<u>Answer:</u> The equilibrium constant for this reaction is 
<u>Explanation:</u>
The equation used to calculate standard Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the standard Gibbs free change of the above reaction is:
![\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))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28CO%29_4%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_%7B%28Ni%28s%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20G%5Eo_%7B%28CO%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-587.4%29%29%5D-%5B%281%5Ctimes%20%280%29%29%2B%284%5Ctimes%20%28-137.3%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D-38.2kJ%2Fmol)
To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

where,
= 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 = ![58^oC=[273+58]K=331K](https://tex.z-dn.net/?f=58%5EoC%3D%5B273%2B58%5DK%3D331K)
= equilibrium constant at 58°C = ?
Putting values in above equation, we get:

Hence, the equilibrium constant for this reaction is 