Respuesta :
Answer: The equilibrium constant for this reaction is [tex]5.85\times 10^{5}[/tex]
Explanation:
The equation used to calculate standard Gibbs free change is of a reaction is:
[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]3H_2(g)+N_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_{(NH_3(g))})]-[(1\times \Delta G^o_{(N_2)})+(3\times \Delta G^o_{(H_2)})][/tex]
We are given:
[tex]\Delta G^o_{(NH_3(g))}=-16.45kJ/mol\\\Delta G^o_{(N_2)}=0kJ/mol\\\Delta G^o_{(H_2)}=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]
To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs free energy = -32.9 kJ/mol = -35900 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]K_{eq}[/tex] = equilibrium constant at 25°C = ?
Putting values in above equation, we get:
[tex]-32900J/mol=-(8.314J/Kmol)\times 298K\times \ln K_{eq}\\\\K_{eq}=e^{13.279}=5.85\times 10^{5}[/tex]
Hence, the equilibrium constant for this reaction is [tex]5.85\times 10^{5}[/tex]
