Chapter 8, Problem 81E

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.

1. a. Use these data to estimate the energy of an S—F bond.
2. b. Compare your calculated value to the value given in Table 8.5. What conclusions can you draw?
3. c. Why are the $\Delta H_t^0$ values for S(g) and F(g) not equal to zero, since sulfur and fluorine are elements?

Interpretation Introduction

Interpretation:

For the given reactions with given standard enthalpies of formation values the energy for $\text{S-F}$ bond, the standard enthalpy data should be compared be compared with the bond energy data and the reason for being elements sulphur and fluorine ${\text{ΔH}}_{\text{f}}$ values are not zero are explained.

Concept Introduction:

Bond energy: It is defined as the strength of the bond means the energy absorbed or released during bond making or bond breaking process.

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions. The standard enthalpy of the reaction is equal to the subtraction of total enthalpy for formation product with the total enthalpy for formation of the reactants which is given as ${\text{ΔH}}^{\circ }{}_{\text{rxn}}=\sum n{\text{ΔH}}^{\circ }{}_f\left(\text{products}\right)-\sum m{\text{ΔH}}^{\circ }{}_f\left(\text{reactants}\right)$.

The total enthalpy change for the molecule is equal to the summation of total bond energy of bonds that break which gets subtracted from the summation of total bond energy of the bonds that formed in the chemical reaction. $\text{ΔH}=\sum n\times \text{D}\left(\text{bonds broken}\right)-\sum n\times \text{D}\left(\text{bonds formed}\right)$.

Ideal Gas equation: It is the equation which equals the pressure and volume product of one mole of gas to the temperature and gas constant which is $\text{PV = nRT}$.

Enthalpy change of combustion: It is the heat energy change involved when one mole of substance gets burned in pressure of oxygen under standard conditions.

Standard enthalpy of formation: It is the heat required when one mole of substance formed from its pure elements in standard state.

Limiting Reagent: A reactant is considered as limiting reagent when it limits the product formation which the product formed depend on the amount of that reactant.

To determine: The $\text{S-F}$ bond energy values, the comparison of these values to the standard enthalpy value finally the sulphur and the fluorine being elements they are much more stable with non-zero ${\text{ΔH}}_{\text{f}}$ values.

Answer to Problem 81E

The energy of $\text{S-F}$ bond in ${\text{SF}}_{\text{4}}{}_{\left(\text{g}\right)}$ is $342.45\text{kJ/mol}$, the $\text{S-F}$ bond energy in ${\text{SF}}_6{}_{\left(\text{g}\right)}$ is $254.63\text{kJ/mol}$ and the $\text{S-F}$ bond energy in third kind of reaction is $296\text{kJ/mol}$.

Explanation of Solution

Given,

Determine the $\text{S-F}$ bond energy value,

$\begin{array}{l}\text{First consider,}\\ {\text{S}}_{\left(\text{g}\right)}{\text{+4F}}_{\left(\text{g}\right)}\to {\text{SF}}_{\text{4}}{}_{\left(\text{g}\right)}\\ \Delta \text{H}\text{=}\text{-775kJ/mol - }\left(278.8\text{kJ/mol}+4\times 79\text{kJ/mol}\right)\\ =-1368.9\text{kJ/mol}\\ \text{4 S-F bond energy = }-1368.9\text{kJ/mol}\\ \text{S-F bond energy}=\frac{1368.9\text{kJ/mol}}{4}=342.45\text{kJ/mol}\\ \end{array}$

$\begin{array}{l}\text{Next consider,}\\ {\text{S}}_{\left(\text{g}\right)}{\text{+6F}}_{\left(\text{g}\right)}\to {\text{SF}}_6{}_{\left(\text{g}\right)}\\ \Delta \text{H}\text{=}\text{-775kJ/mol - }\left(278.8\text{kJ/mol}+6\times 79\text{kJ/mol}\right)\\ =-1527.8\text{kJ/mol}\\ \text{4 S-F bond energy = }-1527.8\text{kJ/mol}\\ \text{S-F bond energy}=\frac{1527.8\text{kJ/mol}}{6}=254.63\text{kJ/mol}\\ \end{array}$

$\begin{array}{l}\text{Finally consider,}\\ {\text{SF}}_{\text{4}}{}_{\left(\text{g}\right)}{\text{+2F}}_{\left(\text{g}\right)}\to {\text{SF}}_6{}_{\left(\text{g}\right)}\\ \Delta \text{H}\text{=}\text{-1209kJ/mol - }\left(-775\text{kJ/mol}+2\times 79\text{kJ/mol}\right)\\ =-592\text{kJ/mol}\\ \text{S-F bond energy}=\frac{592\text{kJ/mol}}{2}=296\text{kJ/mol}\end{array}$

Interpretation Introduction

Interpretation:

For the given reactions with given standard enthalpies of formation values the energy for $\text{S-F}$ bond, the standard enthalpy data should be compared be compared with the bond energy data and the reason for being elements sulphur and fluorine ${\text{ΔH}}_{\text{f}}$ values are not zero are explained.

