Question

The standard heats of formation of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\), \(\mathrm{H}_{2} \mathrm{Se}(\mathrm{g})\), and \(\mathrm{H}_{2} \mathrm{Te}(g)\) are \(-241.8,-20.17,+29.7\), and \(+99.6 \mathrm{~kJ} / \mathrm{mol}\), respectively. The enthalpies necessary to convert the elements in their standard states to one mole of gaseous atoms are \(248,277,227\), and \(197 \mathrm{~kJ} / \mathrm{mol}\) of atoms for \(\mathrm{O}, \mathrm{S}, \mathrm{Se}\), and Te, respectively. The enthalpy for dissociation of \(\mathrm{H}_{2}\) is \(436 \mathrm{~kJ} / \mathrm{mol}\). Calculate the average \(\mathrm{H}-\mathrm{O}, \mathrm{H}-\mathrm{S}, \mathrm{H}-\mathrm{Se}\), and \(\mathrm{H}-\mathrm{Te}\) bond enthalpies, and comment on their trend.

Short answer

The average bond enthalpies are: \(\Delta H_{\text{H-O}} = +221.1 \text{ kJ/mol}\), \(\Delta H_{\text{H-S}} = +346.42 \text{ kJ/mol}\), \(\Delta H_{\text{H-Se}} = +346.35 \text{ kJ/mol}\), and \(\Delta H_{\text{H-Te}} = +366.3 \text{ kJ/mol}\). The trend shows an increase in bond enthalpy as we move down the periodic table from oxygen to tellurium, which can be attributed to the increase in atomic size leading to weaker bonds as the distance between the bonded atoms increases.

Step-by-step solution

Identify Relevant Formulas

To find the bond enthalpy for each molecule, we can use the formula: \( \Delta H_{\text{Bond}} = \frac{1}{2} (\Delta H_{\text{formation}} + \Delta H_{\text{element}} + \Delta H_{\text{H2 dissociation}}) \)

Find the H-O Bond Enthalpy

Using the formula, we can find the H-O bond enthalpy: \( \Delta H_{\text{H-O}} = \frac{1}{2} (- 241.8 \text{ kJ/mol} + 248 \text{ kJ/mol} + 436 \text{ kJ/mol}) \) \( \Delta H_{\text{H-O}} = \frac{1}{2} (442.2 \text{ kJ/mol}) \) \( \Delta H_{\text{H-O}} = +221.1 \text{ kJ/mol} \)

Find the H-S Bond Enthalpy

Using the formula, we can find the H-S bond enthalpy: \( \Delta H_{\text{H-S}} = \frac{1}{2} (- 20.17 \text{ kJ/mol} + 277 \text{ kJ/mol} + 436 \text{ kJ/mol}) \) \( \Delta H_{\text{H-S}} = \frac{1}{2} (692.83 \text{ kJ/mol}) \) \( \Delta H_{\text{H-S}} = +346.42 \text{ kJ/mol} \)

Find the H-Se Bond Enthalpy

Using the formula, we can find the H-Se bond enthalpy: \( \Delta H_{\text{H-Se}} = \frac{1}{2} (+ 29.7 \text{ kJ/mol} + 227 \text{ kJ/mol} + 436 \text{ kJ/mol}) \) \( \Delta H_{\text{H-Se}} = \frac{1}{2} (692.7 \text{ kJ/mol}) \) \( \Delta H_{\text{H-Se}} = +346.35 \text{ kJ/mol} \)

Find the H-Te Bond Enthalpy

Using the formula, we can find the H-Te bond enthalpy: \( \Delta H_{\text{H-Te}} = \frac{1}{2} (+ 99.6 \text{ kJ/mol} + 197 \text{ kJ/mol} + 436 \text{ kJ/mol}) \) \( \Delta H_{\text{H-Te}} = \frac{1}{2} (732.6 \text{ kJ/mol}) \) \( \Delta H_{\text{H-Te}} = +366.3 \text{ kJ/mol} \)

