The standard enthalpies of formation for S(g), F(g), SF 4 (g), and SF 6 (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 Δ H f ° values for S(g) and F(g) not equal to zero, since sulfur and fluorine are elements?

Question

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Answer 1

Textbook Question

Chapter 6, Problem 136CP

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 $Δ H f °$ values for S(g) and F(g) not equal to zero, since sulfur and fluorine are elements?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in $SF6$ is $327 kJ/mol$ .
The SF bond energy in $SF4$ $342.5 kJ$

Explanation of Solution

Record data from given:

$Standard enthalpies of formation S(g) is +278.8 kJ/molStandard enthalpies of formation F(g) is +79.0 kJ/molStandard enthalpies of formation SF4(g) is -775 kJ/molStandard enthalpies of formationSF6(g) is +1209 kJ/mol$

To calculate the SF bond energy in $SF4$ .

$SF4 dissociation reaction : SF4(g)→S(g)+4F(g) ΔH°= 278.8+4(79.0)-(775) ΔH°= 1370 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 4 ×(dissociation energy SF bond). DSF= 1370 kJ4 =342.5 kJ/mol$

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 $SF4$ .
The SF bond energy in $SF4$ $342.5 kJ$ .

To calculate the SF bond energy in $SF6$ .

$SF4 dissociation reaction : SF6(g)→S(g)+6F(g) ΔH°= 278.8+6(79.0)-(1209) ΔH°= 1962 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 6 ×(dissociation energy SF bond). DSF= 1962kJ6 =327.0 kJ/mol$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in table is $SF6$ .

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 $SF6$ .
The SF bond energy in $SF6$ is $327 kJ/mol$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ 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 $327 kJ/mol$ and the calculated bond energies are compared to standard data, the SF bond energy $327 kJ/mol$ is based on the $SF6$ .
The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ not stable so standard enthalpies of formation values are all so not zero..

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Answer 2

Textbook Question

Chapter 6, Problem 136CP

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 $Δ H f °$ values for S(g) and F(g) not equal to zero, since sulfur and fluorine are elements?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in $SF6$ is $327 kJ/mol$ .
The SF bond energy in $SF4$ $342.5 kJ$

Explanation of Solution

Record data from given:

$Standard enthalpies of formation S(g) is +278.8 kJ/molStandard enthalpies of formation F(g) is +79.0 kJ/molStandard enthalpies of formation SF4(g) is -775 kJ/molStandard enthalpies of formationSF6(g) is +1209 kJ/mol$

To calculate the SF bond energy in $SF4$ .

$SF4 dissociation reaction : SF4(g)→S(g)+4F(g) ΔH°= 278.8+4(79.0)-(775) ΔH°= 1370 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 4 ×(dissociation energy SF bond). DSF= 1370 kJ4 =342.5 kJ/mol$

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 $SF4$ .
The SF bond energy in $SF4$ $342.5 kJ$ .

To calculate the SF bond energy in $SF6$ .

$SF4 dissociation reaction : SF6(g)→S(g)+6F(g) ΔH°= 278.8+6(79.0)-(1209) ΔH°= 1962 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 6 ×(dissociation energy SF bond). DSF= 1962kJ6 =327.0 kJ/mol$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in table is $SF6$ .

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 $SF6$ .
The SF bond energy in $SF6$ is $327 kJ/mol$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ 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 $327 kJ/mol$ and the calculated bond energies are compared to standard data, the SF bond energy $327 kJ/mol$ is based on the $SF6$ .
The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ not stable so standard enthalpies of formation values are all so not zero..

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 6, Problem 136CP

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 $Δ H f °$ values for S(g) and F(g) not equal to zero, since sulfur and fluorine are elements?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in $SF6$ is $327 kJ/mol$ .
The SF bond energy in $SF4$ $342.5 kJ$

Explanation of Solution

Record data from given:

$Standard enthalpies of formation S(g) is +278.8 kJ/molStandard enthalpies of formation F(g) is +79.0 kJ/molStandard enthalpies of formation SF4(g) is -775 kJ/molStandard enthalpies of formationSF6(g) is +1209 kJ/mol$

To calculate the SF bond energy in $SF4$ .

