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

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

Step by step solution

01

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. \(\text{H}_2 \text{O}(l) \rightarrow \text{H}_2 \text{O}(g), \Delta H = 40.7 \text{ kJ}\)
02

Reverse the First Reaction

Reverse the first reaction to get \(\text{H}_3 \text{O}^{+}(g) \rightarrow \text{H}^{+}(g) + \text{H}_2 \text{O}(g), \Delta H = 720 \text{ kJ}\)
03

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: \(\begin{aligned}\text{Overall} & \text{ } \Delta H= 720 \text{ kJ} + ( - 1090 \text{ kJ}) + 40.7 \text{ kJ} \Delta H= -329.3 \text{ kJ} \end{aligned}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The electronic transition in Na from \(3 p^{1}\) to \(3 s^{1}\) gives rise to a bright yellow-orange emission at \(589.2 \mathrm{nm}\). What is the energy of this transition?

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A common method for producing a gaseous hydride is to treat a salt containing the anion of the volatile hydride with a strong acid. (a) Write an equation for each of the following examples: ( 1 ) the production of HF from \(\mathrm{CaF}_{2} ;\) (2) the production of \(\mathrm{HCl}\) from \(\mathrm{NaCl} ;\) (3) the production of \(\mathrm{H}_{2} \mathrm{~S}\) from FeS. (b) In some cases, even as weak an acid as water can be used for this preparation if the anion of the salt has a sufficiently strong attraction for protons. An example is the production of \(\mathrm{PH}_{3}\) from \(\mathrm{Ca}_{3} \mathrm{P}_{2}\) and water. Write the equation for this reaction. (c) By analogy, predict the products and write the equation for the reaction of \(\mathrm{Al}_{4} \mathrm{C}_{3}\) with water.

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