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In: Chemistry

3. Use the standard molar entropies in Appendix B in the textbook to calculate ΔS∘ at...

Use the standard molar entropies in Appendix B in the textbook to calculate ΔS∘ at 25 ∘C for each of the following reactions.

Part A

2H2O2(l)→2H2O(l)+O2(g)

Express your answer as an integer and include the appropriate units.

Part B

2SO2(g)+O2(g)→2SO3(g)

Express your answer using one decimal place and include the appropriate units.

Part C

2O3(g)→3O2(g)

Express your answer using one decimal place and include the appropriate units.

Part D

4Al(s)+3O2(g)→2Al2O3(s)

Express your answer using one decimal place and include the appropriate units.

Solutions

Expert Solution

we have:

Sof(H2O2(l)) = 109.6 J/mol.K

Sof(H2O(l)) = 69.91 J/mol.K

Sof(O2(g)) = 205.138 J/mol.K

we have the Balanced chemical equation as:

2 H2O2(l) ---> 2 H2O(l) + O2(g)

deltaSo rxn = 2*Sof(H2O(l)) + 1*Sof(O2(g)) - 2*Sof( H2O2(l))

deltaSo rxn = 2*(69.91) + 1*(205.138) - 2*(109.6)

deltaSo rxn = 125.758 J/K

Answer: 125.758 J/K

we have:

Sof(SO2(g)) = 248.22 J/mol.K

Sof(O2(g)) = 205.138 J/mol.K

Sof(SO3(g)) = 256.76 J/mol.K

we have the Balanced chemical equation as:

2 SO2(g) + O2(g) ---> 2 SO3(g)

deltaSo rxn = 2*Sof(SO3(g)) - 2*Sof( SO2(g)) - 1*Sof(O2(g))

deltaSo rxn = 2*(256.76) - 2*(248.22) - 1*(205.138)

deltaSo rxn = -188.058 J/K

Answer: -188.058 J/K

we have:

Sof(O3(g)) = 238.93 J/mol.K

Sof(O2(g)) = 205.138 J/mol.K

we have the Balanced chemical equation as:

2 O3(g) ---> 3 O2(g)

deltaSo rxn = 3*Sof(O2(g)) - 2*Sof( O3(g))

deltaSo rxn = 3*(205.138) - 2*(238.93)

deltaSo rxn = 137.554 J/K

Answer: 137.554 J/K

we have:

Sof(Al(s)) = 28.3 J/mol.K

Sof(O2(g)) = 205.138 J/mol.K

Sof(Al2O3(s)) = 50.92 J/mol.K

we have the Balanced chemical equation as:

4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)

deltaSo rxn = 2*Sof(Al2O3(s)) - 4*Sof( Al(s)) - 3*Sof(O2(g))

deltaSo rxn = 2*(50.92) - 4*(28.3) - 3*(205.138)

deltaSo rxn = -626.774 J/K

Answer: -626.774 J/K

queen_honey_blossom answered 1 year ago

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