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

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter17: Spontaneity, Entropy, And Free Energy

Section: Chapter Questions

Problem 4ALQ: What types of experiments can be carried out to determine whether a reaction is spontaneous? Does...

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