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Here are the answers for the previous part of this question, it has multiple parts (these...

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Chemistry

Here are the answers for the previous part of this question, ithas multiple parts (these are all at 25 degrees):
A) 2CH4(g)→C2H6(g)+H2(g)
ΔH∘rxn 64.6 kJ
B) 2NH3(g)→N2H4(g)+H2(g)
ΔH∘rxn 187.2 kJ
C) N2(g)+O2(g)→2NO(g)
ΔH∘rxn 182.6 kJ
D) 2KClO3(s)→2KCl(s)+3O2(g)
ΔH∘rxn -77.6 kJ
E) For the reaction in part A calculate ΔS∘rxn at 25∘C.
ΔS∘rxn -12.7

I NEED:
F) For the reaction in part B calculate ΔS∘rxn at 25 ∘C.
G) For the reaction in part C calculate ΔS∘rxn at 25∘C.
H) For the reaction in part D calculate ΔS∘rxn at 25∘C.

Answer & Explanation Solved by verified expert
4.2 Ratings (828 Votes)

You need a table with reference values for So for each reactant and product . The values are in J/mol.K, not kJ/mol.K.

Then use the formula:

ΔSºreaction = ΣnpSº(products) - ΣnrSº(reactants)

B.             

2NH3(g)→N2H4(g)+H2(g)

111                238              131         So values in J/mol.K

dS = 131+238 – 2x111= 147 J/mol.K

C.

N2(g)+O2(g)→2NO(g)

192        205              211

dS = 2x211-205-192 = 25 j/mol.K

D.

2KClO3(s)→2KCl(s)+3O2(g)

143                          83            205

dS = 2x83 + 3x205 – 2x143 = 495 J/mol.K
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