Textbook 4D.23

[Lynne Zhao 2B]

when calculating the standard enthalpy of reaction I got

∆H = 4(91.3 kJ/mol of NO) + 2(-114.1 kJ) + (-110.2 kJ) = 26.8 kJ

26.8 kJ / 2 mol N2O5 = 13.4 kJ / mol

I was wondering if someone could help me check my work and explain how to get 11. 4 kJ/mol

[Chem_Mod]

Hello Lynne,

It may be beneficial to acknowledge the enthalpy of formation of NO is 90.25 kJ/mol, not 91.3 kJ/mol for your remaining calculations, of which should direct you to the answer.

[Kailin Mimaki 2K]

When I calculated the enthalpy of the reaction, I got (-169.2)-2(90.25) which equals 11.3KJ. Like what chem mod said, just make sure that you have the correct values for the enthalpy of formation and I think you should be just fine. Hope this helped!