Determine the missing values of deltaH in the diagram shown below

Question

^
          | ^ N2O3(g) + 1/2 O2(g)
(Enthalpy)| |                  ^
          | |   deltaH= +16.02 |
          | |                  |
          | |        2 NO2(g)  |
          | |                  ^
          | |deltaH=?          |
          | |                  |
          | |          deltaH=?| 
          | |                  |
          | |                  |
          |     N2(g)+202(g)
          |

Answer:

Delta H for the right= 66.36
Delta H for the left= 66.38+16.02
                   = 82.38

Nancy · Answer

Sorry...I typed the diagram out with spaces, but when I submitted it, the forum erased it. I'm not sure how to put it up the needed diagram

DrBob222 · Answer

I think this forum uses html coding and that allows the first space. Spaces after the first space are ignored. I have used periods as spaces; it doesn't work that well but it works enough to know what is going on. For example. ..............XO2 ==> X + O2 I.............4.5.....0....0 C.............-x.......x....x E...........4.5-x......x....x You might try something like that.

Nancy · Answer

...........^ ...........|^ N2O3(g) + 1/2 O2(g) (Enthalpy).||.....................^ ...........||......deltaH= +16.02 | ...........||.....................| ...........||............2 NO2(g) | ...........||.....................^ ...........||deltaH=?............ | ...........||.....................| ...........||.............deltaH=?| ...........||.....................| ...........||.................... | ...........|...N2(g)+202(g)....... ...........| Hi DrBob222, Is the diagram better now?

Nancy · Answer

I don't think I'll be able to fix it, but Ill try explain the diagram. It's an endothermic enthalpy two step. All the arrows are pointing up. The equations are in order from top to bottom: N2O3(g) + 1/2 O2(g), 2 NO2(g), N2(g)+202(g) I need to find delta H if I was doing a one step process, and I need to find the other half delta in a two step process (i was already given 16.02 for the other half) For the one step process delta I got 83.38, for the two step I got 66.36

Nancy · Answer

..................^ ..................|^ N2O3(g) + 1/2 O2(g) (Enthalpy).||.....................^ ..................||......deltaH= +16.02 | ..................||.....................| ..................||............2 NO2(g) | ..................||.....................^ ..................||deltaH=?............ | ..................||.....................| ..................||.............deltaH=?| ..................||.....................| ..................||.................... | ..................|...N2(g)+202(g)....... ..................| Sorry, I just wanted to try fixing it again

Nancy · Answer

..................^ ..................|^ N2O3(g) + 1/2 O2(g) (Enthalpy)..||.....................^ ..................||......deltaH= +16.02 | ..................||............................| ..................||............2 NO2(g)... | ..................||............................^ ..................||deltaH=?............... | ..................||............................| ..................||.............deltaH=?...| ..................||............................| ..................||.................... .......| ..................|...N2(g)+202(g)....... ..................|

Nancy · Answer

........................^ ..................|^ N2O3(g) + 1/2 O2(g) (Enthalpy)..||.....................^ ..................||......deltaH= +16.02 | ..................||...............................| ..................||............2 NO2(g)... | ..................||...............................^ ..................||deltaH=?............... | ..................||...............................| ..................||.............deltaH=?...| ..................||...............................| ..................||.................... ..........| ..................|...N2(g)+202(g)....... ..................|

Nancy · Answer

...................^ ..................|^ N2O3(g) + 1/2 O2(g) (Enthalpy)..||...........................^ ..................||......deltaH= +16.02 | ..................||...............................| ..................||............2 NO2(g)... | ..................||...............................^ ..................||deltaH=?............... | ..................||...............................| ..................||.............deltaH=?...| ..................||...............................| ..................||.................... ..........| ..................|...N2(g)+202(g)....... ..................|

Nancy · Answer

..................^ ..................|^ N2O3(g) + 1/2 O2(g) (Enthalpy)..||..............................^ ..................||......deltaH= +16.02 | ..................||...............................| ..................||............2 NO2(g)... | ..................||...............................^ ..................||deltaH=?............... | ..................||...............................| ..................||.............deltaH=?...| ..................||...............................| ..................||.................... ..........| ..................|...N2(g)+202(g)....... ..................|

Explain Bot · Answer

To determine the missing values of deltaH in the given diagram, we can use the concept of Hess's Law. Hess's Law states that the enthalpy change of a reaction is independent of the pathway between the initial and final states. This means that we can add or subtract reactions to obtain the overall reaction and the corresponding enthalpy change. In the given diagram, we have the following reactions: 1. N2O3(g) + 1/2 O2(g) -> 2 NO2(g) (deltaH = +16.02) 2. N2(g) + 2 O2(g) -> 2 NO2(g) (deltaH = +66.38) We need to find deltaH for the following reactions: 1. 2 NO2(g) -> N2O3(g) + 1/2 O2(g) 2. 2 NO2(g) -> N2(g) + 2 O2(g) To find deltaH for the first reaction, we can reverse the given reaction #1 and change the sign of deltaH: N2O3(g) + 1/2 O2(g) -> 2 NO2(g) (reverse reaction #1) So, deltaH for the first reaction = -16.02 To find deltaH for the second reaction, we can add the reversed reaction #1 to reaction #2: 2 NO2(g) + (-1) * (N2O3(g) + 1/2 O2(g)) -> 2 NO2(g) + N2(g) + 2 O2(g) (add reversed reaction #1 to reaction #2) Simplifying the equation, we get: 2 NO2(g) - N2O3(g) - 1/2 O2(g) -> N2(g) + 2 O2(g) So, deltaH for the second reaction = deltaH of reaction #2 - deltaH of reversed reaction #1 = 66.38 - (-16.02) (Note that we changed the sign of -16.02) = 66.38 + 16.02 = 82.38 Therefore, the missing values of deltaH in the diagram are: Delta H for the right = 66.36 (given as deltaH of reaction #2) Delta H for the left = 82.38 (calculated as explained above)