Question

The standard heats of formation of \(\mathrm{H}_{2} \mathrm{O}(g), \mathrm{H}_{2} \mathrm{S}(g), \mathrm{H}_{2} \mathrm{Se}(g)\) and \(\mathrm{H}_{2} \mathrm{Te}(g)\) are \(-241.8,-20.17,+29.7,\) and \(+99.6 \mathrm{kJ} /\) 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 \(\mathrm{kJ} / \mathrm{mol}\) atoms for \(\mathrm{O}, \mathrm{S},\) 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 \(\mathrm{H}-\) Te bond enthalpies, and comment on their trend.

Short answer

The calculated average bond enthalpies for the compounds are: - H-O bond enthalpy: 459 kJ/mol - H-S bond enthalpy: 579.5 kJ/mol - H-Se bond enthalpy: 564.35 kJ/mol - H-Te bond enthalpy: 534.3 kJ/mol The trend shows that the average bond enthalpy for H-S is the highest, followed by H-Se, H-Te, and H-O. As we move down the periodic table, the bond enthalpy fluctuates, with the H-S bond having the highest enthalpy, possibly due to the decrease in electronegativity down the group affecting bond strengths.

Step-by-step solution

Use standard heat of formation and enthalpies of atomization to calculate bond enthalpy

The standard heat of formation of a compound is calculated as: ΔH_f = m * (Bond enthalpy of the molecule) - (n * Bond enthalpy of H2) where ΔH_f is the standard heat of formation, m is the number of bonds in the molecule, n is the number of hydrogen atoms in the molecule, and the bond enthalpy is the energy required to break the bond. We are given ΔH_f values and enthalpies of atomization. By combining ΔH_f values with enthalpies of atomization, we can obtain bond enthalpies. Bond enthalpy = (ΔH_f + n * (Enthalpy for dissociation of H2) + Enthalpy of atomization) / m For each compound, we will calculate the bond enthalpy using this formula:

Calculate H-O bond enthalpy

In H2O, there are 2 H-O bonds (m = 2), and 2 hydrogen atoms (n = 2). ΔH_f (H2O) = -241.8 kJ/mol Enthalpy of atomization (O) = 248 kJ/mol Enthalpy for dissociation of H2 = 436 kJ/mol Bond enthalpy (H-O) = (ΔH_f(H2O) + 2 * (Enthalpy for dissociation of H2) + Enthalpy of atomization (O)) / 2 = (-241.8 + (2 * 436) + 248) / 2 = 459 kJ/mol

Calculate H-S bond enthalpy

In H2S, there are 2 H-S bonds (m = 2), and 2 hydrogen atoms (n = 2). ΔH_f (H2S) = -20.17 kJ/mol Enthalpy of atomization (S) = 277 kJ/mol Bond enthalpy (H-S) = (ΔH_f(H2S) + 2 * (Enthalpy for dissociation of H2) + Enthalpy of atomization (S)) / 2 = (-20.17 + (2 * 436) + 277) / 2 = 579.5 kJ/mol

Calculate H-Se bond enthalpy

In H2Se, there are 2 H-Se bonds (m = 2), and 2 hydrogen atoms (n = 2). ΔH_f (H2Se) = +29.7 kJ/mol Enthalpy of atomization (Se) = 227 kJ/mol Bond enthalpy (H-Se) = (ΔH_f(H2Se) + 2 * (Enthalpy for dissociation of H2) + Enthalpy of atomization (Se)) / 2 = (29.7 + (2 * 436) + 227) / 2 = 564.35 kJ/mol

Calculate H-Te bond enthalpy

In H2Te, there are 2 H-Te bonds (m = 2), and 2 hydrogen atoms (n = 2). ΔH_f (H2Te) = +99.6 kJ/mol Enthalpy of atomization (Te) = 197 kJ/mol Bond enthalpy (H-Te) = (ΔH_f(H2Te) + 2 * (Enthalpy for dissociation of H2) + Enthalpy of atomization (Te)) / 2 = (99.6 + (2 * 436) + 197) / 2 = 534.3 kJ/mol

Analyze the trend

Here are the calculated average bond enthalpies for the different compounds: - H-O bond enthalpy: 459 kJ/mol - H-S bond enthalpy: 579.5 kJ/mol - H-Se bond enthalpy: 564.35 kJ/mol - H-Te bond enthalpy: 534.3 kJ/mol The trend shows that the average bond enthalpy for H-S is the highest, followed by H-Se, H-Te, and H-O. As we move down the periodic table, the bond enthalpy fluctuates, with the H-S bond having the highest enthalpy. This is possibly due to the decrease in electronegativity down the group, which in turn affects the bond strengths.

Key concepts

Standard Heat of Formation

The standard heat of formation, denoted as \( \Delta H_f^\circ \), is a fundamental thermodynamic quantity representing the change in enthalpy during the formation of one mole of a substance from its elements in their standard states. It's a direct measure of the energy involved in the bonding process and can be negative or positive, indicating whether the formation process is exothermic or endothermic, respectively.

To elucidate its practical application, consider the process to calculate bond enthalpies using \( \Delta H_f^\circ \). In the given exercise, the \( \Delta H_f^\circ \)-values for various hydrides were provided. The calculations demanded understanding that bond enthalpies reflect the strength of chemical bonds. The stronger the bond, the more energy is required to break it, resulting in higher bond enthalpy values. By analyzing the standard heat of formation data, we can infer the energetics of bond formation and compare the strengths of different bonds within a homologous series of compounds.

Enthalpies of Atomization

Enthalpies of atomization (\( \Delta H_{at} \) ) represent the energy required to convert one mole of a substance into its constituent atoms in the gaseous state. It's a critical component when dissecting the energetics of a compound to its basic atomic level. This allows us to understand the energy needed to break the interatomic interactions to form gaseous atoms, which are then available to form different kinds of bonds.

In the context of the exercise, \( \Delta H_{at} \) for each element provided a stepping stone to compute individual bond enthalpies. For instance, the enthalpy of atomization of oxygen was essential in determining the enthalpy of the H-O bond, just as those of sulfur, selenium, and tellurium were to their respective hydrides. This concept emphasizes the significance of considering all energy transactions in an element's cycle from its standard state to forming a bond in a molecule.

Chemical Bonding

Chemical bonding is the foundation of material existence, explaining how atoms connect to form molecules. It involves the interaction between the valence electrons of atoms leading to the formation of a stable electron configuration. The types of chemical bonds including ionic, covalent, and metallic bonds, dictate the properties of compounds.

When dissecting the bond enthalpies, we indirectly assess the nature of chemical bonding in molecules. For example, in our exercise, analyzing the trend in bond enthalpies for hydrides of group 16 elements offered insight into the changing nature of chemical bonds due to differences in electronegativity and bond lengths. The H-S bond exhibited the highest enthalpy, suggesting considerable strength, which can be explained by factors like bond polarity and atomic sizes. Understanding chemical bonding is crucial not just for interpreting bond enthalpies, but for grasping the universal principles that guide the formation and reactivity of molecules.