Concept Introduction:

Bond energy: It is defined as the strength of the bond means the energy absorbed or released during bond making or bond breaking process.

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions. The standard enthalpy of the reaction is equal to the subtraction of total enthalpy for formation product with the total enthalpy for formation of the reactants which is given as ${\text{ΔH}}^{\circ }{}_{\text{rxn}}=\sum n{\text{ΔH}}^{\circ }{}_f\left(\text{products}\right)-\sum m{\text{ΔH}}^{\circ }{}_f\left(\text{reactants}\right)$.

The total enthalpy change for the molecule is equal to the summation of total bond energy of bonds that break which gets subtracted from the summation of total bond energy of the bonds that formed in the chemical reaction. $\text{ΔH}=\sum n\times \text{D}\left(\text{bonds broken}\right)-\sum n\times \text{D}\left(\text{bonds formed}\right)$.

Ideal Gas equation: It is the equation which equals the pressure and volume product of one mole of gas to the temperature and gas constant which is $\text{PV = nRT}$.

Enthalpy change of combustion: It is the heat energy change involved when one mole of substance gets burned in pressure of oxygen under standard conditions.

Standard enthalpy of formation: It is the heat required when one mole of substance formed from its pure elements in standard state.

Limiting Reagent: A reactant is considered as limiting reagent when it limits the product formation which the product formed depend on the amount of that reactant.

To determine: The $\text{S-F}$ bond energy values, the comparison of these values to the standard enthalpy value finally the sulphur and the fluorine being elements they are much more stable with non-zero ${\text{ΔH}}_{\text{f}}$ values.

Answer to Problem 81E

The bond energy for $\text{S-F}$ is $\text{327kJ/mol}$ which is different from the standard enthalpy calculated value in the given table.

Explanation of Solution

Comparing the calculated data with the given bond energy data for $\text{S-F}$ bond shows us that the bond energy value for $\text{S-F}$ is $\text{327kJ/mol}$ which is different from the calculated bond energy values for $\text{S-F}$ bond.

Interpretation Introduction

Interpretation:

For the given reactions with given standard enthalpies of formation values the energy for $\text{S-F}$ bond, the standard enthalpy data should be compared be compared with the bond energy data and the reason for being elements sulphur and fluorine ${\text{ΔH}}_{\text{f}}$ values are not zero are explained.

Concept Introduction:

Bond energy: It is defined as the strength of the bond means the energy absorbed or released during bond making or bond breaking process.

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions. The standard enthalpy of the reaction is equal to the subtraction of total enthalpy for formation product with the total enthalpy for formation of the reactants which is given as ${\text{ΔH}}^{\circ }{}_{\text{rxn}}=\sum n{\text{ΔH}}^{\circ }{}_f\left(\text{products}\right)-\sum m{\text{ΔH}}^{\circ }{}_f\left(\text{reactants}\right)$.

The total enthalpy change for the molecule is equal to the summation of total bond energy of bonds that break which gets subtracted from the summation of total bond energy of the bonds that formed in the chemical reaction. $\text{ΔH}=\sum n\times \text{D}\left(\text{bonds broken}\right)-\sum n\times \text{D}\left(\text{bonds formed}\right)$.

Ideal Gas equation: It is the equation which equals the pressure and volume product of one mole of gas to the temperature and gas constant which is $\text{PV = nRT}$.

Enthalpy change of combustion: It is the heat energy change involved when one mole of substance gets burned in pressure of oxygen under standard conditions.

Standard enthalpy of formation: It is the heat required when one mole of substance formed from its pure elements in standard state.

Limiting Reagent: A reactant is considered as limiting reagent when it limits the product formation which the product formed depend on the amount of that reactant.

To determine: The $\text{S-F}$ bond energy values, the comparison of these values to the standard enthalpy value finally the sulphur and the fluorine being elements they are much more stable with non-zero ${\text{ΔH}}_{\text{f}}$ values.

Answer to Problem 81E

Being elements the sulphur and fluorine enthalpy values are not 0 since they are more stable form that is fluorine in ${\text{F}}_{\text{2}}$ form and sulphur in stable ${\text{S}}_{\text{8}}$ form.

Explanation of Solution

The standard enthalpy values for fluorine and Sulphur are not 0 since they are elements due to the fact that both elements are in their stable form that is fluorine in ${\text{F}}_{\text{2}}$ form and sulphur in stable ${\text{S}}_{\text{8}}$ form owing to have non-zero enthalpy values and that these elements in their non-stable form will results to have zero enthalpy values.