Chapter 7, Problem 126CP

The standard enthalpies of formation for S(g), F(g), SF₄(g), and SF₆(g) are +278.8, +79.0, −775, and +1209 KJ/mol, respectively.

a. Use these data to estimate the energy of an S—F bond.

b. Compare your calculated value to the value given in Table 3-3. What conclusions can you draw?

c. Why are the  $\Delta H_f^{°}$ values for S(g) and F(g) not equal to zero, since sulfur and fluorine are elements?

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of ${\text{S}}_{\left(\text{g}\right)}$ and ${\text{F}}_{\left(\text{g}\right)}$ should be explained.

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that associate with the formation of one mole of a product from its pure elements, with all substances in its standard states is called as a standard enthalpy of formation.
• Formula:

  ${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

• Internal energy change of a reaction is given as,

• $\text{Internal energy change ΔE= Bond broken energy - Bond formation energy}$

  $\text{ΔE}\text{=}{\displaystyle \sum {\text{D}}_{\text{broken}}}\text{-}{\displaystyle \sum {\text{D}}_{\text{formed}}}\mathrm{......}(2)$

• The internal energy change is equal to enthalpy change in the gas phase reactions.
• The dissociation bond energy is equal to enthalpy change in the gas phase reactions

Answer to Problem 126CP

• The SF bond energy in ${\text{SF}}_{\text{6 }}$ is $\text{327}\text{kJ/mol}$.
• The SF bond energy in ${\text{SF}}_{\text{4 }}$ $342.5\text{kJ}$

Explanation of Solution

Record data from given:

$\begin{array}{l}\text{Standard enthalpies of formation S(g)}\text{is}\text{+278}\text{.8}\text{kJ/mol}\\ \text{Standard enthalpies of formation}\text{F(g)}\text{is}\text{+79}\text{.0}\text{kJ/mol}\\ \text{Standard enthalpies of formation}\text{SF4(g)}\text{is}\text{-775}\text{kJ/mol}\\ \text{Standard enthalpies of formationSF6(g)}\text{is}\text{+1209}\text{kJ/mol}\end{array}$

To calculate the SF bond energy in ${\text{SF}}_{\text{4 }}$.

$\begin{array}{l}{\text{SF}}_{\text{4}}\text{dissociation}\text{reaction}\text{:}\\ {\text{SF}}_{\text{4(g)}}\to {\text{S}}_{\text{(g)}}{\text{+4F}}_{\text{(g)}}\\ \text{ΔH°=}\text{278}\text{.8+4(79}\text{.0)-(775)}\\ \text{ΔH°=}\text{1370}\text{kJ}\\ {\text{D}}_{\text{SF}}\text{=}\frac{\text{ΔH°}}{\text{Number}\text{of}\text{S-F}\text{bond}}\\ \text{This enthalpy change is equal to 4 ×}\left(\text{dissociation energy}\text{SF}\text{bond}\right)\text{.}\\ {\text{D}}_{\text{SF}}\text{=}\frac{\text{1370}\text{kJ}}{\text{4}}\\ \text{=342}\text{.5}\text{kJ/mol}\end{array}$

• The given standard enthalpies of formation values are plugging in to above equation 1to given the enthalpy change of the reactions.
• This enthalpy change divide by 4 to give SF bond energy in ${\text{SF}}_{\text{4 }}$.
• The SF bond energy in ${\text{SF}}_{\text{4 }}$ $342.5\text{kJ}$.

To calculate the SF bond energy in ${\text{SF}}_{\text{6 }}$.

$\begin{array}{l}{\text{SF}}_{\text{4}}\text{dissociation}\text{reaction}\text{:}\\ {\text{SF}}_{\text{6(g)}}\to {\text{S}}_{\text{(g)}}{\text{+6F}}_{\text{(g)}}\\ \text{ΔH°=}\text{278}\text{.8+6(79}\text{.0)-(1209)}\\ \text{ΔH°=}\text{1962}\text{kJ}\\ {\text{D}}_{\text{SF}}\text{=}\frac{\text{ΔH°}}{\text{Number}\text{of}\text{S-F}\text{bond}}\\ \text{This enthalpy change is equal to 6 ×}\left(\text{dissociation energy}\text{SF}\text{bond}\right)\text{.}\\ {\text{D}}_{\text{SF}}\text{=}\frac{\text{1962kJ}}{\text{6}}\\ \text{=327}\text{.0}\text{kJ/mol}\end{array}$

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of ${\text{S}}_{\left(\text{g}\right)}$ and ${\text{F}}_{\left(\text{g}\right)}$ should be explained.

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that associate with the formation of one mole of a product from its pure elements, with all substances in its standard states is called as a standard enthalpy of formation.
• Formula:

  ${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

• Internal energy change of a reaction is given as,

• $\text{Internal energy change ΔE= Bond broken energy - Bond formation energy}$

  $\text{ΔE}\text{=}{\displaystyle \sum {\text{D}}_{\text{broken}}}\text{-}{\displaystyle \sum {\text{D}}_{\text{formed}}}\mathrm{......}(2)$

• The internal energy change is equal to enthalpy change in the gas phase reactions.
• The dissociation bond energy is equal to enthalpy change in the gas phase reactions

Answer to Problem 126CP

• The SF bond energy in table is ${\text{SF}}_{\text{6 }}$.

Explanation of Solution

• The standard enthalpy of formation values are plugging in to equation (1) to get enthalpy change of the reaction.
• This enthalpy change divide by 6 to give SF bond energy in ${\text{SF}}_{\text{6 }}$.
• The SF bond energy in ${\text{SF}}_{\text{6 }}$ is $\text{327}\text{kJ/mol}$.

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of ${\text{S}}_{\left(\text{g}\right)}$ and ${\text{F}}_{\left(\text{g}\right)}$ should be explained.

Hess's Law:

Standard enthalpy of formation:

• The change in enthalpy that associate with the formation of one mole of a product from its pure elements, with all substances in its standard states is called as a standard enthalpy of formation.
• Formula:

  ${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

• Internal energy change of a reaction is given as,

• $\text{Internal energy change ΔE= Bond broken energy - Bond formation energy}$

  $\text{ΔE}\text{=}{\displaystyle \sum {\text{D}}_{\text{broken}}}\text{-}{\displaystyle \sum {\text{D}}_{\text{formed}}}\mathrm{......}(2)$

• The internal energy change is equal to enthalpy change in the gas phase reactions.
• The dissociation bond energy is equal to enthalpy change in the gas phase reactions

Answer to Problem 126CP

• The standard state of Sulfur is and Fluorine are ${\text{S}}_{8\left(\text{g}\right)}$ and ${\text{F}}_{2\left(\text{g}\right)}$ states so the given states  are ${\text{S}}_{\left(\text{g}\right)}$ and ${\text{F}}_{\left(\text{g}\right)}$ not stable so standard enthalpies of formation values are all so not zero.

Explanation of Solution

• In the standard data table the given SF bond energy is $\text{327}\text{kJ/mol}$ and the calculated bond energies are compared to standard data, the SF bond energy $\text{327}\text{kJ/mol}$ is based on the ${\text{SF}}_{\text{6 }}$.
• The standard state of Sulfur is and Fluorine are ${\text{S}}_{8\left(\text{g}\right)}$ and ${\text{F}}_{2\left(\text{g}\right)}$ states so the given states  are ${\text{S}}_{\left(\text{g}\right)}$ and ${\text{F}}_{\left(\text{g}\right)}$ not stable so standard enthalpies of formation values are all so not zero..