Question

At \(25^{\circ} \mathrm{C}\), the standard enthalpy of formation of \(\mathrm{HF}(a q)\) is \(-320.1 \mathrm{~kJ} / \mathrm{mol} ;\) of \(\mathrm{OH}^{-}(a q),\) it is \(-229.6 \mathrm{~kJ} / \mathrm{mol} ;\) of \(\mathrm{F}^{-}(a q)\) it is \(-329.1 \mathrm{~kJ} / \mathrm{mol} ;\) and of \(\mathrm{H}_{2} \mathrm{O}(l),\) it is \(-285.8 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the standard enthalpy of neutralization of \(\mathrm{HF}(a q)\) \(\mathrm{HF}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{F}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) Using the value of \(-56.2 \mathrm{~kJ}\) as the standard enthalpy change for the reaction \(\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) calculate the standard enthalpy change for the reaction \(\mathrm{HF}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\)

Short answer

(a) -65.2 kJ/mol; (b) -9.0 kJ/mol

Step-by-step solution

Write the Reaction Equation

The given neutralization reaction is: \( \mathrm{HF}(\text{aq}) + \mathrm{OH}^- (\text{aq}) \rightarrow \mathrm{F}^- (\text{aq}) + \mathrm{H}_2\mathrm{O}(l) \).

Determine the Standard Enthalpy Change

The enthalpy change \( \Delta H \) for a reaction is calculated as the difference between the sum of the enthalpies of formation of products and the sum of the enthalpies of formation of reactants. The formula is \( \Delta H^\circ = \sum H^\circ_f(\text{products}) - \sum H^\circ_f(\text{reactants}) \).

Apply Values to Reaction (a)

For the reaction provided: - Products: \( \mathrm{F}^-: -329.1 \, \mathrm{kJ/mol}, \mathrm{H}_2\mathrm{O}: -285.8 \, \mathrm{kJ/mol} \)- Reactants: \( \mathrm{HF}: -320.1 \, \mathrm{kJ/mol}, \mathrm{OH}^-: -229.6 \, \mathrm{kJ/mol} \)- Calculate: \[ \Delta H^\circ = [(-329.1) + (-285.8)] - [(-320.1) + (-229.6)] \]

Calculate the Enthalpy Change for Reaction (a)

Perform the calculations: \[ \Delta H^\circ = (-329.1 - 285.8) - (-320.1 - 229.6) \]\[ \Delta H^\circ = (-614.9) - (-549.7) \]\[ \Delta H^\circ = -65.2 \, \mathrm{kJ/mol} \]

Use Given Enthalpy of Hydrogen Ion Reaction

The given enthalpy change for \( \mathrm{H}^+(\text{aq}) + \mathrm{OH}^-(\text{aq}) \rightarrow \mathrm{H}_2\mathrm{O}(l) \) is \(-56.2 \, \mathrm{kJ} \) per mole.

Apply Hess's Law for Reaction (b)

Using Hess's Law: \( \mathrm{HF}(\text{aq}) \rightarrow \mathrm{H}^+(\text{aq}) + \mathrm{F}^-(\text{aq}) \) can be written as: \( \Delta H_{\mathrm{final}} = \Delta H_{\mathrm{desires}} - \Delta H_{\mathrm{given}} \).Substitute known values: - Desires (part a): \(-65.2 \, \mathrm{kJ/mol} \)- Given: \(-56.2 \, \mathrm{kJ/mol} \)\[ \Delta H_{\mathrm{HF \rightarrow H^+ + F^-}} = -65.2 + 56.2 = -9.0 \, \mathrm{kJ/mol} \]Thus, the reaction enthalpy for \( \mathrm{HF}(\text{aq}) \rightarrow \mathrm{H}^+(\text{aq}) + \mathrm{F}^-(\text{aq}) \) is \(-9.0 \, \mathrm{kJ/mol} \).

Key concepts

Standard Enthalpy of Formation

Standard enthalpy of formation refers to the change in enthalpy when one mole of a compound is formed from its elements in their standard states under standard conditions, namely 25°C and 1 atm pressure. This concept is pivotal in chemistry because it provides a reference point to calculate overall enthalpy changes in reactions.
To clarify, let’s consider a simple reaction where a compound like hydrogen fluoride (HF) is formed from its elements hydrogen and fluorine. Here, the standard enthalpy of formation would measure the heat change that occurs under standard conditions. For HF, this value is given as \(-320.1 \text{ kJ/mol}\).

• HF is a compound formed from H₂ and F₂. The standard enthalpy of formation is the energy change for forming HF from these elements.
• All standard enthalpy values are given at a reference state, so consistency is maintained during calculations.

Knowing the standard enthalpy of formation allows chemists to predict the heat exchange during chemical reactions and perform calculations critical to understanding the energy dynamics of reactions.

Hess's Law

Hess's Law is an essential principle in thermodynamics, stating that the total enthalpy change for a chemical reaction is the same regardless of the pathway taken, provided the initial and final conditions are the same. This law essentially underlines that enthalpy is a state function and, therefore, independent of the route taken.
Using Hess's Law allows for the calculation of enthalpy changes even when they cannot be directly measured. It works by breaking complex reactions into a series of steps for which the enthalpy changes are known.
For example, in the exercise, the enthalpy change for the reaction \( \mathrm{HF}(\text{aq}) \rightarrow \mathrm{H}^+(\text{aq}) + \mathrm{F}^-(\text{aq}) \) is calculated using known reactions. Part of it involves the standard enthalpy of formation values, and another already known process, namely the neutralization of hydrogen ions by hydroxide ions, which has a given enthalpy change value of \(-56.2 \text{ kJ/mol}\).

• Hess's Law makes use of principles from energy conservation—energy input/output over a cycle cancels out.
• This is extremely useful for experimentally difficult reactions where direct enthalpy measurement isn’t possible.

By employing Hess's Law correctly, chemists can accurately predict enthalpy changes, assisting in controlling reactions and processes in laboratory and industrial contexts.

Enthalpy Change Calculation

Enthalpy change calculation is a fundamental process in understanding the energy aspects of chemical reactions. It aids in determining whether reactions are endothermic (heat absorbing) or exothermic (heat releasing).
To calculate the standard enthalpy change of a reaction, one uses the formula: \[\Delta H^\circ = \sum H^\circ_f(\text{products}) - \sum H^\circ_f(\text{reactants})\].
This equation shows the difference between the sum of standard enthalpies of formation for the products and reactants. Here are steps highlighted from the exercise:

• List given standard enthalpy of formation values for each component (both reactants and products).
• Plug these values into the enthalpy change formula, ensuring correct signs for absorption or release of energy.
• The final calculation gives the net result of energy exchanged during a chemical reaction.

This exercise used the enthalpy of formation values of HF, OH^-, F^-, and H₂O to determine the enthalpy change in a neutralization reaction. This allows students to appreciate the use of standard reference values in performing real-world thermodynamic calculations.