Question

\(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \longrightarrow \mathrm{HS}(\mathrm{g})+\mathrm{H}(\mathrm{g}), \Delta \mathrm{H}^{\circ}=\mathrm{x}_{1}\) \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\right]=\mathrm{x}_{2}, \Delta \mathrm{H}_{\mathrm{f}}^{\circ}[\mathrm{H}(\mathrm{g})]=\mathrm{x}_{3}\) hence, \(\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{a}}(\mathrm{HS})\) is (a) \(\mathrm{x}_{1}+\mathrm{x}_{2}-\mathrm{x}_{3}\) (b) \(x_{3}-x_{1}-x_{2}\) (c) \(\mathrm{x}_{1}-\mathrm{x}_{2}-\mathrm{x}_{3}\) (d) \(\mathrm{x}_{3}-\mathrm{x}_{1}+\mathrm{x}_{2}\)

Short answer

The correct answer is (d): \( \mathrm{x}_3 - \mathrm{x}_1 + \mathrm{x}_2 \).

Step-by-step solution

Understand the Reaction and Given Data

The given reaction is the decomposition of \( \mathrm{H}_2 \mathrm{S}(\mathrm{g}) \) into \( \mathrm{HS}(\mathrm{g}) \) and \( \mathrm{H}(\mathrm{g}) \) with an enthalpy change of \( \Delta \mathrm{H}^{\circ} = \mathrm{x}_1 \). Additionally, we have the standard enthalpy of formation for \( \mathrm{H}_2 \mathrm{S}(\mathrm{g}) \) as \( \mathrm{x}_2 \) and for \( \mathrm{H}(\mathrm{g}) \) as \( \mathrm{x}_3 \). We need to find \( \Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{a}}(\mathrm{HS}) \).

Apply Hess's Law

According to Hess's Law, the enthalpy change of a reaction can be calculated by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products. This is expressed as:\[\Delta \mathrm{H}_{\text{reaction}} = \Sigma \Delta \mathrm{H}_{\mathrm{f}, \text{products}} - \Sigma \Delta \mathrm{H}_{\mathrm{f}, \text{reactants}}\]

Set Up the Enthalpy Change Equation

For the given reaction:- The product \( \mathrm{HS}(\mathrm{g}) \) has an unknown enthalpy of formation that we need to determine \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HS}) \).- The product \( \mathrm{H}(\mathrm{g}) \) has an enthalpy of formation \( \mathrm{x}_3 \).- The reactant \( \mathrm{H}_2 \mathrm{S}(\mathrm{g}) \) has an enthalpy of formation \( \mathrm{x}_2 \).Thus, we have:\[\Delta \mathrm{H}^{\circ} = \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HS}) + \mathrm{x}_3 - \mathrm{x}_2\]

Solve for \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HS}) \)

Rearrange the equation from Step 3 to solve for \( \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HS}) \):\[\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HS}) = \Delta \mathrm{H}^{\circ} - \mathrm{x}_3 + \mathrm{x}_2\]Substituting \( \Delta \mathrm{H}^{\circ} = \mathrm{x}_1 \), we get:\[\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HS}) = \mathrm{x}_1 - \mathrm{x}_3 + \mathrm{x}_2\]

Compare with the Given Options

Now compare the expression \( \mathrm{x}_1 - \mathrm{x}_3 + \mathrm{x}_2 \) with the given options:- (a) \( \mathrm{x}_1 + \mathrm{x}_2 - \mathrm{x}_3 \)- (b) \( x_{3} - x_{1} - x_{2} \)- (c) \( \mathrm{x}_1 - \mathrm{x}_2 - \mathrm{x}_3 \)- (d) \( \mathrm{x}_3 - \mathrm{x}_1 + \mathrm{x}_2 \)The correct option is (d): \( \mathrm{x}_3 - \mathrm{x}_1 + \mathrm{x}_2 \).

Key concepts

Enthalpy Change

Enthalpy change is a crucial concept in chemistry that reflects the energy exchange during a chemical reaction. It occurs when bonds are broken and formed as reactants transform into products.
In our exercise, it involves the breakdown (or decomposition) of hydrogen sulfide gas (\(\mathrm{H}_2 \mathrm{S}(\mathrm{g})\)) into hydrosulfide radicals (\(\mathrm{HS}(\mathrm{g})\)) and hydrogen atoms (\(\mathrm{H}(\mathrm{g})\)). The symbol \(\Delta \mathrm{H}^{\circ}\) represents this energy change under standard conditions, indicating how much energy is absorbed or released.
Understanding enthalpy changes helps us anticipate the behavior of substances in reactions. For example, if \(\Delta \mathrm{H}^{\circ}\) is positive, the reaction requires energy input (endothermic), while a negative value suggests energy is released (exothermic). This insight is vital for anticipating how a reaction will proceed and what conditions might be necessary to achieve desired results.
Key takeaways include:

• Enthalpy change is denoted by \(\Delta \mathrm{H}\) and can be either positive or negative.
• Positive \(\Delta \mathrm{H}\) values indicate energy absorption, while negative values indicate energy release.
• Enticingly, Hess's Law allows us to compute overall enthalpy changes using known values for simpler reactions.

Standard Enthalpy of Formation

The standard enthalpy of formation is a specialized enthalpy value that represents the energy change when one mole of a compound is formed from its pure elements under standard conditions (1 bar pressure and commonly at 25°C or 298 K). Notably, the standard enthalpy of formation for any element in its most stable form is defined as zero.
In this exercise, we're given the standard enthalpy of formation for \(\mathrm{H}_2 \mathrm{S}(\mathrm{g})\) as \(\mathrm{x}_2\) and for \(\mathrm{H}(\mathrm{g})\) as \(\mathrm{x}_3\). These values are crucial for applying Hess's Law in calculating unknown enthalpy changes by arranging and solving the enthalpy change equation for a specific product or reactant.
The concept of standard enthalpy of formation is powerful because it allows us to construct complex reaction enthalpies from tabulated data, offering a standardized way to predict how energy is distributed during chemical transformations.
Key points to remember include:

• Standard enthalpy of formation deals with forming one mole of a substance from its elements.
• These values are essential for using Hess's Law to solve enthalpy problems.
• Elements in their standard state have a standard enthalpy of formation defined as zero.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. Typically, this involves breaking bonds within the compound, necessitating an energy input (often making these reactions endothermic).
In the context of our exercise, the decomposition of \(\mathrm{H}_2 \mathrm{S}(\mathrm{g})\) into \(\mathrm{HS}(\mathrm{g})\) and \(\mathrm{H}(\mathrm{g})\) represents a classic case of decomposition. Such reactions can occur in a variety of environments and are fundamental in understanding larger chemical processes.
Characteristics of decomposition reactions that help in identifying them include:

• One reactant breaks into two or more products.
• Generally require an energy input, classified as endothermic.
• Can be influenced by physical conditions such as temperature and pressure.

Recognizing and understanding decomposition reactions is particularly useful in industrial and laboratory settings where these reactions might be harnessed for synthesis or breaking down complex mixtures into simpler forms.