Calculate the enthalpy of the reaction: 2NO(g) + O2(g) --> 2NO2(g)? Calculate the enthalpy of the reaction: 2NO(g) + O2(g) --> 2NO2(g)given the following reactions and enthalpies of formation: 1/2 N2(g) + O2(g) --> NO2(g), delta H *A = 33.2 kJ 1/2 N2(g) + 1/2O2(g) --> NO(g), delta H *B= 90.2 kJ 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) delta H *A = +2035 kJ 2B(s) + 3H2(g) --> B2H6(g) delta H *B = +36 kJ H2(g) + 1/2O2(g) --> H2O(l) delta H *C = -285 kJ H2O(l) --> H2O(g) delta H *D = +44 kJ Can someone show me step by step? I would be very grateful because I do not understand Hesss law very well.

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
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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

Plz Approve!

Plz Approve!

Plz Approve!