Problem 88 - Chemistry The Molecular Science

Question

Consider the decomposition of ammonium hydrogen sulfide: $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ In a sealed flask at \(25^{\circ} \mathrm{C}\) are \(10.0 \mathrm{~g} \mathrm{NH}_{4} \mathrm{HS},\) ammonia with a partial pressure of \(0.692 \mathrm{~atm}\), and \(\mathrm{H}_{2} \mathrm{~S}\) with a partial pressure of \(0.0532 \mathrm{~atm}\). When equilibrium is established, it is found that the partial pressure of ammonia has increased by \(12.4 \% .\) Calculate \(K_{\mathrm{P}}\) for the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

\(K_p = 0.1080 \text{ atm}^2\).

Step 1: Initial Partial Pressures

The initial partial pressure of ammonia, \( P_{\text{NH}_3, \text{initial}} \), is given as \(0.692 \, \text{atm}\), and for \( \text{H}_2\text{S} \), \( P_{\text{H}_2\text{S}, \text{initial}} \), it is \(0.0532 \, \text{atm}\). These are the pressures before any change due to the equilibrium shift.

Step 2: Calculate Change in Partial Pressure

We know that when equilibrium is reached, the partial pressure of ammonia \( P_{\mathrm{NH}_3} \) has increased by \(12.4\%\). Calculate the increase by \(\Delta P_{\text{NH}_3} = 0.124 \times 0.692 \, \text{atm} = 0.0857 \, \text{atm}\). Therefore, the equilibrium total pressure for \( \mathrm{NH}_3 \) is \(0.692 + 0.0857 = 0.7777 \, \text{atm}\).

Step 3: Equilibrium Partial Pressures

As the reaction is balanced (1:1 stoichiometry), a rise of \(0.0857 \, \text{atm}\) in \( P_{\text{NH}_3} \) also corresponds to an equivalent rise in \( P_{\text{H}_2\text{S}} \). Thus, \( P_{\text{H}_2\text{S}, \text{eq}} = 0.0532 + 0.0857 = 0.1389 \, \text{atm} \).

Step 4: Calculate the Equilibrium Constant \(K_p\)

The equilibrium constant \(K_p\) for the reaction is given by the product of the equilibrium partial pressures: \( K_p = P_{\text{NH}_3,\text{eq}} \times P_{\text{H}_2\text{S}, \text{eq}} \). Substitute the equilibrium partial pressures: \( K_p = 0.7777 \, \text{atm} \times 0.1389 \, \text{atm} = 0.1080 \, \text{atm}^2 \).

Key Concepts

Decomposition Reactions

Decomposition reactions involve the breaking down of a single compound into two or more simpler substances. In the context of chemical equilibrium, these reactions can be represented as reversible processes. For example, ammonium hydrogen sulfide (\(\mathrm{NH}_4 \mathrm{HS}\) decomposes into ammonia \(\mathrm{NH}_3\) and hydrogen sulfide \(\mathrm{H}_2 \mathrm{S}\) when heated or placed in a closed container.
A key characteristic of decomposition reactions is that they are often endothermic. This means they require energy input to break apart the bonds within the initial compound. In equilibrium processes, both the forward (decomposition) and reverse (reformation) reactions occur simultaneously.
It's crucial to understand the role of dynamic equilibrium in these reactions. Initially, the substances produced by decomposition build up, but as they reach a specific concentration, the reverse reaction begins to compete, establishing a state where the rate of decomposition equals the rate of reformation. Thus, decomposition reactions contribute significantly to various chemical processes and maintaining equilibrium in natural systems.

Equilibrium Constant (Kp)

The equilibrium constant, \(K_p\), is a vital concept in understanding chemical equilibria involving gases. It represents the ratio of the products' partial pressures to the reactants' partial pressures, raised to the power of their stoichiometric coefficients.
For gases, \(K_p\) is calculated using the partial pressures rather than concentrations, which is ideal under constant temperature and pressure. In our example of ammonium hydrogen sulfide (\(\mathrm{NH}_4 \mathrm{HS}\), the equilibrium constant \(K_p\) indicates how far the decomposition reaction proceeds before reaching equilibrium.
Mathematically, \(K_p\) can be expressed in terms of the gaseous products' equilibrium pressures: \(K_p = P_{\mathrm{NH}_3,\text{eq}} \times P_{\mathrm{H}_2\mathrm{S},\text{eq}}\). If the value of \(K_p\) is significantly greater than 1, the reaction favors the formation of products at equilibrium. If smaller, it favors reactants. The value of \(K_p\) gives insight into the relative proportions of reactants and products and helps predict the extent of a reaction under given conditions.

Partial Pressure Calculation

Partial pressures are essential for understanding gas behavior in chemical reactions. They describe the pressure that each gas in a mixture would exert if it occupied the entire volume alone. Calculating these values requires careful consideration of changes during a reaction.
In our problem, the partial pressure of ammonia \(P_{\mathrm{NH}_3, \text{initial}}\) is initially given. With equilibrium established, the pressure increases by a certain percentage, which affects corresponding components like hydrogen sulfide. This percentage change is added to the initial values to find the equilibrium partial pressures.
To calculate the equilibrium constant \(K_p\), understanding these partial pressures is crucial. Use the relation \( P_{\text{eq}} = P_{\text{initial}} + \Delta P_{\text{increase}} \), where \(\Delta P\) is the change calculated from the percentage given. From these pressures, \(K_p\) can then be derived, illustrating how the system's pressure dynamics influence the chemical equilibrium.

• Partial pressure is a measure of concentration for gases.
• Helps to compute equilibrium constants using \(K_p\) for gaseous reactions.