Replace each question mark with the correct information:

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 $\Delta S_{rxn}$ and $\Delta G_{rxn}$ are negative. Therefore, comment: Spontaneous at all T.

By definition, $\Delta G_{rxn}^{°}=\Delta H_{rxn}^{°}-T\cdot \Delta S_{rxn}^{°}$Knowing that enthalpy is 0,

Then, if $\Delta G_{rxn}^{°}<0$

$\begin{array}{l}\Delta G_{rxn}^{°}=-T\cdot \Delta S_{rxn}^{°}<0-T\cdot \Delta S_{rxn}^{°}<0\\ \Delta S_{rxn}^{°}>0\end{array}$

As a result, the reaction's entropy is positive, $\Delta S_{rxn}^{°}{:}^{+}$

c)

Due to the non-spontaneous nature of the reaction, Gibb's energy change is negative,$\Delta G_{\text{rxn}}{:}^{-}$

d)

Again, looking at the equation:

$\Delta G_{rxn}^{°}=\Delta H_{rxn}^{°}-T\cdot \Delta S_{rxn}^{°}$

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

$\begin{array}{l}\Delta G_{rxn}^{°}=\Delta H_{rxn}^{°}-T\cdot 0\\ \Delta G_{rxn}^{°}=\Delta H_{rxn}^{°}\end{array}$

Thus, since$\Delta G_{rxn}^{°}<0$ , the enthalpy is negative also,$\Delta H_{rxn}^{°}{\text{-}}^{-}$

e)

Again, looking at the equation:

$\Delta G_{rxn}^{°}=\Delta H_{rxn}^{°}-T\cdot \Delta S_{rxn}^{°}$

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

$\begin{array}{l}\Delta G_{rxn}^{°}=0-T\cdot \Delta S_{rxn}^{°}\\ \Delta G_{rxn}^{°}=-T\cdot \Delta S_{rxn}^{°}\end{array}$

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

$\Delta S_{rxn}^{°}$ :-Comment: non-spontaneous

f)

Again, looking at the equation:

$\Delta G_{rxn}^{°}=\Delta H_{rxn}^{°}-T\cdot \Delta S_{rxn}^{°}$

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,

$\begin{array}{l}\Delta H_{rxn}^{°}-T\cdot \Delta S_{rxn}^{°}<0\\ T\cdot \Delta S_{rxn}^{°}>\Delta H_{rxn}^{°}0\end{array}$

The reaction is spontaneous at high temperatures only.

$\Delta G_{r:n}^{°}:-$Comment: At high temperatures, if$T\cdot \Delta S_{rxn}^{°}>\Delta H_{rxn}^{°}0$ , the reaction is spontaneous.