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Chapter 8 Question 65

Posted: Sun Jan 22, 2017 7:13 pm
by Kira_Maszewski_1B
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?

Re: Chapter 8 Question 65

Posted: Sun Jan 22, 2017 7:54 pm
by Rosaline_Chow_2L
Hi,

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!

Re: Chapter 8 Question 65

Posted: Sun Jan 22, 2017 8:08 pm
by Kira_Maszewski_1B
Rosaline_Chow_2L wrote:Hi,

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!


Oh! Thanks so much, I forgot that the enthalpy for O2 is 0 kJ.