In: Chemistry
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), ?H?A=33.2 kJ
12N2(g)+12O2(g)?NO(g), ?H?B=90.2 kJ
Express your answer with the appropriate units.
|
|||
?H? = |
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), ?H?A=+2035 kJ
2B(s)+3H2(g)?B2H6(g), ?H?B=+36 kJ
H2(g)+12O2(g)?H2O(l), ?H?C=?285 kJ
H2O(l)?H2O(g), ?H?D=+44 kJ
Express your answer with the appropriate units.
|
|||
?H? = |
The Hess law is used to discover the reaction's entalphy and
doing it, you must cancel the same substances, if they are in
differents sides, to make the given reaction. Remember that if you
multiply or divide the reaction, the same happens with the entalphy
and when you invert the reaction, the entalphy's sign must be also
inverted. The first one is easy, but the second one took me some
work.
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