Chapter 8 Question 65
Posted: Sun Jan 22, 2017 7:13 pm
Can someone explain why -169.2 kJ (the enthalpy of the reaction 2NO + 3/2 O2 ---> N2O5) equals the enthalpy of formation of N2O5 minus twice the enthalpy of formation of NO?
Created by Dr. Laurence Lavelle
After getting the enthalpy of the reaction to be -169.2 kJ, it should be easier to understand if you write out the whole process. The enthalpy of the reaction should equal the enthalpy of the products minus the reactants of the equation 2NO + 3/2O2 --> N2O5. Thus, you get -169.2 kJ = deltaHfN2O5 - [2*deltaHfNO + 3/2*deltaHfO2]. However the enthalpy of formation of O2 is 0 kJ, so you can eliminate the 3/2*deltaHfO2. That's why -169.2 kJ = deltaHN2O5 - 2*deltaHfNO. Hope my explanation was understandable!