## Problem 8.65

Andy Liao 1B
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am

### Problem 8.65

Problem:
Calculate the standard enthalpy of formation of dinitrogen pentoxide from the following data,
2NO (g) + O2 (g) --> 2NO2 (g) DH = -114.1 kJ
4NO2 (g) + O2 (g) --> 2N2O5 (g) DH = -110.2 kJ
and from the standard enthalpy of formation of nitric oxide, NO (see Appendix 2A).

I was wondering why "The reaction we want is N2 (g) + (5/2)O2 (g) --> N2O5 (g)." Also, what is the purpose of the reaction 2NO (g) + (3/2)O2 (g) --> N2O5 (g)? This is from the solutions manual. Can someone please explain?

Rana YT 2L
Posts: 49
Joined: Thu Jul 27, 2017 3:01 am

### Re: Problem 8.65

that is the reaction you want because that forms one mole of N2O5 and the equation asks for the enthalpy of formation for N2O5 and that refers to the amount of energy needed for one mole of N2O5 when it is formed by elements in their natural form.
the purpose of the reaction you mentioned is to allow you to cancel out the 2NO is the first equation they give you.

Cam Bear 2F
Posts: 60
Joined: Thu Jul 27, 2017 3:01 am

### Re: Problem 8.65

2NO (g) + (3/2)O2 (g) --> N2O5 (g) is crucial to solving the problem.

When you multiply the second reaction 4NO2 (g) + O2 (g) --> 2N2O5 (g) by 0.5 you get 2NO (g) + (1/2)O2 (g) --> N2O5 (g). You then add this to the first reaction 2NO (g) + O2 (g) --> 2NO2 (g) to get 2NO (g) + (3/2)O2 (g) --> N2O5 (g).

The $\Delta H^{\circ }$ for this new reaction is the enthalpy of the first reaction plus the enthalpy of the second reaction multiplied by 0.5. $\Delta H^{\circ }$= -114.1 kJ+0.5( -110.2 kJ)= -169.2 kJ.

We know from the book that $\Delta H^{\circ }$=$\sum n*\Delta H^{\circ}_{f}$ of the products- $\sum n*\Delta H^{\circ}_{f}$ of the reactants.

So, using the reaction 2NO (g) + (3/2)O2 (g) --> N2O5 (g) and the new value of $\Delta H^{\circ }$=-169.2 kJ we can plug in the numbers. -169.2 kJ=$\Delta H^{\circ}_{f}$(N2O5)-2*$\Delta H^{\circ}_{f}$(NO). (we find the $\Delta H^{\circ}_{f}$ of NO by looking it up in the Appendix and O2(g) is a pure element in its most stable form so its $\Delta H^{\circ}_{f}$ is 0. Now you can solve for $\Delta H^{\circ}_{f}$(N2O5).

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