Respuesta :
Answer: The Gibbs free energy of the given reaction is [tex]1.379\times 10^3kJ[/tex]
Explanation:
The equation used to calculate Gibbs free energy change is of a reaction is:
[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}][/tex]
For the given chemical reaction:
[tex]2CH_3OH(g)+3O_2(g)\rightarrow 2CO_2(g)+4H_2O(g)[/tex]
The equation for the Gibbs free energy change of the above reaction is:
[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(CO_2(g))})+(4\times \Delta G^o_f_{(H_2O(g))})]-[(2\times \Delta G^o_f_{(CH_3OH(g))})+(3\times \Delta G^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta G^o_f_{(H_2O(g))}=-228.57kJ/mol\\\Delta G^o_f_{(CO_2(g))}=-394.36kJ/mol\\\Delta G^o_f_{(CH_3OH(g))}=-161.96kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=[(2\times (-394.36))+(4\times (-228.57))]-[(2\times (-161.96))+(3\times (0))]\\\\\Delta G^o_{rxn}=-1379.08kJ/mol[/tex]
The equation used to Gibbs free energy of the reaction follows:
[tex]\Delta G=\Delta G^o+RT\ln Q_{p}[/tex]
where,
[tex]\Delta G[/tex] = free energy of the reaction
[tex]\Delta G^o[/tex] = standard Gibbs free energy = -1379.08 kJ/mol = -1379080 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/K mol
T = Temperature = [tex]25^oC=[273+25]K=298K[/tex]
[tex]Q_{p}[/tex] = Ratio of concentration of products and reactants = [tex]\frac{(p_{CO_2})^2(p_{H_2O})^4}{(p_{CH_3OH})^2(p_{O_2})^3}[/tex]
[tex]p_{CO_2}=5.29atm\\p_{H_2O}=3.89atm\\p_{CH_3OH}=3.82atm\\p_{O_2}=7.56atm[/tex]
Putting values in above expression, we get:
[tex]\Delta G=-1379080J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(5.29)^2\times (3.89)^4}{(3.82)^2\times (7.56)^3}))\\\\\Delta G=-1379039J=1379.039kJ=1.379\times 10^3kJ[/tex]
Hence, the Gibbs free energy of the given reaction is [tex]1.379\times 10^3kJ[/tex]
