Problem 100 - Chemistry The Central Science In Si Units

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 \(197 \mathrm{~kJ} / \mathrm{mol}\) 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.

Step-by-Step Solution

Answer

The bond enthalpies are: 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 trend shows increasing bond enthalpy down the group.

Step 1: Understanding the Reaction and Required Data

We need to calculate the average bond enthalpies for the molecules \( \mathrm{H}_{2} \mathrm{O}(g), \mathrm{H}_{2} \mathrm{S}(g), \mathrm{H}_{2} \mathrm{Se}(g), \mathrm{H}_{2} \mathrm{Te}(g) \). For this, we need the standard heats of formation and the enthalpies required to form gaseous atoms from their standard states. The relevant heats of formation are: \(-241.8, -20.17, +29.7, +99.6\) kJ/mol, and the atomization enthalpies for \(\mathrm{O}, \mathrm{S}, \mathrm{Se}, \mathrm{Te}\) are respectively: \(248, 277, 227, 197\) kJ/mol. We also need the enthalpy for \(\mathrm{H}_2\) dissociation, which is \(436\) kJ/mol.

Step 2: Write and Analyze the Reaction Schemes

The formation of each molecule \( \mathrm{H}_2Y \) can be represented as: \[ \mathrm{Y}(g) + \mathrm{H}_2(g) \rightarrow \mathrm{H}_2Y(g) \]Using Hess's Law, this reaction can be expressed via these steps:1. \( \mathrm{Y}(s) \rightarrow \mathrm{Y}(g) \) (atomization of \(\mathrm{Y}\))2. \( \mathrm{H}_2(g) \rightarrow 2\mathrm{H}(g) \) (dissociation of \( \mathrm{H}_2 \))3. \( 2\mathrm{H}(g) + \mathrm{Y}(g) \rightarrow \mathrm{H}_2Y(g) \) (bond formation)Calculate the energy required for each molecule using:\[\Delta H = \Delta H^\circ_f + 2 \cdot B_H - \Delta H_{atomization}^{Y} - \Delta H_{dissociation}^{H_2}\]Where \( \Delta H^\circ_f \) is the standard heat of formation, \( B_H \) is the bond enthalpy, \( \Delta H_{atomization}^{Y} \) and \( \Delta H_{dissociation}^{H_2} \) are given enthalpies.

Step 3: Calculate the \(\mathrm{H}-\mathrm{O}\) Bond Enthalpy

For \( \mathrm{H}_2\mathrm{O} \), the given data is: - Heat of formation = \(-241.8\) kJ/mol- Atomization enthalpy of \(\mathrm{O}\) = 248 kJ/mol - \(\Delta H_{dissociation}^{H_2}\) = 436 kJ/mol.Plugging values into the bond energy equation:\[-241.8 = 2 \cdot B_H - 248 - 436 \]Solving gives:\[2 \cdot B_H = 248 + 436 - 241.8 = 442.2\]\[B_H = \frac{442.2}{2} = 221.1\; \text{kJ/mol}\]

Step 4: Calculate the \(\mathrm{H}-\mathrm{S}\) Bond Enthalpy

For \( \mathrm{H}_2\mathrm{S} \), the given data is:- Heat of formation = \(-20.17\) kJ/mol- Atomization enthalpy of \(\mathrm{S}\) = 277 kJ/mol.Plugging values into the bond energy equation:\[-20.17 = 2 \cdot B_H - 277 - 436 \]Solving gives:\[2 \cdot B_H = 277 + 436 - 20.17 = 692.83\]\[B_H = \frac{692.83}{2} = 346.42\; \text{kJ/mol}\]

Step 5: Calculate the \(\mathrm{H}-\mathrm{Se}\) Bond Enthalpy

For \( \mathrm{H}_2\mathrm{Se} \), the given data is:- Heat of formation = \(+29.7\) kJ/mol- Atomization enthalpy of \(\mathrm{Se}\) = 227 kJ/mol.Plugging values into the bond energy equation:\[29.7 = 2 \cdot B_H - 227 - 436 \]Solving gives:\[2 \cdot B_H = 227 + 436 + 29.7 = 692.7\]\[B_H = \frac{692.7}{2} = 346.35\; \text{kJ/mol}\]