Evaluate the Trend

Now, we compare the bond enthalpies and comment on the trend: - \(\Delta H_{\text{H-O}} = +221.1 \text{ kJ/mol}\) - \(\Delta H_{\text{H-S}} = +346.42 \text{ kJ/mol}\) - \(\Delta H_{\text{H-Se}} = +346.35 \text{ kJ/mol}\) - \(\Delta H_{\text{H-Te}} = +366.3 \text{ kJ/mol}\) The trend shows that the H-O bond has the lowest bond enthalpy, and the bond enthalpy increases as we move down the periodic table from oxygen to tellurium. This can be explained by the increase in atomic size, which leads to weaker bonds as the distance between the bonded atoms increases. As a result, more energy is required to break the bonds as we move down the group.

Key concepts

Heats of Formation

In chemistry, heats of formation are a crucial concept for understanding how much energy is involved in forming a compound from its elements in their most stable forms. When we talk about the standard heat of formation, we refer to the energy change that occurs when one mole of a compound forms from its elements at standard conditions \(1 \,\text{atm}, 298.15 \,\text{K}\).

For example, the standard heat of formation of water vapor \(-241.8 \,\text{kJ/mol}\) means energy is released when hydrogen and oxygen combine to form water. Negative values imply the formation releases energy, whereas positive values indicate energy is absorbed.

These values are vital for calculating reaction energies, predicting reaction spontaneity, and comparing the stability of compounds. They're a fundamental building block for engaging with other thermodynamic concepts like bond enthalpies. Knowledge of heats of formation allows us to understand how different substances interact at a basic molecular level.

Enthalpy of Dissociation

The enthalpy of dissociation is related to the energy required to break a bond in a molecule to form individual atoms. It is crucial for understanding chemical kinetics and reaction mechanisms, as breaking bonds is often the initial step of chemical reactions.

Consider the hydrogen molecule \(H_2\). It takes \(436 \,\text{kJ/mol}\) to dissociate its bond, which means strong forces hold the molecule together. This is important in reactions involving hydrogen gas, where initial bond breaking can significantly impact the rate and energy profile of the reaction.

Calculating bond enthalpies using dissociation enthalpies provides insights into how molecules behave under various conditions. It helps chemists predict how readily a molecule may react and what conditions may be necessary to initiate a reaction. In broader terms, understanding dissociation entails understanding the stability and reactivity of various chemical bonds.

Periodic Trend

Periodic trends describe patterns within the periodic table, providing insights into atomic and molecular behaviors. In the context of bond enthalpies, a key trend is how bond strength changes as you move down a group.

For instance, in the case of \(H-O\), \(H-S\), \(H-Se\), and \(H-Te\) bonds, there is a clear trend: bond enthalpies increase from oxygen to tellurium. This occurs because atomic size increases down the group, expanding the distance between bonded atoms which requires less energy to break apart. However, in this exercise, the calculated bond enthalpies show an increase as you go down the group, an anomaly associated with other factors like molecular environments and specific bond structures in real-world conditions.

Understanding such trends helps predict properties such as melting and boiling points, reactivity, and bond strength across different elements. These periodic trends form the foundation of modern chemistry, underpinning the behavior of elements and their compounds.

Chemical Bonding

Chemical bonding is the force that holds atoms together to form molecules and compounds. The type and strength of these bonds determine a substance's properties, such as its melting point, boiling point, and reactivity. There are primarily three types of bonds: ionic, covalent, and metallic, with covalent bonds playing a key role in molecular compounds discussed here.

In the exercise, covalent bonds are critical. Hydrogen bonds with other elements like oxygen, sulfur, selenium, and tellurium. Covalent bonds involve the sharing of electrons and the strength of these bonds can vary based on the elements involved.

Factors like atomic size and electronegativity differences influence bond strength. Smaller atoms with higher electronegativity typically form stronger bonds. Understanding bond nature helps us anticipate chemical behavior, predict reactions, and synthesize new compounds effectively. The study of chemical bonding remains integral to uncovering the intricacies of molecular interactions and their practical applications in fields like material science and pharmaceuticals.