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Question: In \(\Delta G\) = -177Kcal for, (1) \(2Fe(s)+\dfrac{3}{2}{{O}_{2}}(g)\to F{{e}_{2}}{{O}_{3}}(s)\) ...

In ΔG\Delta G = -177Kcal for,
(1) 2Fe(s)+32O2(g)→Fe2O3(s)2Fe(s)+\dfrac{3}{2}{{O}_{2}}(g)\to F{{e}_{2}}{{O}_{3}}(s) and, ΔG\Delta G= -19 Kcal for
(2) 4Fe2O3(s)+Fe(s)→3Fe3O4(s)4F{{e}_{2}}{{O}_{3}}(s)+Fe(s)\to 3F{{e}_{3}}{{O}_{4}}(s)
What is the Gibbs free energy of the formation of Fe3O4F{{e}_{3}}{{O}_{4}} ?
(A) +296kcal/mol
(B) -242.3kcal/mol
(C) -727 kcal/mol
(D) -229.6 kcal/mol

Explanation

Solution

Gibbs free energy which combines enthalpy and entropy into a single value is denoted as G. the sum of enthalpy with a product of temperature and the entropy of the system indicates the change in free energy ΔG\Delta G . The Gibbs free energy change can be predicting the direction of the chemical reactions at constant temperature and constant pressure.

Complete step by step solution:
The energy which is associated with a chemical reaction that can be used to do work is Gibbs free energy of a system, which is the sum of enthalpy, the product of temperature, and the entropy of the system.
G=H–TSG = H – TS
A spontaneous reaction is considered to be natural because this reaction occurs by itself without any external actions towards it and a non-spontaneous reaction needs constant energy during the process.
In a chemical reaction involving the change in thermodynamic quantities, the above equation changes into,
ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta S
Given chemical reactions,
2Fe(s)+32O2(g)→Fe2O3(s)2Fe(s)+\dfrac{3}{2}{{O}_{2}}(g)\to F{{e}_{2}}{{O}_{3}}(s), ΔG\Delta G = -177Kcal --- (1)
4Fe2O3(s)+Fe(s)→3Fe3O4(s)4F{{e}_{2}}{{O}_{3}}(s)+Fe(s)\to 3F{{e}_{3}}{{O}_{4}}(s), ΔG\Delta G = -19 Kcal --- (2)
Multiplying Equation (1) with 43\dfrac{4}{3} ,
83Fe(s)+2O2(g)→43Fe2O3(s),ΔG=−43(177)=−236Kcal\dfrac{8}{3}Fe(s)+2{{O}_{2}}(g)\to \dfrac{4}{3}F{{e}_{2}}{{O}_{3}}(s),\Delta G=-\dfrac{4}{3}(177)=-236Kcal ------ (3)
Multiplying equation (2) with 13\dfrac{1}{3} ,
43Fe2O3(s)+13Fe(s)→Fe3O4(s),ΔG=−193=−6.34Kcal\dfrac{4}{3}F{{e}_{2}}{{O}_{3}}(s)+\dfrac{1}{3}Fe(s)\to F{{e}_{3}}{{O}_{4}}(s),\Delta G=-\dfrac{19}{3}=-6.34Kcal ------ (4)
Equation (3) + equation (4)
3Fe(s)+2O2(g)→Fe3O4,ΔG=−242.3Kcal3Fe(s)+2{{O}_{2}}(g)\to F{{e}_{3}}{{O}_{4}},\Delta G=-242.3Kcal
Hence, the Gibbs free energy of the formation of Fe3O4F{{e}_{3}}{{O}_{4}}= -242.3Kcal

The correct answer is option B.

Note: If the Gibbs free energy is positive for a reaction is nonspontaneous which means an input of external energy is required for the reaction and if Gibbs free energy is negative for a reaction is spontaneous which means the reaction occurred without external energy requirement.