$SF4 dissociation reaction : SF4(g)→S(g)+4F(g) ΔH°= 278.8+4(79.0)-(775) ΔH°= 1370 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 4 ×(dissociation energy SF bond). DSF= 1370 kJ4 =342.5 kJ/mol$

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 $SF4$ .
The SF bond energy in $SF4$ $342.5 kJ$ .

To calculate the SF bond energy in $SF6$ .

$SF4 dissociation reaction : SF6(g)→S(g)+6F(g) ΔH°= 278.8+6(79.0)-(1209) ΔH°= 1962 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 6 ×(dissociation energy SF bond). DSF= 1962kJ6 =327.0 kJ/mol$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in table is $SF6$ .

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 $SF6$ .
The SF bond energy in $SF6$ is $327 kJ/mol$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ 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 $327 kJ/mol$ and the calculated bond energies are compared to standard data, the SF bond energy $327 kJ/mol$ is based on the $SF6$ .
The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ not stable so standard enthalpies of formation values are all so not zero..

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Answer 4

Textbook Question

Chapter 6, Problem 136CP

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 $Δ H f °$ values for S(g) and F(g) not equal to zero, since sulfur and fluorine are elements?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in $SF6$ is $327 kJ/mol$ .
The SF bond energy in $SF4$ $342.5 kJ$

Explanation of Solution

Record data from given:

$Standard enthalpies of formation S(g) is +278.8 kJ/molStandard enthalpies of formation F(g) is +79.0 kJ/molStandard enthalpies of formation SF4(g) is -775 kJ/molStandard enthalpies of formationSF6(g) is +1209 kJ/mol$

To calculate the SF bond energy in $SF4$ .

$SF4 dissociation reaction : SF4(g)→S(g)+4F(g) ΔH°= 278.8+4(79.0)-(775) ΔH°= 1370 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 4 ×(dissociation energy SF bond). DSF= 1370 kJ4 =342.5 kJ/mol$

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 $SF4$ .
The SF bond energy in $SF4$ $342.5 kJ$ .

To calculate the SF bond energy in $SF6$ .

$SF4 dissociation reaction : SF6(g)→S(g)+6F(g) ΔH°= 278.8+6(79.0)-(1209) ΔH°= 1962 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 6 ×(dissociation energy SF bond). DSF= 1962kJ6 =327.0 kJ/mol$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in table is $SF6$ .

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 $SF6$ .
The SF bond energy in $SF6$ is $327 kJ/mol$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ 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 $327 kJ/mol$ and the calculated bond energies are compared to standard data, the SF bond energy $327 kJ/mol$ is based on the $SF6$ .
The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ not stable so standard enthalpies of formation values are all so not zero..

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in $SF6$ is $327 kJ/mol$ .
The SF bond energy in $SF4$ $342.5 kJ$

Explanation of Solution

Record data from given:

$Standard enthalpies of formation S(g) is +278.8 kJ/molStandard enthalpies of formation F(g) is +79.0 kJ/molStandard enthalpies of formation SF4(g) is -775 kJ/molStandard enthalpies of formationSF6(g) is +1209 kJ/mol$

To calculate the SF bond energy in $SF4$ .

$SF4 dissociation reaction : SF4(g)→S(g)+4F(g) ΔH°= 278.8+4(79.0)-(775) ΔH°= 1370 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 4 ×(dissociation energy SF bond). DSF= 1370 kJ4 =342.5 kJ/mol$

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 $SF4$ .
The SF bond energy in $SF4$ $342.5 kJ$ .

To calculate the SF bond energy in $SF6$ .

$SF4 dissociation reaction : SF6(g)→S(g)+6F(g) ΔH°= 278.8+6(79.0)-(1209) ΔH°= 1962 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 6 ×(dissociation energy SF bond). DSF= 1962kJ6 =327.0 kJ/mol$

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in table is $SF6$ .

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 $SF6$ .
The SF bond energy in $SF6$ is $327 kJ/mol$ .

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ 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 $327 kJ/mol$ and the calculated bond energies are compared to standard data, the SF bond energy $327 kJ/mol$ is based on the $SF6$ .
The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ not stable so standard enthalpies of formation values are all so not zero..

