Chapter 7. Nitrogen
Answers to  Questions

1.  Li3N(s) + 3 D2O(aq)    ND3(aq) + 3 LiOD

2. 15.8 M

3. 416 L

4. This is the procedure:

1. Use DG298 and DH to calculate DS
2. Use DH and  DS to calculate DG4000
3. Use DG4000 and DG = -RTlnKp to calculate Kp
4. Kp = 4.13 x 10-7


5. All structures have separation of charge.  CO2 is an example of a molecule that would be stable and yet have the same Lewis structures.

 

6. The oxidation half involves 3 electron loss per N:

2 NH4+  N2 + 6 e-

Because the oxidation half involves 3 electrons per NH4NO3 formula, the reaction must involve the same number, 3 electrons.  At high temperatures, lots of products are possible, but one that requires only three electrons is:

NO3- + 3 e-  NO
The above reaction and the oxidation half reaction can be added together:

2 NH4NO3(s)  N2(g) + 2 NO(g) + 2 H2O(g)

Other products are possilbe under these high-energy conditions.  For example, NO(g) is thermodynamically less stable than O2 and N2.

7. from the text:

NH3(aq) + OCl-(aq)  OH-(aq) + H2NCl(aq)

2 NH3(aq) + H2NCl(aq)  N2H4(aq) + NH4Cl(aq)


8.  DH = -172 kJ

9.  Here are the two forms.  The left one is far superior.

 

10.

N2O3 + H2O  2 HNO2

N2O5 + H2O  2 HNO3

NH4NO3   N2O + 2 H2O (below 260 oC)

NH4NO3   N2 + 2 H2O  (above 260 oC)


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NO: (ss)2(ss*)2(sp)2(p)4(p*)1;  BO = 2.5; paramagnetic

NO+: (ss)2(ss*)2(sp)2(p)4(p*)0;  BO = 3.0; diamagnetic


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18. In NF3 the center of electron density is clost to the center of mass creating a small molecular dipole.  In NH3 the center of mass and center of electron density are separated.

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