At high temperatures, coke reduces silica impurities in iron are to elemental si
ID: 558092 • Letter: A
Question
At high temperatures, coke reduces silica impurities in iron are to elemental silicon.
SiO2(s) + 2 C(s) arrow sign Si(s) + CO(g)
Use the table to calculate the values of AH degrees, AS degrees, and AG degrees for this reaction at 25 degrees C.
AH (kJ mol-1 AG (kJ mol-1) AS (J mol-1k-1)
SiO 2(s) -910.7 856.3 41.5
C(s) 0 0 5.7
Si(s) 0 0 18.8
Co(g) 0 -137.2 197.6
Gibbs free energy for the reaction is_____________kJ
Explanation / Answer
1)
we have:
Gof(SiO2(s)) = 856.3 KJ/mol
Gof(C(s)) = 0.0 KJ/mol
Gof(Si(s)) = 0.0 KJ/mol
Gof(Co(g)) = -137.2 KJ/mol
we have the Balanced chemical equation as:
SiO2(s) + 2 C(s) ---> Si(s) + 2 Co(g)
deltaGo rxn = 1*Gof(Si(s)) + 2*Gof(Co(g)) - 1*Gof( SiO2(s)) - 2*Gof(C(s))
deltaGo rxn = 1*(0.0) + 2*(-137.2) - 1*(856.3) - 2*(0.0)
deltaGo rxn = -1130.7 KJ
Answer: deltaGo rxn = -1130.7 KJ
2)
we have:
Hof(SiO2(s)) = -910.7 KJ/mol
Hof(C(s)) = 0.0 KJ/mol
Hof(Si(s)) = 0.0 KJ/mol
Hof(Co(g)) = 0.0 KJ/mol
we have the Balanced chemical equation as:
SiO2(s) + 2 C(s) ---> Si(s) + 2 Co(g)
deltaHo rxn = 1*Hof(Si(s)) + 2*Hof(Co(g)) - 1*Hof( SiO2(s)) - 2*Hof(C(s))
deltaHo rxn = 1*(0.0) + 2*(0.0) - 1*(-910.7) - 2*(0.0)
deltaHo rxn = 910.7 KJ
Answer: deltaHo rxn = 910.7 KJ
3)
we have:
Sof(SiO2(s)) = 41.5 J/mol.K
Sof(C(s)) = 5.7 J/mol.K
Sof(Si(s)) = 18.8 J/mol.K
Sof(Co(g)) = 197.6 J/mol.K
we have the Balanced chemical equation as:
SiO2(s) + 2 C(s) ---> Si(s) + 2 Co(g)
deltaSo rxn = 1*Sof(Si(s)) + 2*Sof(Co(g)) - 1*Sof( SiO2(s)) - 2*Sof(C(s))
deltaSo rxn = 1*(18.8) + 2*(197.6) - 1*(41.5) - 2*(5.7)
deltaSo rxn = 361.1 J/K
Answer: deltaSo rxn = 361.1 J/K
Feel free to comment below if you have any doubts or if this answer do not work. I will correct it and submit again if you let me know
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