Question

The species \(\mathrm{H}_{3}^{+}\) is the simplest polyatomic ion. The geometry of the ion is that of an equilateral triangle. (a) Draw three resonance structures to represent the ion. (b) Given the following information $$ \begin{aligned} 2 \mathrm{H}+\mathrm{H}^{+} \longrightarrow & \mathrm{H}_{3}^{+} & & \Delta H^{\circ}=-849 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \longrightarrow & 2 \mathrm{H} & & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}^{+}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{3}^{+} $$

Short answer

\( \Delta H^{\circ} = -1285.4 \mathrm{~kJ/mol} \) for \( \mathrm{H}^+ + \mathrm{H}_2 \rightarrow \mathrm{H}_3^+ \).

Step-by-step solution

Understand Resonance Structures

Since \( \mathrm{H}_3^+ \) has an equilateral triangle geometry, we can depict each hydrogen bond being shared among all the hydrogen atoms. However, for resonance structures, we show different positions for bonds or charges among the same atoms.

Draw Resonance Structures

The three resonance structures for \( \mathrm{H}_3^+ \) can each have a positive charge on a different hydrogen atom, with covalent bonds between the remaining atoms:1. \[ \begin{array}{c}\begin{array}{c} \hline \begin{align*} & \overset{+}{\text{H}} - \text{H} - \text{H} \\end{align*}\hline\end{array}\end{array} \]2. \[ \begin{array}{c}\begin{array}{c} \hline \begin{align*} & \text{H} - \overset{+}{\text{H}} - \text{H} \\end{align*}\hline\end{array}\end{array} \]3. \[ \begin{array}{c}\begin{array}{c} \hline \begin{align*} & \text{H} - \text{H} - \overset{+}{\text{H}} \\end{align*}\hline\end{array}\end{array} \]

Understand Reaction Enthalpies

We are given two reactions with their enthalpy changes and asked to find the enthalpy change for a third. Recall that chemical equations can be manipulated in the same manner as algebraic equations.

Reverse the Given Reactions

For the goal reaction, \( \mathrm{H}^+ + \mathrm{H}_2 \rightarrow \mathrm{H}_3^+ \), we'll reverse one of the given reactions:1. Reverse \( \mathrm{H}_2 \rightarrow 2\mathrm{H} \): - \( 2\mathrm{H} \rightarrow \mathrm{H}_2 \) - \( \Delta H^{\circ} = -436.4 \mathrm{~kJ/mol} \)2. Use the other: \( 2\mathrm{H} + \mathrm{H}^+ \rightarrow \mathrm{H}_3^+ \) has \( \Delta H^{\circ} = -849 \mathrm{~kJ/mol} \)

Combine the Equations

Add the reversed \( 2\mathrm{H} \rightarrow \mathrm{H}_2 \) reaction to \( 2\mathrm{H} + \mathrm{H}^+ \rightarrow \mathrm{H}_3^+ \):- Combine: \( 2\mathrm{H} + \mathrm{H}^+ + 2\mathrm{H} \rightarrow \mathrm{H}_2 + \mathrm{H}_3^+ \) gives:- Simplify to \( \mathrm{H}^+ + \mathrm{H}_2 \rightarrow \mathrm{H}_3^+ \)Since the second \( 2\mathrm{H} \) cancels out, the final enthalpy is:\[ \Delta H^{\circ} = -849 + (-436.4) = -1285.4 \mathrm{~kJ/mol} \]

Calculate the Final Enthalpy Change

By combining the manipulated equations, we find the total enthalpy change for \( \mathrm{H}^+ + \mathrm{H}_2 \rightarrow \mathrm{H}_3^+ \) is:\[ \Delta H^{\circ} = -1285.4 \mathrm{~kJ/mol} \]

Key concepts

Polyatomic Ions

Polyatomic ions are ions that consist of more than one atom. Unlike monatomic ions that have a single atom with a positive or negative charge, polyatomic ions are composed of a group of covalently bonded atoms that carry an overall charge. The charge on a polyatomic ion arises because the number of electrons does not equal the number of protons in the entire group. This can result from either gaining or losing electrons. A common example is the sulfate ion (\( ext{SO}_4^{2-}\)), which consists of one sulfur atom and four oxygen atoms, carrying a negative charge overall.
In our exercise, we examined the simplest polyatomic ion, \( ext{H}_3^+\), which has three hydrogen atoms sharing one positive charge. Despite its simplicity, this ion is an excellent example to understand the sharing and distribution of charge among atoms in polyatomic ions. An important characteristic of polyatomic ions is their ability to exhibit resonance. This means the actual structure is a hybrid of multiple possible structures, which helps in spreading out the charge more evenly across the atoms.

Enthalpy Change

Enthalpy change (\(\Delta H^{\circ}\)) is the heat change that occurs during a chemical reaction at constant pressure. It helps us understand the energy changes associated with chemical processes. This value can either be positive, indicating that the system absorbs heat (endothermic), or negative, indicating that the system releases heat (exothermic).
Chemical reactions involve breaking old bonds and forming new ones. The energy required to break bonds is often different from the energy released during the formation of new bonds, leading to an overall enthalpy change for the reaction. In our exercise, we need to calculate the enthalpy change for the conversion of hydrogen ion (\( ext{H}^+\)) and diatomic hydrogen (\( ext{H}_2\)) to the \( ext{H}_3^+\) ion.
We were given two initial reactions with known enthalpy changes. By reversing and combining these reactions, we can derive the desired reaction. The final calculated enthalpy change of \(-1285.4 \, \text{kJ/mol}\) indicates that the formation of \( ext{H}_3^+\) from \( ext{H}^+\) and \( ext{H}_2\) is highly exothermic.

Equilateral Triangle Geometry

Geometry in chemistry often refers to the spatial arrangement of atoms in a molecule. An equilateral triangle geometry implies that the positions of the atoms form a triangle where all sides and angles are equal. This is particularly interesting when discussing molecular symmetry and resonance in polyatomic ions.
The \( ext{H}_3^+\) ion, as mentioned in the exercise, forms an equilateral triangle. This configuration allows each hydrogen atom to be equidistant from the others, offering a balanced distribution of electrons and charges. Such symmetry often results in the stabilization of the ion.
Being aware of the molecular geometry is also crucial when sketching resonance structures. Because all three hydrogen atoms are identical and equally bonded, it forms an excellent illustration of how resonance can shift the location of bonds or charge among equivalent positions without changing the overall energy of the system. This understanding aids students in visualizing how electrons can delocalize, which contributes to the stability and reactivity of the ion.