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Use the values from the table and calculate AGºrxn at 306 K in kJ. Enter the result no decimal places and no units. 2A + 1B →
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Answer #1

Given:

Hof(A) = -78.0 KJ/mol

Hof(B) = -82.0 KJ/mol

Hof(C) = -80.0 KJ/mol

Hof(D) = -75.0 KJ/mol

Balanced chemical equation is:

2 A + B ---> 2 C + 3 D

ΔHo rxn = 2*Hof(C) + 3*Hof(D) - 2*Hof( A) - 1*Hof(B)

ΔHo rxn = 2*(-80.0) + 3*(-75.0) - 2*(-78.0) - 1*(-82.0)

ΔHo rxn = -147 KJ

Given:

Sof(A) = 194.0 J/mol.K

Sof(B) = 183.0 J/mol.K

Sof(C) = 137.0 J/mol.K

Sof(D) = 200.0 J/mol.K

Balanced chemical equation is:

2 A + B ---> 2 C + 3 D

ΔSo rxn = 2*Sof(C) + 3*Sof(D) - 2*Sof( A) - 1*Sof(B)

ΔSo rxn = 2*(137.0) + 3*(200.0) - 2*(194.0) - 1*(183.0)

ΔSo rxn = 303 J/K

We have:

ΔHo = -147.0 KJ

ΔSo = 303 J/K

= 0.303 KJ/K

T = 306 K

use:

ΔGo = ΔHo - T*ΔSo

ΔGo = -147.0 - 306.0 * 0.303

ΔGo = -239.718 KJ

Answer: -240 KJ

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