Question

In: Chemistry

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

Here are the answers for the previous part of this question, it has 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.

Solutions

Expert Solution

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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