Part A Calculate the enthalpy of the reaction 2NO(g)+O2(g)2NO2(g) given the foll

Question

Part A

Calculate the enthalpy of the reaction

2NO(g)+O2(g)2NO2(g)

given the following reactions and enthalpies of formation:

12N2(g)+O2(g)NO2(g),   HA=33.2 kJ

12N2(g)+12O2(g)NO(g),  HB=90.2 kJ

Express your answer with the appropriate units.

Part B

Calculate the enthalpy of the reaction

4B(s)+3O2(g)2B2O3(s)

given the following pertinent information:

B2O3(s)+3H2O(g)3O2(g)+B2H6(g),    HA=+2035 kJ

2B(s)+3H2(g)B2H6(g),                            HB=+36 kJ

H2(g)+12O2(g)H2O(l),                HC=285 kJ

H2O(l)H2O(g),                                          HD=+44 kJ

Express your answer with the appropriate units.

Explanation / Answer

1)
The reaction you have is: 2 NO + O2 -> 2 NO2
It means that 2 NO and O2 must be on the left side and 2 NO2, on the right one, so you will look for the entalphies that were given and invert and/or multiply to copy this order, look:

1/2 N2(g) + O2(g) --> NO2(g), deltaH = 33.2 kJ (should be multiplied by 2)

N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ

1/2 N2(g) + 1/2O2(g) --> NO(g), delta H = 90.2 kJ (should be multiplied by 2 and inverted)

2 NO -> N2 + O2..............deltaH = -180,4 kJ

Now you have:
N2 + 2 O2 -> 2 NO2.........deltaH = +66,4 kJ
2 NO -> N2 + O2..............deltaH = -180,4 kJ
--------------------------------------...
canceling both N2 and 2 O2 with O2, you will find the given reaction:
2 NO + O2 -> 2 NO2........deltaH = 66,4 - 180,4 = -114,0 kJ


2)
In this one, you must know that you cannot cancel, for example, H2O(l) with H2O(g). The substances must be in the same physical state.

The reaction you have is: 4 B(s) + 3 O2(g) --> 2 B2O3(s)
It means that you have 4B(s) and 3 O2(g) on the left side and 2 B2O3(s) on the right side.
I will start for the second given entalphy reaction.

2 B(s) + 3 H2(g) --> B2H6(g), deltaH = +36 kJ (you need 4 B, so you should multiply this by 2)

4 B(s) + 6 H2(g) -> 2 B2H6(g).......deltaH = +72 kJ

B2O3(s) + 3 H2O(g) --> 3 O2(g) + B2H6(g), deltaH = +2035 kJ
at first look you need just to invert because of the 3 O2, but you have another equation that involves oxygen, so you will see that here you need to invert AND multiply by 2 (you can also look at the B2H6, where in the reaction above you have 2, so now you need 2 to cancel, as they are on different sides)

6 O2(g) + 2 B2H6(g) -> 2 B2O3(s) + 6 H2O(g)..deltaH = -4070 kJ

H2(g) + 1/2 O2(g) --> H2O(l), deltaH = -285 kJ
Since you have 6 O2 and you must have only 3 O2, you should now multiply by 6 and invert this reaction:

6 H2O(l) -> 6 H2(g) + 3 O2(g)........deltaH = +1710 kJ

H2O(l) --> H2O(g), deltaH = +44 kJ
To cancel both H2O(l), you should multiply it by 6 and also invert

6 H2O(g) -> 6 H2O(l) ......deltaH = -264 kJ

Now you have:
6 O2(g) + 2 B2H6(g) -> 2 B2O3(s) + 6 H2O(g)..deltaH = -4070 kJ
4 B(s) + 6 H2(g) -> 2 B2H6(g)............................delta... = +72 kJ
6 H2O(l) -> 6 H2(g) + 3 O2(g)...........................deltaH = +1710 kJ
6 H2O(g) -> 6 H2O(l)..................................... = -264 kJ
--------------------------------------...
doing the same way as we did before, you will find the same given reaction:
4 B(s) + 3 O2(g) -> 2 B2O3(s)
deltaH = 72 + 1710 - 4070 - 264 = -2552 kJ

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