Problem 32 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Estimate the average S-F bond energy in \(\mathrm{SF}_{6}\). The values of standard enthalpy of formation of \(\mathrm{SF}_{6}(\mathrm{~g}), \mathrm{S}(\mathrm{g})\) and \(\mathrm{F}(\mathrm{g})\) are \(:-1100,275\) and \(80 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively.

Step-by-Step Solution

Answer

The average S-F bond energy is approximately 248 kJ/mol.

Step 1: Identify the Reaction

The formation reaction for \(\mathrm{SF}_{6}\) from its elements in their gaseous forms can be written as: \[\mathrm{S}(\mathrm{g}) + 3\mathrm{F}_2(\mathrm{g}) \rightarrow \mathrm{SF}_{6}(\mathrm{g})\] This reaction indicates the formation of one mole of \(\mathrm{SF}_{6}\) from sulfur and fluorine.

Step 2: Utilize Enthalpy of Formation Formula

The standard enthalpy of formation \(\Delta H_f^\circ\) of a compound is the enthalpy change when one mole of the compound forms from its elements in their standard states. This can be expressed as: \[\Delta H_f^\circ(\mathrm{SF}_{6}) = \Delta H_f^\circ(\mathrm{SF}_{6}) - \Delta H_f^\circ(\mathrm{S}) - 3 \times \Delta H_f^\circ(\mathrm{F}_{2})\] Since \(\mathrm{F}_2\) as a diatomic molecule has \(\Delta H_f^\circ = 0\), the equation simplifies to: \[\Delta H_f^\circ(\mathrm{SF}_{6}) = -1100 \, \text{kJ}/\text{mol}\] and \[\Delta H = \Delta H_f^\circ(\mathrm{S}) + 3 \times \Delta H_f^\circ(\mathrm{F}) = 275 + 3 \times 80 \, \text{kJ}/\text{mol}\]

Key Concepts

Enthalpy of Formation

The standard enthalpy of formation (\(\Delta H_f^\circ\)) represents the heat absorbed or evolved during the formation of one mole of a compound from its pure elements under standard conditions (usually1 bar pressure and25°C temperature).For our exercise, this is thechange in enthalpy when\(\mathrm{SF}_{6}(\mathrm{g})\)forms from\(\mathrm{S}(\mathrm{g})\)and\(\mathrm{F}_{2}(\mathrm{g})\).Key points to remember:

• \(\Delta H_f^\circ\)is essentially a measure of energy content. Lower values indicate a more stable compound.
• Elements in their standard state, like\(\mathrm{F}_{2}\)(grey area in this exercise) have an enthalpy of formation value of zero.
• Calculations involving\(\Delta H_f^\circ\)help in understanding the energetics of chemical reactions.

By using given values and rearranging our equation, we find that the difference in these enthalpy values leads us to the energy content of the bonds formed in the product.

Chemical Bond

Chemical bonds are the glue that hold atoms together within a molecule. They form due to the interactions between electrons of atoms. In the context of our exercise, we focus on the sulfur-fluorine (S-F) bond. Understanding bond energy is crucial because it informs us how much energy is needed to break bonds and how much is released when they form.When sulfur and fluorine combine:

• Their atomic orbitals interact to form strongS-F bonds.
• The strength of these bonds, indicated by bond energy, reflects how stable\(\mathrm{SF}_{6}\) is as a molecule.

Bond energy plays a significant role in chemical reactions:

• Higher bond energy implies a more stable bond.
• If more energy is released during bond formation than absorbed in bond-breaking, the reaction is exothermic.

Moreover, the structure of\(\mathrm{SF}_{6}\)is a key factor for its robust S-F bonds, contributing to the overall stability and low reactivity of the compound.

Thermochemistry

Thermochemistry is the study of heat change in chemical reactions. It helps understand whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). Exploring the formation of\(\mathrm{SF}_{6}\)provides insight into these thermochemical principles.Every chemical reaction involves energy transitions:

• It is governed by thefirst law of thermodynamics, which states energy can't be created or destroyed.
• Instead, energy is transferred in the form of heat and work.

In our exercise, we identify that the formation of\(\mathrm{SF}_{6}\)is exothermic:

• The enthalpy change (\(\Delta H\)) as calculated indicates that more energy is released during the bond-making process than is consumed during the bond-breaking process.
• This overall decrease in enthalpy suggests the stability of\(\mathrm{SF}_{6}\).

Thermochemistry not only predicts the heat change but also provides a glimpse into the feasibility and spontaneity of chemical reactions under specific conditions.