91Ó°ÊÓ

It is used in chemistry to indicate a change in enthalpy as well as the addition of heat to the reaction. A triangle symbol is used to represent it.

a)

The reaction is always spontaneous at any temperature since both $\mathrm{Δ}{S}_{rxn}$ and $\mathrm{Δ}{G}_{rxn}$ are negative. Therefore, comment: Spontaneous at all T.

By definition, $\mathrm{Δ}{G}_{rxn}^{°}=\mathrm{Δ}{H}_{rxn}^{°}−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}$Knowing that enthalpy is 0,

Then, if $\mathrm{Δ}{G}_{rxn}^{°}<0$

$\begin{array}{l}\mathrm{Δ}{G}_{rxn}^{°}=−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}<0−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}<0\\ \mathrm{Δ}{S}_{rxn}^{°}>0\end{array}$

As a result, the reaction's entropy is positive, $\mathrm{Δ}{S}_{rxn}^{°}{:}^{+}$

c)

Due to the non-spontaneous nature of the reaction, Gibb's energy change is negative,$\mathrm{Δ}{G}_{\text{rxn}}{:}^{\text{ }−}$

d)

Again, looking at the equation:

$\mathrm{Δ}{G}_{rxn}^{°}=\mathrm{Δ}{H}_{rxn}^{°}−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}$

If the entropy is zero, it can be rearranged as follows:

$\begin{array}{l}\mathrm{Δ}{G}_{rxn}^{°}=\mathrm{Δ}{H}_{rxn}^{°}−Tâ‹\dots 0\\ \mathrm{Δ}{G}_{rxn}^{°}=\mathrm{Δ}{H}_{rxn}^{°}\end{array}$

Thus, since$\mathrm{Δ}{G}_{rxn}^{°}<0$ , the enthalpy is negative also,$\mathrm{Δ}{H}_{rxn}^{°}{\text{-}}^{−}$

e)

Again, looking at the equation:

$\mathrm{Δ}{G}_{rxn}^{°}=\mathrm{Δ}{H}_{rxn}^{°}−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}$

If the enthalpy is zero, it can be rearranged as follows:

$\begin{array}{l}\mathrm{Δ}{G}_{rxn}^{°}=0−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}\\ \mathrm{Δ}{G}_{rxn}^{°}=−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}\end{array}$

As a result, the entropy is negative since $\mathrm{Δ}{G}_{rxn}^{°}>0$. The reaction is non-spontaneous because Gibb's energy is positive.

$\mathrm{Δ}{S}_{rxn}^{°}$ :-Comment: non-spontaneous

f)

Again, looking at the equation:

$\mathrm{Δ}{G}_{rxn}^{°}=\mathrm{Δ}{H}_{rxn}^{°}−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}$

If both enthalpy and entropy are positive, the spontaneity of the reaction will be determined by the reaction temperature, as Gibb's energy for a spontaneous reaction should be negative. Then,

$\begin{array}{l}\mathrm{Δ}{H}_{rxn}^{°}−Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}<0\\ Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}>\mathrm{Δ}{H}_{rxn}^{°}0\end{array}$

The reaction is spontaneous at high temperatures only.

$\mathrm{Δ}{G}_{r:n}^{°}:−$Comment: At high temperatures, if$Tâ‹\dots \mathrm{Δ}{S}_{rxn}^{°}>\mathrm{Δ}{H}_{rxn}^{°}0$ , the reaction is spontaneous.