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sys Example: Calculate the AG˚ve for N₂+ 3H2 → 2NH3 at 25°C and 225°C S° N2(g) 191.5 J/mol K = S° H2(g) = 130.6 J/mol K = S° NH3(g) 192.5 J/mol K o AH, N2(g) = 0 kJ/mol • AH H2(g) = 0 kJ/mol • AH NH3(g) = -46.19 kJ/mol = AG sys AH sys - TAS sys

sys Example: Calculate the AG˚ve for N₂+ 3H2 → 2NH3 at 25°C and 225°C S° N2(g) 191.5 J/mol K = S° H2(g) = 130.6 J/mol K = S° NH3(g) 192.5 J/mol K o AH, N2(g) = 0 kJ/mol • AH H2(g) = 0 kJ/mol • AH NH3(g) = -46.19 kJ/mol = AG sys AH sys - TAS sys

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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sys Example: Calculate the AG˚ve for N₂+ 3H2 → 2NH3 at 25°C and 225°C S° N2(g) 191.5 J/mol K = S° H2(g) = 130.6 J/mol K = S° NH3(g) 192.5 J/mol K o AH, N2(g) = 0 kJ/mol • AH H2(g) = 0 kJ/mol • AH NH3(g) = -46.19 kJ/mol = AG sys AH sys - TAS sys
Transcribed Image Text:sys Example: Calculate the AG˚ve for N₂+ 3H2 → 2NH3 at 25°C and 225°C S° N2(g) 191.5 J/mol K = S° H2(g) = 130.6 J/mol K = S° NH3(g) 192.5 J/mol K o AH, N2(g) = 0 kJ/mol • AH H2(g) = 0 kJ/mol • AH NH3(g) = -46.19 kJ/mol = AG sys AH sys - TAS sys
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