Respuesta :
Answer: The Gibbs free energy of the given reaction is -40 kJ
Explanation:
The given chemical equation follows:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
The equation for the standard Gibbs free change of the above reaction is:
[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NH_3(g))})]-[(1\times \Delta G^o_f_{(N_2(g))})+(3\times \Delta G^o_f_{(H_2(g))})][/tex]
We are given:
[tex]\Delta G^o_f_{(NH_3(l))}=-16.45kJ/mol\\\Delta G^o_f_{(H_2(g))}=0kJ/mol\\\Delta G^o_f_{(N_2(g))}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol[/tex]
The equation used to Gibbs free energy of the reaction follows:
[tex]\Delta G=\Delta G^o+RT\ln Q_{eq}[/tex]
where,
[tex]\Delta G[/tex] = free energy of the reaction
[tex]\Delta G^o[/tex] = standard Gibbs free energy = -32.9 kJ/mol = -32900 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314J/K mol
T = Temperature = [tex]25^oC=[273+25]K=298K[/tex]
[tex]Q_{eq}[/tex] = Ratio of concentration of products and reactants at any time = [tex]\frac{(p_{NH_3})^2}{(p_{H_2})^3\times p_{N_2}}[/tex]
[tex]p_{NH_3}=2.11atm[/tex]
[tex]p_{N_2}=7.92atm[/tex]
[tex]p_{H_2}=2.02atm[/tex]
Putting values in above equation, we get:
[tex]\Delta G=-32900J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(2.11)^2}{(2.02)^3\times 7.92}))\\\\\Delta G=-39553.04J/mol=--39.55kJ=-40kJ[/tex]
Hence, the Gibbs free energy of the given reaction is -40 kJ
