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Question

Consider the following reaction:


N2(g) + 2 H2O(g)  2 NO(g) + 2

H2(g)



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

Top Answer

The correct option is b... View the full answer

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