Respuesta :
Answer: The equilibrium constant for this reaction is [tex]1.068\times 10^{6}[/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]Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)[/tex]
The equation for the standard Gibbs free change of the above reaction is:
[tex]\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})][/tex]
We are given:
[tex]\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol[/tex]
To calculate the equilibrium constant (at 58°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 = -38.2 kJ/mol = -38200 J/mol (Conversion factor: 1 kJ = 1000 J )
R = Gas constant = 8.314 J/K mol
T = temperature = [tex]58^oC=[273+58]K=331K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant at 58°C = ?
Putting values in above equation, we get:
[tex]-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}[/tex]
Hence, the equilibrium constant for this reaction is [tex]1.068\times 10^{6}[/tex]
