Question

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 \(\mathrm{S}-\) F bond. b. Compare your calculated value to the value given in Table \(8.5 .\) What conclusions can you draw? c. Why are the \(\Delta H_{f}^{\circ}\) values for \(\mathrm{S}(g)\) and \(\mathrm{F}(g)\) not equal to zero, since sulfur and fluorine are elements?

Short answer

The energy of an S-F bond is estimated to be -217 kJ/mol, calculated using the given standard enthalpies of formation and Hess's Law. The comparison with Table 8.5 is not possible here. The non-zero enthalpies of formation for S(g) and F(g) are because sulfur and fluorine are not in their standard states (solid for sulfur and diatomic gas for fluorine).

Step-by-step solution

a. Estimating the energy of an S-F bond using standard enthalpies of formation

We'll first consider the formation reactions for \(\mathrm{SF}_{4}(g)\) and \(\mathrm{SF}_{6}(g)\): For \(\mathrm{SF}_{4}(g)\): \[S(g) + 4F(g) \rightarrow SF_4(g)\] For \(\mathrm{SF}_{6}(g)\): \[S(g) + 6F(g) \rightarrow SF_6(g)\] Now, we can use Hess's Law by subtracting the formation reaction of \(\mathrm{SF}_{4}(g)\) from the formation reaction of \(\mathrm{SF}_{6}(g)\) and equating their enthalpy change to find a reaction that gives the energy of a single S-F bond. Subtracting the equations by considering \(SF_{4}(g)\) as a reactant: \[SF_6(g) - SF_4(g) \rightarrow 2F(g)\] Applying Hess's Law for enthalpy change: \[\Delta H_6 - \Delta H_4 = \Delta H_{S-F}\] Using provided enthalpies: \[(-1209) - (-775) = \Delta H_{S-F}\] Calculate the energy: \[\Delta H_{S-F} = -434 \mathrm{kJ/mol}\] Now to find the energy of a single S-F bond, we must divide the calculated value by 2 as there are 2 additional F moieties in the reaction: \[Energy_{S-F} = \frac{-434}{2} = -217 \mathrm{kJ/mol}\]

b. Comparing the calculated value with the given value in Table 8.5

We are unable to compare our calculated S-F bond energy with the one given in Table 8.5, as the table is not provided. If the table value were given, we would compare the calculated and the tabulated bond energy values and analyze the difference between the two. Further conclusions could be drawn based on how close our calculated value is to the value in the table.

c. Explaining the non-zero enthalpies of formation for S(g) and F(g)

The standard enthalpy of formation, \(\Delta H_{f}^{\circ}\), is defined as the change in enthalpy during the formation of 1 mole of a substance from its constituent elements in their standard states. For solid or liquid elements and diatomic elements in the gas state (like O2, H2), the standard enthalpies of formation are equal to zero because they are in their most stable and elemental form at standard conditions. However, for non-diatomic elements like S and F in their gaseous state, the enthalpy of formation is non-zero because they are not in their standard state. The standard state for sulfur (S) is solid, and for fluorine (F) is diatomic (\(F_2\)) gas. Thus, the given enthalpies of formation are for converting elemental sulfur and diatomic fluorine gas to form gaseous, atomic S and F, and therefore, these values are not equal to zero.

Key concepts

Hess's Law

Hess's Law is a core principle in chemistry. It states that the total enthalpy change for a chemical reaction is the same, no matter how many steps the reaction is divided into. This is because enthalpy is a state function, meaning it depends only on the initial and final states, not on how the reaction proceeds in between.
In the context of calculating bond energies, Hess's Law allows us to simplify complex calculations. Consider the formation of sulfur hexafluoride (\(\mathrm{SF}_{6}\)). We first form sulfur tetrafluoride (\(\mathrm{SF}_{4}\)). By evaluating the enthalpy changes in these processes, we calculate the energy of a single sulfur-fluorine (S-F) bond.

• Using the formation reactions for \(\mathrm{SF}_{6}\) and \(\mathrm{SF}_{4}\), subtract the enthalpies to obtain the binding energy.
• This calculated energy relates to the additional fluorines being bonded in \(\mathrm{SF}_{6}\) compared to \(\mathrm{SF}_{4}\).
• The principle ensures that these bonds' energy calculations remain accurate regardless of the different reaction pathways chosen.

Using Hess's Law simplifies evaluating reactions and deriving key pieces of data, such as enthalpy changes, without experimenting with every single reaction.

bond energy calculation

Bond energy calculation involves determining the energy required to break a bond in a molecule. This is crucial for understanding reactions and their energy requirements.
For the S-F bond energy, we calculate using enthalpy changes supplied by the formation of \(\mathrm{SF}_{4}\) and \(\mathrm{SF}_{6}\). The enthalpies given are: \(-775 \mathrm{kJ/mol}\) for \(\mathrm{SF}_{4}\) and \(-1209 \mathrm{kJ/mol}\) for \(\mathrm{SF}_{6}\).

• Subtract \(\Delta H_{\mathrm{SF}_4}\) from \(\Delta H_{\mathrm{SF}_6}\) to derive the total energy difference regarding added fluorine bonds.
• The result of \(-434 \mathrm{kJ/mol}\) represents the energy for adding two fluorine atoms.
• Since there are two more S-F bonds in \(\mathrm{SF}_{6}\), divide this by two to obtain the bond energy for one S-F bond, resulting in \(-217 \mathrm{kJ/mol}\).

This calculation assumes that the enthalpy change is evenly divided among the bonds, which often approximates well enough for theoretical predictions and comparisons.

chemical thermodynamics

Chemical thermodynamics is the study of heat and energy transformations in chemical reactions. In terms of the enthalpy of formation, it helps explain why certain elemental gases have non-zero values.
The enthalpy of formation, \(\Delta H_{f}^{\circ}\), is the enthalpy change when one mole of a compound is formed from its elements in their standard states. This is crucial for comprehending why gases like sulfur and fluorine in their monoatomic forms have non-zero formation enthalpies.

• The standard state for sulfur is its solid form. For fluorine, it's the diatomic \(F_2\) gas. Hence \(\Delta H_{f}^{\circ}\) values for \(\mathrm{S}(g)\) and \(\mathrm{F}(g)\) are non-zero because they represent the enthalpy change from their standard to gaseous states.
• In essence, these values account for converting sulfur from solid and fluorine from diatomic gas to their monoatomic gaseous forms.
• This concept underlines the importance of reference states in assigning standard enthalpy values and assessing chemical processes' energy demands.

Chemical thermodynamics, thus, bridges the comprehension of energy changes not only during reactions but also when considering phase and state transitions.