Question

The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to \(\mathrm{H}^{+}\) ions; that is, \(\Delta H_{\mathrm{f}}^{\mathrm{o}}\left[\mathrm{H}^{+}(a q)\right]=0 .\) (a) For the following reaction \(\begin{aligned} \mathrm{HCl}(g) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q) & \Delta H^{\circ}=-74.9 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for the \(\mathrm{Cl}^{-}\) ions. \((\mathrm{b})\) Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{OH}^{-}\) ions is \(-229.6 \mathrm{~kJ} / \mathrm{mol},\) calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as \(\mathrm{HCl}\) ) is titrated by \(1 \mathrm{~mole}\) of a strong base \((\) such as \(\mathrm{KOH})\) at \(25^{\circ} \mathrm{C}\).

Short answer

\(-167.2 \text{ kJ/mol}\) for \(\mathrm{Cl}^{-}\); neutralization: \(-56.2 \text{ kJ/mol}\).

Step-by-step solution

Write the Given Information

We know that \( \Delta H^{\circ} = -74.9 \text{ kJ/mol} \) for the reaction: \( \mathrm{HCl}(g) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{Cl}^{-}(aq) \). Also, \( \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{H}^{+}(aq) \right]=0 \).

Use Enthalpy of Formation Equation

The enthalpy change for a reaction is given by the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. For the reaction: \( \mathrm{HCl}(g) \rightarrow \mathrm{H}^{+}(aq) + \mathrm{Cl}^{-}(aq) \), we have:\[ \Delta H^{\circ} = \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{H}^{+}(aq) \right] + \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{Cl}^{-}(aq) \right] - \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{HCl}(g) \right] \]

Solve for Enthalpy of Formation of Cl⁻

Rearrange the equation from Step 2 to solve for \( \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{Cl}^{-}(aq) \right] \):\[ \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{Cl}^{-}(aq) \right] = \Delta H^{\circ} + \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{HCl}(g) \right] \] We need \( \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{HCl}(g) \right] \), which is commonly \(-92.3 \text{ kJ/mol} \). Substitute the known values:\[ \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{Cl}^{-}(aq) \right] = -74.9 \text{ kJ/mol} + (-92.3 \text{ kJ/mol}) \]\[ \Delta H_{\mathrm{f}}^{\circ} \left[ \mathrm{Cl}^{-}(aq) \right] = -167.2 \text{ kJ/mol} \]

Determine Enthalpy of Neutralization Reaction

The enthalpy of neutralization for the reaction between a strong acid and a strong base is calculated using: \( \mathrm{H}^{+}(aq) + \mathrm{OH}^{-}(aq) \rightarrow \mathrm{H}_2\mathrm{O}(l) \). The known value of this reaction is typically about \(-55.8 \text{ kJ/mol} \). To verify, using formation enthalpies: \[ \Delta H_{\mathrm{f}}^{\circ}[\mathrm{H}_2\mathrm{O}(l)] - \Delta H_{\mathrm{f}}^{\circ}[\mathrm{H}^{+}(aq)] - \Delta H_{\mathrm{f}}^{\circ}[\mathrm{OH}^{-}(aq)] = -285.8 \text{ kJ/mol} - 0 \text{ kJ/mol} - (-229.6 \text{ kJ/mol}) \]\[ = -285.8 + 229.6 \text{ kJ/mol} = -56.2 \text{ kJ/mol} \] \( \Delta H_{\text{neutralization}} = -56.2 \text{ kJ/mol} \).

Key concepts

Standard Enthalpy Change

Standard enthalpy change is a key concept in thermodynamics, often denoted by \( \Delta H^\circ \). It represents the heat change that occurs in a reaction at standard conditions, which are 1 bar pressure and a specified temperature, commonly \( 25^\circ \text{C} \). This measure helps standardize data so chemists and engineers can easily compare different reactions.
To fully understand this concept, remember:

• The sign of \( \Delta H^\circ \) is important. If it's negative, the reaction releases heat and is exothermic. If positive, it absorbs heat and is endothermic.
• Standard conditions ensure consistency, allowing for meaningful comparisons.

In essence, standard enthalpy change is like a common language for reactions, helping scientists accurately analyze and predict reaction behaviors under controlled circumstances.

Neutralization Reaction

Neutralization reactions are special types of chemical reactions where an acid and a base react to form water and a salt. During this process, the \( \text{H}^+ \) ions from the acid combine with the \( \text{OH}^- \) ions from the base to produce water \((\text{H}_2\text{O})\).
This type of reaction is incredibly significant in both practical and experimental chemistry. Here's why:

• Neutralization is foundational in titration, which is a means of determining the concentration of an unknown acid or base.
• These reactions are typically exothermic, releasing heat. This heat release is the enthalpy of neutralization.
• Understanding this process helps explain everyday occurrences such as antacids neutralizing stomach acid.

By grasping the fundamentals of neutralization reactions, we gain insights into wider chemical processes and applications, both in the lab and in real-world scenarios.

Enthalpy Calculations

Enthalpy calculations are pivotal in determining the energy changes during chemical reactions. These calculations help us understand how energy is absorbed or released. The basic equation used in these calculations is:

\[ \Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \]
This formula is a powerful tool because:

• It allows calculation of reaction enthalpies using known standard enthalpies of formation.
• Gives insights into the energy efficiency of reactions, which has implications for industrial processes.
• Provides a clear pathway to determining whether a reaction is feasible based on energy change.

Enthalpy calculations are not just about balancing equations; they are about unlocking the energy secrets of reactions, turning abstract chemistry into usable science.