Animated Solution for Chemistry - Chemical Thermodynamics: The average S—F bond energy in ${\text{kJ mol}}^{-1}$ of ${\text{SF}}_6$ is ............ (Rounded off to the nearest integer) [Given, the values of standard enthalpy of formation of ${\text{SF}}_6(g)$, $\text{S}(g)$ and $\text{F}(g)$ are $-1100$, $275$ and $80{\text{ kJ mol}}^{-1}$ respectively. ]

Have you ever wondered how much energy it takes to rip a molecule apart? In this problem, we are tasked with finding the average S-F bond energy in the sulfur hexafluoride (${\text{SF}}_6$) molecule. It sounds like a daunting task, but with the power of Hess's Law, it becomes an elegant puzzle of energy conservation.

Imagine the ${\text{SF}}_6$ molecule. It features a central sulfur atom symmetrically surrounded by six fluorine atoms in an octahedral geometry. To find the "average" bond energy, we need to determine the total energy required to break all six of these S-F bonds and then simply divide by six.

But how do we find that total energy? We can't just put a single molecule in a microscopic stretching machine. Instead, we use thermochemistry and standard enthalpies of formation.

Hess's Law tells us that the total enthalpy change of a reaction is independent of the pathway taken. We can envision the formation of ${\text{SF}}_6(g)$ from its standard state elements ($\text{S}(s)$ and ${\text{F}}_2(g)$) in two distinct ways:

Path 1: Direct formation from standard states. The energy change here is simply the standard enthalpy of formation of ${\text{SF}}_6(g)$.

Path 2: A two-step detour. First, we atomize the standard state elements into gaseous atoms: $\text{S}(g)$ and $6\text{F}(g)$. The energy required is the enthalpy of formation of $\text{S}(g)$ plus six times the enthalpy of formation of $\text{F}(g)$. Second, we allow these gaseous atoms to snap together to form the six S-F bonds. Since forming bonds releases energy, this step has an enthalpy change of $-6\times E_{\text{S-F}}$.

Equating the two paths gives us our master equation:

Now, it's just a matter of plugging in the numbers provided in the problem:

Let's simplify the atomization energy:

Rearranging to solve for the bond energy term:

Finally, we divide by six to find the energy of a single bond:

Rounding off to the nearest integer, we get $309{\text{ kJ mol}}^{-1}$.

This beautiful application of Hess's Law shows how macroscopic thermodynamic data can give us profound insights into the microscopic strength of chemical bonds!