Question

Given the following information, $$\begin{array}{rlrr}\mathrm{H}^{+}(g)+\mathrm{H}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(g) & \Delta H=-720 \mathrm{~kJ} \\\ \mathrm{H}^{+}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q) & & \Delta H=-1090 \mathrm{~kJ} \\ \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H= & 40.7 \mathrm{~kJ} \end{array}$$ calculate the heat of solution of the hydronium ion: $$\mathrm{H}_{3} \mathrm{O}^{+}(g) \stackrel{\mathrm{H}_{3} \mathrm{O}}{\longrightarrow} \mathrm{H}_{3} \mathrm{O}^{+}(a q)$$

Short answer

\( \Delta H = -329.3 \text{ kJ} \)

Step-by-step solution

Write the Given Reactions

Write down the reactions and their respective enthalpy changes: 1. \(\text{H}^{+}(g) + \text{H}_2 \text{O}(g) \rightarrow \text{H}_3 \text{O}^{+}(g), \Delta H = -720 \text{ kJ}\)2. \(\text{H}^{+}(g) + \text{H}_2 \text{O}(l) \rightarrow \text{H}_3 \text{O}^{+}(aq), \Delta H = -1090 \text{ kJ}\)3. \(\text{H}_2 \text{O}(l) \rightarrow \text{H}_2 \text{O}(g), \Delta H = 40.7 \text{ kJ}\)

Reverse the First Reaction

Reverse the first reaction to get \(\text{H}_3 \text{O}^{+}(g) \rightarrow \text{H}^{+}(g) + \text{H}_2 \text{O}(g), \Delta H = 720 \text{ kJ}\)

Combine Reactions

Combine the reversed first reaction, the second reaction, and the third reaction to form the overall reaction. The overall reaction is: \(\text{H}_3 \text{O}^{+}(g) \rightarrow \text{H}_3 \text{O}^{+}(aq)\).\Sum the enthalpy changes: \(\begin{aligned}\text{Overall} & \text{ } \Delta H= 720 \text{ kJ} + ( - 1090 \text{ kJ}) + 40.7 \text{ kJ} \Delta H= -329.3 \text{ kJ} \end{aligned}\)

Key concepts

Enthalpy Change

Enthalpy change (∆H) refers to the heat change that occurs at constant pressure during a chemical reaction. It signifies the difference between the enthalpy of the products and the enthalpy of the reactants. In chemical reactions, enthalpy changes are essential because:

• They help predict whether reactions are exothermic (release heat) or endothermic (absorb heat).
• They aid in understanding reaction spontaneity and energy requirements.

In the exercise, we see enthalpy changes associated with different transitions involving hydronium ions and water in various states. Such calculations are helpful in predicting the energy behavior of reactions involving dissolving ionic compounds in water.

Thermodynamics

Thermodynamics is the branch of science that deals with heat, work, and the associated energy transformations that occur in chemical processes. It is underpinned by four main laws, the first two being most relevant to this context:

• The First Law of Thermodynamics: Energy cannot be created or destroyed, only converted from one form to another. This principle forms the basis of enthalpy calculations.
• The Second Law of Thermodynamics: In any spontaneous process, the total entropy of a system and its surroundings always increases.

In the given exercise, thermodynamic principles help us determine the heat of solution for hydronium ions by combining multiple reactions and their respective enthalpy changes. This ultimately respects energy conservation and intrinsic reaction directions dictated by entropy.

Chemical Reactions

Chemical reactions involve breaking and forming chemical bonds, leading to the transformation of reactants into products. During this process, the energy balance is crucial, affecting whether a reaction absorbs or releases heat. Reactions can be categorized as:

• Exothermic Reactions: These reactions release heat to the surroundings, resulting in a negative ∆H.
• Endothermic Reactions: These reactions absorb heat from the surroundings, leading to a positive ∆H.

In the exercise, the enthalpy changes for different reactions (formation and transformation of hydronium ions) illustrate these principles. For accurate calculations, it’s essential to correctly reverse reactions (as done in Step 2) and sum their enthalpy changes, providing the overall heat change for the solution process.