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Question: If \(\Delta G = - 177KCal\)for \[2Fe(s) + \dfrac{3}{2}{O_2}(g) \to F{e_2}{O_3}(s)\] and \(\Delta G =...

If ΔG=−177KCal\Delta G = - 177KCalfor 2Fe(s)+32O2(g)→Fe2O3(s)2Fe(s) + \dfrac{3}{2}{O_2}(g) \to F{e_2}{O_3}(s) and ΔG=−19KCal\Delta G = - 19KCalfor 4Fe2O3(s)+Fe(s)→3Fe3O4(s)4F{e_2}{O_3}(s) + Fe(s) \to 3F{e_3}{O_4}(s).what is Gibbs free energy of formation of Fe3O4(s)F{e_3}{O_4}(s)
A. +229.6 Kcal/mol + 229.6{\text{ }}Kcal/mol
B. −242.3 Kcal/mol - 242.3{\text{ }}Kcal/mol
C. −727 Kcal/mol - 727{\text{ }}Kcal/mol
D. −229.6 Kcal/mol - 229.6{\text{ }}Kcal/mol

Explanation

Solution

In this given question we have to calculate the Gibbs free energy of a given reaction which will take place in between iron and oxygen. The given energy of the complete reaction is considerably used to calculate the free energy formation of compound Fe3O4(s)F{e_3}{O_4}(s).

Complete step by step solution: As we know that Gibbs free energy the energy associated with a chemical reaction that can be used to carry out work and the free energy of a system is the sum of its enthalpy (H)\left( H \right) plus the product of the temperature (Kelvin) and the entropy (S)\left( S \right) of the system.
So, the equation occur will be-
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
Where
ΔG\Delta G is free energy
ΔH\Delta H is enthalpy
ΔS\Delta S is entropy
As It is a thermodynamic process which can be used to calculate the maximum of a reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.
So, we have, Given the two reaction that are-
2Fe(s)+32O2(g)→Fe2O3(s)2Fe(s) + \dfrac{3}{2}{O_2}(g) \to F{e_2}{O_3}(s) has ΔG=−177KCal\Delta G = - 177KCal….(1)(1)
4Fe2O3(s)+Fe(s)→3Fe3O4(s)4F{e_2}{O_3}(s) + Fe(s) \to 3F{e_3}{O_4}(s)has ΔG=−19KCal\Delta G = - 19KCal….(2)(2)
Now by dividing equation (2)(2)by 3 we get;
43Fe2O3(s)+13Fe(s)→33Fe3O4(s)\dfrac{4}{3}F{e_2}{O_3}(s) + \dfrac{1}{3}Fe(s) \to \dfrac{3}{3}F{e_3}{O_4}(s)has ΔG=−193KCal\Delta G = - \dfrac{{19}}{3}KCal
After calculation we get;
43Fe2O3(s)+13Fe(s)→Fe3O4(s)\dfrac{4}{3}F{e_2}{O_3}(s) + \dfrac{1}{3}Fe(s) \to F{e_3}{O_4}(s)has ΔG=−193KCal\Delta G = - \dfrac{{19}}{3}KCal....(3)(3)
Similarly, multiply by 43\dfrac{4}{3}in equation (1)(1)we get;
83Fe(s)+43×32O2(g)→43Fe2O3(s)\dfrac{8}{3}Fe(s) + \dfrac{4}{3} \times \dfrac{3}{2}{O_2}(g) \to \dfrac{4}{3}F{e_2}{O_3}(s)has ΔG=−177×43KCal\Delta G = - 177 \times \dfrac{4}{3}KCal
After calculation we get;
83Fe(s)+2O2(g)→43Fe2O3(s)\dfrac{8}{3}Fe(s) + 2{O_2}(g) \to \dfrac{4}{3}F{e_2}{O_3}(s)has ΔG=−177×43KCal\Delta G = - 177 \times 43KCal….(4)(4)
Now by adding equation (3)(3)had (4)(4)we get;
3Fe(s)+2O2(g)→Fe3O4(s)3Fe(s) + 2{O_2}(g) \to F{e_3}{O_4}(s)has ΔG=(−177×43−193)KCal\Delta G = ( - 177 \times 43 - \dfrac{{19}}{3})KCal
Thus, ΔG=−242.3KCal/Mol\Delta G = - 242.3KCal/Mol

Hence, option B is the correct answer.

Note: Gibbs free energy was accidently determined by an American scientist Josiah Willard Gibbs. Its value will be usually expressed in Joules or Kilojoules. Gibbs free energy can also be defined as the maximum amount of work that can be extracted from a closed system of a thermodynamic process. There are certain laws of thermodynamics which are based on the concepts of entropy, enthalpy and spontaneity of a reaction.