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Consider the reaction: 2HBR(g) >H2(g) + Br2() Using standard thermodynamic data at 298K, calculate the free energy change wheAHof (kJ/mol) AG°F (kJ/mol) s° (J/mol K) Beryllium Вe(s) 0 9.5 -569.0 ВeO(s) -599.0 14.0 Be(ОН)2(s) -902.5 -815.0 51.9 AH°f (Other Tables Molecular Structure AH°F (kJ/mol) AG°f (kJ/mol) s° (J/mol K) Bromine Br(g) 175.0 111.9 82.4 Br2() 152.2 0 0 Br2(

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Answer #1

Given:

Gof(HBr(g)) = -53.5 KJ/mol

Gof(H2(g)) = 0.0 KJ/mol

Gof(Br2(l)) = 0.0 KJ/mol

Balanced chemical equation is:

2 HBr(g) ---> H2(g) + Br2(l)

ΔGo rxn = 1*Gof(H2(g)) + 1*Gof(Br2(l)) - 2*Gof( HBr(g))

ΔGo rxn = 1*(0.0) + 1*(0.0) - 2*(-53.5)

ΔGo rxn = 107 KJ

This is when 2 mol of HBr reacts.

For 1.51 mol,

ΔGo rxn = 107*1.51/2 = 80.8 KJ

Answer: 80.8 KJ

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