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Problem Explication # 57

Problem Explication # 57. Tricia, Kayla, and Tiffanie Group 1. Given the following data:. H 2(g) + 1/2O 2(g) H 2 O (l) Δ H = -285.8 kJ N 2 O 5(g) + H 2 O (l) 2HNO 3(l) Δ H = -76.6 kJ 1/2N 2(g) + 3/2O 2(g) + 1/2H 2(g) HNO 3(l) Δ H = -174.1 kJ

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Problem Explication # 57

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  1. Problem Explication # 57 Tricia, Kayla, and Tiffanie Group 1

  2. Given the following data: • H2(g) + 1/2O2(g) H2O(l) ΔH = -285.8 kJ • N2O5(g) + H2O(l) 2HNO3(l) ΔH = -76.6 kJ • 1/2N2(g) + 3/2O2(g) + 1/2H2(g) HNO3(l) ΔH = -174.1 kJ Calculate the ΔH for the reaction: 2N2(g) + 5O2(g) 2N2O5(g)

  3. 2N2(g) + 5O2(g) 2N2O5(g) • H2(g) + 1/2O2(g) H2O(l)ΔH = -285.8 kJ Switch and multiply by 2. Then you change your sign . • N2O5(g) + H2O(l) 2HNO3(l)ΔH = -76.6 kJ Switch and multiply by 2. Then you change your sign . • 1/2N2(g) + 3/2O2(g) + 1/2H2(g) HNO3(l)ΔH = -174.1 kJ Just multiply by 4. This time don’t change your sign .

  4. The Problem ENDS • 2H2O(l) 2H2(g) + O2(g) ΔH = +571.6kJ • 4HNO3(l) 2N2O5(g) + 2H2O(l) ΔH = +153.2kJ • 2N2(g) + 602(g)+ 2H2(g) 4HNO3(l) ΔH = -696.4kJ 2N2(g) + 5O2(g) 2N2O5(g) ΔH = 28.4kJ

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