Apply Hess\'s Law Hess\'s law states that \"the heat released or in a chemical p

Question

Apply Hess's Law Hess's law states that "the heat released or in a chemical process is the same whether the process takes place in one or in several steps 2N0(g) + O2 (g) 2NO2 (g) It is important to recall the following rules: given the following reactions and enthalpies of formation 1. When two reactions are added, their enthalpy 2. When a reaction is reversed, the sign of its enthalpy 3. When the coefficients of a reaction are multiplied by 1.IN2(g)+O2(g)NO2(g), =33.2 kJ values are added value changes. a factor, the enthalpy value is multiplied by that Express your answer with the appropriate units. same factor H'. 1 Value Units PartB Calculate the enthalpy of the reaction 4B(s) + 30, (g)2B,03 (s) given the following pertinent information: / 1 B2O3(s) + 3H:00g)3O3(g) + B2H6(g), 3. 4. H20(l)H2O(g), =+2035 kJ View Available Hint s)

Explanation / Answer

Part A

Reaction : 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,

Now,

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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Now adding the above equations and 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

Part B

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.
Now,

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
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, so we multiply by 2 and invert.

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 B2H2(g) -> 2 B2O3(s) + 6 H2O(g)..deltaH = -4070 kJ
4 B(s) + 6 H2(g) -> 2 B2H6(g)............................deltaH = +72 kJ
6 H2O(l) -> 6 H2(g) + 3 O2(g)...........................deltaH = +1710 kJ
6 H2O(g) -> 6 H2O(l)........................................deltaH = -264 kJ
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Now adding the above equations and cancelling terms on LHS and RHS, you will find the same given reaction:
4 B(s) + 3 O2(g) -> 2 B2O3(s)
deltaH = 72 + 1710 - 4070 - 264 = -2552 kJ

Thank You

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