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For the following reaction: 2C3H18(1) + 2502(g) → 16C02(g) + 18H20(1) Compound AH°F (kJ moll) Sº (3 mol-1 K-1) C8H18 ( 259.30

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Answer #1


Given:
Hof(C8H18(l)) = -259.3 KJ/mol
Hof(O2(g)) = 0.0 KJ/mol
Hof(CO2(g)) = -393.51 KJ/mol
Hof(H2O(l)) = -285.83 KJ/mol

Balanced chemical equation is:
2 C8H18(l) + 25 O2(g) ---> 16 CO2(g) + 18 H2O(l)

ΔHo rxn = 16*Hof(CO2(g)) + 18*Hof(H2O(l)) - 2*Hof( C8H18(l)) - 25*Hof(O2(g))
ΔHo rxn = 16*(-393.51) + 18*(-285.83) - 2*(-259.3) - 25*(0.0)
ΔHo rxn = -10922.5 KJ

Given:
Sof(C8H18(l)) = 328.0 J/mol.K
Sof(O2(g)) = 205.7 J/mol.K
Sof(CO2(g)) = 213.74 J/mol.K
Sof(H2O(l)) = 69.91 J/mol.K

Balanced chemical equation is:
2 C8H18(l) + 25 O2(g) ---> 16 CO2(g) + 18 H2O(l)

ΔSo rxn = 16*Sof(CO2(g)) + 18*Sof(H2O(l)) - 2*Sof( C8H18(l)) - 25*Sof(O2(g))
ΔSo rxn = 16*(213.74) + 18*(69.91) - 2*(328.0) - 25*(205.7)
ΔSo rxn = -1120.28 J/K

At equilibrium,
ΔGo = 0.0 KJ/mol

We have:
ΔHo = -10922.5 KJ/mol
ΔSo = -1120.28 J/mol.K
= -1.12028 KJ/mol.K

use:
ΔGo = ΔHo - T*ΔSo
0.0 = -10922.5 - T *(-1.1203)
T = 9749.79 K
Answer: 9749.79 K

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