Question

Use the values provided in the table below to calculate ΔGrxn in kJ at 298 K for the combustion of 1 mole of sucrose, C12H22O11. Substance ΔHf°, kJ/mol Sf°, J/mol∙K

C12H22O11 (s) -2222 360

O2 (g) 0 205

CO2 (g) -394 214

H2O (g) -242 189

Answer 1

Given:

Hof(C12H22O11(s)) = -2222.0 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Hof(CO2(g)) = -394.0 KJ/mol

Hof(H2O(g)) = -242.0 KJ/mol

Balanced chemical equation is:

C12H22O11(s) + 12 O2(g) ---> 12 CO2(g) + 11 H2O(g)

ΔHo rxn = 12*Hof(CO2(g)) + 11*Hof(H2O(g)) - 1*Hof( C12H22O11(s)) - 12*Hof(O2(g))

ΔHo rxn = 12*(-394.0) + 11*(-242.0) - 1*(-2222.0) - 12*(0.0)

ΔHo rxn = -5168 KJ/mol

Given:

Sof(C12H22O11(s)) = 360.0 J/mol.K

Sof(O2(g)) = 205.0 J/mol.K

Sof(CO2(g)) = 214.0 J/mol.K

Sof(H2O(g)) = 189.0 J/mol.K

Balanced chemical equation is:

C12H22O11(s) + 12 O2(g) ---> 12 CO2(g) + 11 H2O(g)

ΔSo rxn = 12*Sof(CO2(g)) + 11*Sof(H2O(g)) - 1*Sof( C12H22O11(s)) - 12*Sof(O2(g))

ΔSo rxn = 12*(214.0) + 11*(189.0) - 1*(360.0) - 12*(205.0)

ΔSo rxn = 1827 J/mol.K

WE HAVE:

ΔH = -5168.0KJ/mol

ΔS = 1827J/mol.k

= 1.827 KJ/mol.K

T = 298 K

use:

ΔG = ΔH - T*ΔS

ΔG = -5168.0 - 298.0 * 1.827

ΔG = -5712.4 KJ/mol

Answer: -5712.4 KJ/mol