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Os > 3/2 02 ( O2 ( → 20 (g). NO (0 + O (1) → NO2 ) NO + Os (e → NO2 ( + O2 ( AH° = -198.9 kJ AH° = -142.3 kJ AH° = +495.0 kJ AH° = ? NO + Os () → NO2 + O2 ( AH° = -198.9 kJ 3/2 02 + → O3 (4) 2 ( 20 ()→ 02 () AH° = +142.3 kJ 2 (AH° = -495.0 kJ) NO + Os → NO2 ( + O2 AH° = -198.9 kJ AH° = +142.3 kJ AH° = -247.5 kJ 3/2 02 (+ O3 ( (g) NO (10 + O ud NO2 (6). AH° = -304.1 kJ

Os > 3/2 02 ( O2 ( → 20 (g). NO (0 + O (1) → NO2 ) NO + Os (e → NO2 ( + O2 ( AH° = -198.9 kJ AH° = -142.3 kJ AH° = +495.0 kJ AH° = ? NO + Os () → NO2 + O2 ( AH° = -198.9 kJ 3/2 02 + → O3 (4) 2 ( 20 ()→ 02 () AH° = +142.3 kJ 2 (AH° = -495.0 kJ) NO + Os → NO2 ( + O2 AH° = -198.9 kJ AH° = +142.3 kJ AH° = -247.5 kJ 3/2 02 (+ O3 ( (g) NO (10 + O ud NO2 (6). AH° = -304.1 kJ

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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Explain the step-by-step calculation of the HESS law in the picture.

O3 ) → 3/2 02 (e) O2 () → 20 (8) NO (4 + O (g→ NO2 (5) NO + O3 (e) → NO2 (2) + O2 ( AH° = -198.9 kJ AH° = -142.3 kJ AH° = +495.0 kJ AH° = ? NO + O3 () → NO2 (2) + O2 ( AH° = -198.9 kJ 3/2 02 ( + → 03 (g) ¥2 ( 20 (g) → 02 (). AH° = +142.3 kJ 2 (AH° = -495.0 kJ) 3/2 02 ( + → O3 (g) → ½ O2 (g. NO + O → NO + O3 () → NO2 (2) + O2 (e) AH° = -198.9 kJ AH° = +142.3 kJ AH° = -247.5 kJ (g > NO2 (K) AH° = -304.1 kJ
Transcribed Image Text:O3 ) → 3/2 02 (e) O2 () → 20 (8) NO (4 + O (g→ NO2 (5) NO + O3 (e) → NO2 (2) + O2 ( AH° = -198.9 kJ AH° = -142.3 kJ AH° = +495.0 kJ AH° = ? NO + O3 () → NO2 (2) + O2 ( AH° = -198.9 kJ 3/2 02 ( + → 03 (g) ¥2 ( 20 (g) → 02 (). AH° = +142.3 kJ 2 (AH° = -495.0 kJ) 3/2 02 ( + → O3 (g) → ½ O2 (g. NO + O → NO + O3 () → NO2 (2) + O2 (e) AH° = -198.9 kJ AH° = +142.3 kJ AH° = -247.5 kJ (g > NO2 (K) AH° = -304.1 kJ
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