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in $SF6$ is $327 kJ/mol$ .
The SF bond energy in $SF4$ $342.5 kJ$

Explanation of Solution

Record data from given:

$Standard enthalpies of formation S(g) is +278.8 kJ/molStandard enthalpies of formation F(g) is +79.0 kJ/molStandard enthalpies of formation SF4(g) is -775 kJ/molStandard enthalpies of formationSF6(g) is +1209 kJ/mol$

To calculate the SF bond energy in $SF4$ .

$SF4 dissociation reaction : SF4(g)→S(g)+4F(g) ΔH°= 278.8+4(79.0)-(775) ΔH°= 1370 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 4 ×(dissociation energy SF bond). DSF= 1370 kJ4 =342.5 kJ/mol$

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 $SF4$ .
The SF bond energy in $SF4$ $342.5 kJ$ .

To calculate the SF bond energy in $SF6$ .

$SF4 dissociation reaction : SF6(g)→S(g)+6F(g) ΔH°= 278.8+6(79.0)-(1209) ΔH°= 1962 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 6 ×(dissociation energy SF bond). DSF= 1962kJ6 =327.0 kJ/mol$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 13

Answer to Problem 136CP

The SF bond energy in $SF6$ is $327 kJ/mol$ .
The SF bond energy in $SF4$ $342.5 kJ$

Answer 14

Explanation of Solution

Record data from given:

$Standard enthalpies of formation S(g) is +278.8 kJ/molStandard enthalpies of formation F(g) is +79.0 kJ/molStandard enthalpies of formation SF4(g) is -775 kJ/molStandard enthalpies of formationSF6(g) is +1209 kJ/mol$

To calculate the SF bond energy in $SF4$ .

$SF4 dissociation reaction : SF4(g)→S(g)+4F(g) ΔH°= 278.8+4(79.0)-(775) ΔH°= 1370 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 4 ×(dissociation energy SF bond). DSF= 1370 kJ4 =342.5 kJ/mol$

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 $SF4$ .
The SF bond energy in $SF4$ $342.5 kJ$ .

To calculate the SF bond energy in $SF6$ .

$SF4 dissociation reaction : SF6(g)→S(g)+6F(g) ΔH°= 278.8+6(79.0)-(1209) ΔH°= 1962 kJ DSF= ΔH°Number of S-F bondThis enthalpy change is equal to 6 ×(dissociation energy SF bond). DSF= 1962kJ6 =327.0 kJ/mol$

Answer 15

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The SF bond energy in table is $SF6$ .

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 $SF6$ .
The SF bond energy in $SF6$ is $327 kJ/mol$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 20

Answer to Problem 136CP

The SF bond energy in table is $SF6$ .

Answer 21

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 $SF6$ .
The SF bond energy in $SF6$ is $327 kJ/mol$ .

Answer 22

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 136CP

The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ 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 $327 kJ/mol$ and the calculated bond energies are compared to standard data, the SF bond energy $327 kJ/mol$ is based on the $SF6$ .
The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ not stable so standard enthalpies of formation values are all so not zero..

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Interpretation Introduction

Interpretation:

The S- F bond energy should be calculated and standard enthalpies of formation values of $S(g)$ and $F(g)$ 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:
$ΔHf = ∑npΔH°f(products)-∑nrΔH°f(reactants)......(1)$
Internal energy change of a reaction is given as,

$Internal energy change ΔE= Bond broken energy - Bond formation energy$

$ΔE = ∑Dbroken-∑Dformed......(2)$
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 27

Answer to Problem 136CP

The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ not stable so standard enthalpies of formation values are all so not zero.

Answer 28

Explanation of Solution

In the standard data table the given SF bond energy is $327 kJ/mol$ and the calculated bond energies are compared to standard data, the SF bond energy $327 kJ/mol$ is based on the $SF6$ .
The standard state of Sulfur is and Fluorine are $S8(g)$ and $F2(g)$ states so the given states are $S(g)$ and $F(g)$ not stable so standard enthalpies of formation values are all so not zero..