Step 6: Calculate the \(\mathrm{H}-\mathrm{Te}\) Bond Enthalpy

For \( \mathrm{H}_2\mathrm{Te} \), the given data is:- Heat of formation = \(+99.6\) kJ/mol- Atomization enthalpy of \(\mathrm{Te}\) = 197 kJ/mol.Plugging values into the bond energy equation:\[99.6 = 2 \cdot B_H - 197 - 436 \]Solving gives:\[2 \cdot B_H = 197 + 436 + 99.6 = 732.6\]\[B_H = \frac{732.6}{2} = 366.3\; \text{kJ/mol}\]

Step 7: Observe and Comment on the Trend

The calculated bond enthalpies are as follows:- \(\mathrm{H}-\mathrm{O}\): 221.1 kJ/mol- \(\mathrm{H}-\mathrm{S}\): 346.42 kJ/mol- \(\mathrm{H}-\mathrm{Se}\): 346.35 kJ/mol- \(\mathrm{H}-\mathrm{Te}\): 366.3 kJ/molAs we move down the group from \( \mathrm{H}_2\mathrm{O} \) to \( \mathrm{H}_2\mathrm{Te} \), the bond enthalpy increases, except for two values \( \mathrm{H}-\mathrm{S} \) and \( \mathrm{H}-\mathrm{Se} \) which are almost similar. This is primarily due to the differences in atomic size and the resulting bond strengths.

Key Concepts

Heats of Formation

The heat of formation, or standard enthalpy of formation (\( \Delta H^\circ_f \)), is a measure of energy released or absorbed during the formation of one mole of a compound from its elements in their standard states. Standards states refer to the most stable form of an element at 1 atmosphere and a specified temperature, usually 25°C.

The heats of formation are calculated using tables that provide standard values for numerous compounds. By knowing these values, chemists can predict the energy change involved in forming a particular compound. For example:

• Negative heat of formation implies that the compound formation is exothermic, releasing energy to the surroundings, as seen with \( \text{H}_2\text{O(g)} \) (-241.8 kJ/mol).
• Positive heat of formation indicates an endothermic process, where energy is absorbed. This is the case with \( \text{H}_2\text{Te(g)} \) (+99.6 kJ/mol).

Understanding heats of formation helps in predicting the stability of compounds and the feasibility of chemical reactions.

Atomization Enthalpy

Atomization enthalpy refers to the amount of energy required to convert a mole of a given element in its standard state into individual gaseous atoms. This process involves breaking chemical bonds to generate isolated atoms, and it is always an endothermic process due to the necessity of adding energy to separate atoms.

Each specific element has a unique atomization enthalpy value. For instance:

• The atomization enthalpy of oxygen to form \( ext{O(g)} \) is 248 kJ/mol, indicating substantial energy is needed due to its strong double bond in \( ext{O}_2 \).
• Sulfur's atomization enthalpy is 277 kJ/mol, also relatively high reflecting its strong bonds in solid form.

Understanding atomization enthalpy is critical for chemists, as it helps to determine the energy required to break bonds within diatomic or more complex molecules in simulations and experiments.

Hess's Law

Hess's Law is a fundamental principle in chemistry, which states that the total change in enthalpy for a given chemical reaction is the same, no matter how it is achieved, i.e., it is path-independent. This allows us to break down a complex chemical reaction into a series of steps, each with its own enthalpy change, and sum these to find the overall reaction enthalpy.

In the exercise, Hess's Law is used to calculate the bond enthalpies in molecules like \(\text{H}_2\text{O}\) and \(\text{H}_2\text{S}\). By applying Hess’s Law:

• We can express a chemical reaction in terms of formation, dissociation, and atomization enthalpies.
• These processes can be individually calculated using known values, and the results can be summed to calculate unknown bond enthalpies.

Hess’s Law is an invaluable tool in thermochemistry, allowing scientists to use known enthalpy changes from tabulated values to deduce unknowns, thus illustrating the power of constructing a thermochemical cycle.