Problem 7 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The decomposition at constant volume of \(1 \mathrm{mol}\) of gaseous krypton difluoride \(\left(\mathrm{KrF}_{2}\right)\) to its elements at \(298 \mathrm{K}\) gives out \(59.4 \mathrm{kJ}\) of heat. Calculate the standard enthalpy change of formation, \(\Delta_{t} H_{298}^{\circ}\), of solid \(\mathrm{KrF}_{2}\). (The enthalpy change of sublimation of solid \(\mathrm{KrF}_{2}\) is \(+41 \mathrm{kJmol}^{-1}\). ) (Section \(13.3)\).

Step-by-Step Solution

Answer

The standard enthalpy change of formation for solid \(\text{KrF}_2\) is -100.4 kJ/mol.

Step 1: Understand the Problem

We are given a reaction where 1 mol of gaseous krypton difluoride (\(\text{KrF}_2\)) decomposes into its elements (solid krypton and fluorine gas) at 298 K, releasing 59.4 kJ of heat. We need to find the standard enthalpy change of formation for solid \(\text{KrF}_2\). The enthalpy change of sublimation from solid to gas is given as 41 kJ/mol.

Step 2: Write the Decomposition Reaction

The decomposition reaction of gaseous \(\text{KrF}_2\) is given as:\[ \text{KrF}_2(g) \rightarrow \text{Kr}(s) + \text{F}_2(g) \]This reaction releases 59.4 kJ of heat, meaning it is exothermic, with \( \Delta H = -59.4 \text{kJ/mol} \).

Step 3: Relate Enthalpy Changes

The enthalpy change for the formation of gaseous \(\text{KrF}_2\) from its elements can be determined using Hess’s law. Combine the given enthalpy changes:\[ \Delta H_{sublimation} = +41 \text{kJ/mol} \]Since we need the change in enthalpy for the formation of solid \(\text{KrF}_2\), we'll reverse the sublimation process and the decomposition reaction to find the desired enthalpy change.

Step 4: Calculate Enthalpy of Formation

To find \(\Delta_{f} H_{298}^{\circ}(\text{KrF}_2(s))\), use the following relationship derived from Hess’s law:\[ \text{formation}\: \Delta H = \Delta H_{decomposition} + \Delta H_{sublimation} \]Substitute the values:\[ \Delta H_{formation} = -59.4 \text{ kJ/mol} + (-41 \text{ kJ/mol}) = -100.4 \text{ kJ/mol} \]

Step 5: Conclude the Calculation

The standard enthalpy change of formation \(\Delta_{f} H_{298}^{\circ}\) for solid \(\text{KrF}_2\) from its elements is -100.4 kJ/mol.

Key Concepts

Krypton Difluoride

Krypton difluoride, abbreviated as KrF\(_2\), is a chemical compound composed of one krypton atom bonded to two fluorine atoms. This compound is unique because krypton is part of the noble gases, a group known for their lack of reactivity. Typically, noble gases do not form compounds easily due to their filled electron shell that makes them chemically inert.
Yet, by using extreme conditions such as high pressure and electrical discharge, scientists can produce krypton difluoride. Once formed, KrF\(_2\) is a powerful oxidizing agent and can be used in various chemical applications including fluorination reactions.
Understanding the behavior of compounds like krypton difluoride helps in exploring the chemistry of noble gases and extending our knowledge of chemical bonding beyond conventional reactions. When it thermally decomposes, it breaks down into its constituent elements, krypton (solid under many conditions) and fluorine gas.

Enthalpy of Sublimation

The enthalpy of sublimation refers to the energy required to convert a solid into a gas without passing through a liquid phase. It is an important thermodynamic concept because it represents the amount of heat needed to overcome intermolecular forces holding the solid together. For example, if you have solid krypton difluoride, to turn it into gaseous krypton difluoride, you will require an input of energy known as sublimation enthalpy.
In the problem at hand, the enthalpy of sublimation for KrF\(_2\) is given as +41 kJ/mol. The positive sign indicates that energy is absorbed from the surroundings to convert solid KrF\(_2\) into its gaseous form, fitting the general rule that phase transitions from solid to gas are endothermic.
Understanding sublimation and its associated enthalpy helps us quantify and predict changes during phase transitions, crucial for processes in both industrial applications and laboratory settings.

Hess's Law

Hess's law is a fundamental principle in thermodynamics that allows us to calculate the total enthalpy change of a chemical reaction by breaking it into a series of simpler steps. The core idea of Hess's law is that the total enthalpy change is the same regardless of the pathway taken, as long as the initial and final conditions are the same. This means that even if a chemical reaction occurs in several stages, the sum of the enthalpy changes of each stage equals the overall enthalpy change for the reaction.
In the given exercise, Hess's law is used to determine the standard enthalpy of formation of solid KrF\(_2\). By summing the enthalpy changes of sublimation and decomposition of KrF\(_2\), Hess's law enables a calculation of the overall energy change when KrF\(_2\) forms from its elements solid krypton and fluorine gas. Expressly, Hess's law helps us find that the enthalpy of formation of KrF\(_2\) is -100.4 kJ/mol by combining the energy released in decomposition with the energy required for sublimation.
Learning how to apply Hess’s law is crucial for solving complex chemical thermodynamics problems and provides insight into the energetic changes that govern chemical reactions.