same conditions the entropy change for the reaction is Sorxn = 113.24 J/mol-K. Calculate the value of Gorxn at a temperature of 484 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 = -5.41e+04 kJ/mol.
b) At this temperature Gorxn = 6.10e+02 kJ/mol.
c) At this temperature Gorxn = 6.98e+02 kJ/mol.
d) At this temperature Gorxn = -3.31e+04 kJ/mol.
e) At this temperature Gorxn = -6.10e+02 kJ/mol.
correct answer is b...I got a but it is wrong, please explain