1. 2NO(g) + O2(g)→ 2NO (g) DH° = -112 kJ/mol Absolute Entropy at 298 K (J/(K mol)) NO(9) 211 O2(9) 205 NO2(g) 240 a. Explain the sign of the entropy change based on the chemical equation. b. Is the reaction thermodynamically favorable at 298K? Justify your answer based on the calculation of the value and sign of DGº. c. Is the value of the equilibrium constant, K, for the reaction greater than 1, or less than 1? Justify your answer.

1. 2NO(g) + O2(g)→ 2NO (g) DH° = -112 kJ/mol Absolute Entropy at 298 K (J/(K mol)) NO(9) 211 O2(9) 205 NO2(g) 240 a. Explain the sign of the entropy change based on the chemical equation. b. Is the reaction thermodynamically favorable at 298K? Justify your answer based on the calculation of the value and sign of DGº. c. Is the value of the equilibrium constant, K, for the reaction greater than 1, or less than 1? Justify your answer.

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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