The standard enthalpies of formation for S(g), F(g), SF4(g), and SF6(g) are +278.8,+79.0,-775, a

The standard enthalpies of formation for S(g), F(g), SF4(g),...

The standard enthalpies of formation for $\mathrm{S}(g), \mathrm{F}(g), \mathrm{SF}_{4}(g)$, and $\mathrm{SF}_{6}(g)$ are $+278.8,+79.0,-775$, and $-1209 \mathrm{~kJ} / \mathrm{mol}$, respectively. a. Use these data to estimate the energy of an $\mathrm{S}-\mathrm{F}$ bond. b. Compare your calculated value to the value given in Table $8.4$. What conclusions can you draw? c. Why are the $\Delta H_{\mathrm{f}}^{\circ}$ values for $\mathrm{S}(g)$ and $\mathrm{F}(g)$ not equal to zero, since sulfur and fluorine are elements?

The standard enthalpies of formation for $\mathrm{S}(g), \mathrm{F}(g), \mathrm{SF}_{4}(g)$, and $\mathrm{SF}_{6}(g)$ are $+278.8,+79.0,-775$, and $-1209 \mathrm{~kJ} / \mathrm{mol}$, respectively.
a. Use these data to estimate the energy of an $\mathrm{S}-\mathrm{F}$ bond.
b. Compare your calculated value to the value given in Table $8.4$. What conclusions can you draw?
c. Why are the $\Delta H_{\mathrm{f}}^{\circ}$ values for $\mathrm{S}(g)$ and $\mathrm{F}(g)$ not equal to zero, since sulfur and fluorine are elements?

Key Concepts

Standard Enthalpy of Formation

This is the heat change associated with the formation of one mole of a compound from its elements in their standard states. It serves as a reference point in thermodynamics for calculating and comparing the energy changes of various chemical reactions, even when the elements involved might not be in their standard states under the conditions of a particular calculation.

Bond Energy

Bond energy (or bond enthalpy) refers to the amount of energy required to break a particular type of bond in one mole of gaseous molecules. It is a critical concept in understanding molecular stability and reactivity, and it can be estimated by examining the overall energy changes involved in forming and breaking chemical bonds within a molecule.

Hess's Law

Hess's Law is the principle stating that the total enthalpy change for a reaction is the same, regardless of whether it occurs in one step or in several steps. This law allows for the calculation of bond energies by constructing thermochemical cycles with known enthalpy changes, such as the standard enthalpies of formation for various substances.

Reference State Considerations

In thermodynamics, standard enthalpies of formation are defined for elements in their standard states. However, when elements exist in a different physical state than the reference state (for example, as isolated atoms in the gas phase rather than as their most stable form), their formation enthalpies do not equal zero. This difference arises because additional energy is involved in converting the element from its standard state to the state used in the calculation.

Transcript

00:01 Okay, so for this problem we are asked to identify or get the bond energy for the sf bond for two different compounds, sulfur tetrafluoride or sf4 and sulfur hexafluoride or sf6.

00:16 So what i wrote here are the two reactions corresponding to the breakdown of sf4 and sf6 into the single atoms sulfur and fluoride.

00:26 The reason i chose to use these reactions is because those are the enthalpies of formation that we are given, so it's going to be useful to use these reactions given those enthalpies of formation.

00:40 Now let's look at the structures of sf4 and sf6 and we can see that sf4 has four single covalent bonds between s and f and sf6 has six single covalent bonds between s and f.

00:59 Let's also remember that there are two ways of calculating the total energy or enthalpy of a reaction.

01:06 One is by looking at the energies, the bond energies for the compounds and for that we can use equation number one that says that the total enthalpy of the reaction equals the energy input to break bonds minus the energy output for the bonds that are formed.

01:27 Notice that for the products we do not have any bonds that are formed because these are the single atoms, so we're only going to have the energy input for the bonds that are broken if we use equation one.

01:43 For the second equation we know that we can also calculate the enthalpies of formation of the products summed up minus the enthalpies of formation for the reactants also summed up.

01:57 So we're going to use both equations to answer this question...

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