1 / 4

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

melanie-mccoy
Télécharger la présentation

Problem Explication # 57

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related