Q20.83 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Replace each question mark with the correct information:

Step-by-Step Solution

Answer

The results are presented in Table format below, but full explanations are provided below.

Step 1: Definition of Delta

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.

Step 2: Replace each question mark with the correct information given in a)

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.

Step 3: Replace each question mark with the correct information given in b)

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, $\Delta S_{rxn}^{°}{:}^{+}$

Step 4: Replace each question mark with the correct information given in c)

c) 

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

Step 5: Replace each question mark with the correct information given in d)

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, $\Delta H_{rxn}^{°}{\text{ - }}^{-}$

Step 6: Replace each question mark with the correct information given in e)

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

Step 7: Replace each question mark with the correct information given in f)

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.