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Problem Explication # 57

This problem involves calculating the change in enthalpy (ΔH) for the reaction producing dinitrogen pentoxide from nitrogen and oxygen. Starting with given reactions and their respective enthalpy changes, the values are manipulated through switching and multiplying as needed. The process utilizes Hess's Law to find the ΔH for the target reaction: 2N2(g) + 5O2(g) → 2N2O5(g). With proper conversions of ΔH values, the final enthalpy change is calculated to determine the energy changes involved in the reaction, facilitating insight into thermochemical processes.

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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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