Question

Which one of the following correctly represents the physical significance of Gibbs Energy Change
A.$- \Delta G = {W_{{\text{compression}}}}$
B.$\Delta G = {W_{{\text{expansion}}}}$
C.$\Delta G = - {W_{{\text{expansion}}}} = - {W_{{\text{non - expansion}}}}$
D.$- \Delta G = {W_{{\text{expansion}}}}$

Answer 1

To answer this question, you should recall the concept of Gibbs free energy. Gibbs free energy, also known as the Gibbs function, Gibbs energy, or free enthalpy, is a quantity that is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant.

Complete step by step answer:
Gibbs equation helps us to predict the spontaneity of a reaction based on enthalpy and entropy values directly. When the reaction is exothermic, enthalpy of the system is negative, thus making Gibbs free energy negative. Hence, we can say that the reaction will proceed in the forward direction due to a positive value of the equilibrium constant. This law can be scaled for two different reactions taking place in a system too.
The overall reaction will occur if and only if the net $\Delta G$of the two possible reactions is negative. The decrease in Gibbs energy results in useful work done by the system.
Therefore, $- \Delta G = {W_{{\text{expansion}}}}$ ​is the correct expression.

Hence, the correct answer to this question is option D.

Note:
The concept of Gibbs Free energy is used in the development of the Ellingham diagram. You should know about the Ellingham diagram is a plot between ${\Delta _f}{G^o}$ and T for the formation of oxides of metals.
A general reaction expressing oxidation is given by: $2xM\left( s \right){\text{ }} + {\text{ }}{O_2}\left( g \right){\text{ }} \to {\text{ }}2{M_x}O\left( s \right)$.
As is evident from the reaction, the gaseous amount of reactant is decreasing from left to right as the product formed is solid metal oxide on the right side. Hence, we can say that molecular randomness is also decreasing from left to right. Thus, $\Delta {\text{S}}$ is negative and $\Delta {G^o}$ shifts towards higher side despite rising T. Hence, for most of the reactions shown above for the formation of ${M_x}O{\text{ }}\left( s \right),$the curve is positive. Except for the processes in which change of phase takes place, each plot is a straight line. This temperature at which change of phase takes place is indicated by a positive increase in the slope. For example, the melting is indicated by an abrupt change in the curve in $Zn,{\text{ }}ZnO$the plot. The metal oxide is stable at the point in a curve below which $\Delta {G^o}$ is negative. Above this point, the metal oxide is unstable and decomposes on its own. Feasibility of reductions of the oxide of the upper line by the element represented by the lower line is determined by the difference in the two ${\Delta _r}{G^0}$ values after the point of intersection in the Ellingham diagram.