Question

The standard heats of formation of \(\mathrm{H}_{2} \mathrm{O}(g), \mathrm{H}_{2} S(g)\), \(\mathrm{H}_{2} \mathrm{Se}(g)\), and \(\mathrm{H}_{2} \mathrm{Te}(g)\) are \(-241.8,-20.17,+29.7\), and \(+99.6 \mathrm{~kJ} / \mathrm{mol}\), 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 \(\mathrm{O}, \mathrm{S}\), 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 H-O, H-S, H-Se, and H-Te are as follows: - H-O: 221.1 kJ/mol - H-S: 346.42 kJ/mol - H-Se: 346.35 kJ/mol - H-Te: 366.3 kJ/mol The bond enthalpy increases from H-O to H-S, remains almost constant between H-S and H-Se, and increases again from H-Se to H-Te. The trends in bond enthalpy can be attributed to the increasing size and decreasing electronegativity of the atoms along the periodic table from oxygen to tellurium, leading to weaker and longer H-X bonds with higher bond enthalpies.

Step-by-step solution

Calculate H-O bond enthalpy

To calculate the H-O bond enthalpy, we use the formula mentioned above: H-O bond enthalpy = (Standard heat of formation of H₂O + Enthalpy for converting O to gaseous atoms + Enthalpy for dissociation of H₂) / 2 H-O bond enthalpy = (-241.8 kJ/mol + 248 kJ/mol + 436 kJ/mol) / 2 H-O bond enthalpy = \( (442.2 \textrm{ kJ/mol})/2 = 221.1 \textrm{ kJ/mol} \)

Calculate H-S bond enthalpy

Repeat the formula for the H-S bond: H-S bond enthalpy = (Standard heat of formation of H₂S + Enthalpy for converting S to gaseous atoms + Enthalpy for dissociation of H₂) / 2 H-S bond enthalpy = (-20.17 kJ/mol + 277 kJ/mol + 436 kJ/mol) / 2 H-S bond enthalpy = \( (692.83 \textrm{ kJ/mol})/2 = 346.42 \textrm{ kJ/mol} \)

Calculate H-Se bond enthalpy

Repeat the formula for the H-Se bond: H-Se bond enthalpy = (Standard heat of formation of H₂Se + Enthalpy for converting Se to gaseous atoms + Enthalpy for dissociation of H₂) / 2 H-Se bond enthalpy = (29.7 kJ/mol + 227 kJ/mol + 436 kJ/mol) / 2 H-Se bond enthalpy = \( (692.7 \textrm{ kJ/mol})/2 = 346.35 \textrm{ kJ/mol} \)

Calculate H-Te bond enthalpy

Repeat the formula for the H-Te bond: H-Te bond enthalpy = (Standard heat of formation of H₂Te + Enthalpy for converting Te to gaseous atoms + Enthalpy for dissociation of H₂) / 2 H-Te bond enthalpy = (99.6 kJ/mol + 197 kJ/mol + 436 kJ/mol) / 2 H-Te bond enthalpy = \( (732.6 \textrm{ kJ/mol})/2 = 366.3 \textrm{ kJ/mol} \)

Analyze the trend in bond enthalpies

We have calculated the average bond enthalpies for H-O, H-S, H-Se, and H-Te: - H-O: 221.1 kJ/mol - H-S: 346.42 kJ/mol - H-Se: 346.35 kJ/mol - H-Te: 366.3 kJ/mol The bond enthalpy increases from H-O to H-S, remains almost constant between H-S and H-Se, and increases again from H-Se to H-Te. The trends in bond enthalpy can be attributed to the increasing size and decreasing electronegativity of the atoms along the periodic table from oxygen to tellurium. As the size of the non-hydrogen atom increases, the H-X bond (X being O, S, Se, or Te) becomes weaker and has a higher bond enthalpy due to its increased length and reduced bond strength.

Key concepts

Standard Heat of Formation

The standard heat of formation, or enthalpy of formation, is a crucial concept in thermodynamics and chemistry. It is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (usually 1 bar of pressure and 25°C or 298 K). The standard heat of formation can be used to calculate the energy change that occurs during a chemical reaction. For example, to determine the H-O bond enthalpy, we consider the standard heat of formation of H2O(g), which is -241.8 kJ/mol, indicating that forming water from hydrogen and oxygen releases energy, making it an exothermic process.

Enthalpy of Atomization

Moving on to the enthalpy of atomization, this term refers to the amount of energy required to convert one mole of a substance into individual gaseous atoms. This is a key part of the process for calculating bond enthalpies because it accounts for the energy needed to separate the elements before they form bonds. In our example, when calculating the bond enthalpy of the H-O bond, the enthalpy of atomization of oxygen is considered, which is 248 kJ/mol. This indicates the considerable energy required to break the stable O2 molecule into individual oxygen atoms.

H-X Bond Energy

The H-X bond energy, also known as the bond enthalpy, is the energy required to break one mole of the bond in the gas phase, resulting in separate atoms. It's an average value taken over a number of compounds, as individual bond energies vary with the molecular environment. The bond enthalpy is indicative of the bond's strength; the higher the bond enthalpy, the stronger the bond. For the H-O bond calculated in the exercise, the value came out to be 221.1 kJ/mol, which is a measure of the strength of the bond between hydrogen and oxygen in water.

Significance in Calculations

Understanding H-X bond energy is essential in practical chemistry, as it allows scientists to predict the heat changes and the stability of chemical compounds. By calculating the H-X bond energies for H-S, H-Se, and H-Te as well, we learn about the comparative strengths of these bonds and gain insights into their chemical behavior.

Periodic Trends

The periodic trends observed in bond enthalpies are largely due to the periodic nature of the elements involved. For example, as we progress from oxygen to tellurium in the periodic table, we notice an increase in atomic size and a decrease in electronegativity. These changes affect the bond length and bond strength, influencing the bond enthalpy. Larger atoms like tellurium will form bonds that are generally longer and weaker compared to smaller atoms such as oxygen, which is why we see an increasing trend in bond enthalpies from H-O to H-Te.

Implications on Bond Strength

This periodic trend provides valuable insights into the reactivity and stability of compounds. Chemists can predict how a change in bonding partners might alter a molecule's properties, which is important for designing new chemicals and materials. The exercise demonstrates a clear pattern where the bond enthalpy increases with increasing atomic number, also corresponding to a weaker H-X bond.