The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement ? Support your opinion with example.

Explanation

Yes, The choice of a reducing agent in a particular case depends on thermodynamic factor. Example : The free energies of formation `(Delta_(f) G^(Ө))` of most sulphides are greater than those or `CS_(2)` and `H_(2)S`. Carbon disulphide is, in fact, an endothermic compound.
Therefore, neither carbon nor hydrogen is a suitable reducing agent for metal sulphides.