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

Answer 1

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