Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) The equilibrium pressure at \(25^{\circ} \mathrm{C}\) is \(0.660 \mathrm{~atm}\). What is \(\mathrm{K}_{\mathrm{P}}\) for the reaction? (a) \(0.109\) (b) \(0.218\) (c) \(1.89\) (d) \(2.18\)

Step-by-Step Solution

Answer

\(\mathrm{K}_{\mathrm{P}} = 0.109\), which corresponds to option (a).

Step 1: Understanding the Reaction

The reaction given is \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\). Here, the solid \(\mathrm{NH}_{4}\mathrm{HS}\) dissociates into two gases: ammonia (\(\mathrm{NH}_3\)) and hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)). The equilibrium involves these gas components.

Step 2: Setting up the Equilibrium Expression

For the equilibrium \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\), \(\mathrm{K}_{\mathrm{P}}\) is defined only in terms of gaseous products/reactants. The equilibrium expression is \(\mathrm{K}_{\mathrm{P}} = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2\mathrm{S}}\). Since solids do not appear in the \(\mathrm{K}_{\mathrm{P}}\) expression, only the pressures of the gases are considered.

Step 3: Using Equilibrium Conditions

At equilibrium, the total pressure is given as \(0.660\, \mathrm{atm}\). Since \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2}\mathrm{S}\) are produced at a 1:1 ratio, the pressure of each gas is the same. Let \(P\) be the pressure of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2}\mathrm{S}\). Thus, the total pressure \(0.660 = P_{\mathrm{NH}_3} + P_{\mathrm{H}_2\mathrm{S}} = P + P = 2P\).

Step 4: Solve for Individual Pressures

From \(2P = 0.660\), solve for \(P\): \[ P = \frac{0.660}{2} = 0.330\, \mathrm{atm}. \] So, \(P_{\mathrm{NH}_3} = 0.330\, \mathrm{atm}\) and \(P_{\mathrm{H}_2\mathrm{S}} = 0.330\, \mathrm{atm}\).

Step 5: Calculating \(\mathrm{K}_{\mathrm{P}}\)

Plug the values into the expression for \(\mathrm{K}_{\mathrm{P}}\):\[ \mathrm{K}_{\mathrm{P}} = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2\mathrm{S}} = 0.330 \times 0.330 = 0.109. \]Thus, \(\mathrm{K}_{\mathrm{P}}\) for the reaction is 0.109.

Key Concepts

Equilibrium Constant (Kp)

In chemical equilibrium, reactions reach a state where the concentrations or pressures of reactants and products remain constant over time. For reactions involving gases, the equilibrium constant is denoted as \( K_p \), which is specific to the pressure conditions of the experiment. This constant helps describe the ratio of the product of the pressures of the gases involved in the reaction.

The equation used to define \( K_p \) only considers the gaseous components of the reaction. For the given reaction, \( \text{NH}_4 \text{HS(s)} \rightleftharpoons \text{NH}_3(\text{g}) + \text{H}_2\text{S}(\text{g}) \), the expression for \( K_p \) is formulated as:

• \( K_p = P_{\text{NH}_3} \times P_{\text{H}_2\text{S}} \)

Since solids do not affect the equilibrium of gases, they are omitted from the calculation. Only the partial pressures of ammonia and hydrogen sulfide contribute to this equilibrium constant. Calculating \( K_p \) requires knowledge of these partial pressures, which in this case can be determined from the total equilibrium pressure given.

Gaseous Reactions

Gaseous reactions are characterized by interactions between compounds in their gaseous state. These reactions often involve changes in pressure and volume, making them quite sensitive to external conditions such as temperature and pressure.

For a gaseous reaction like the decomposition of ammonium hydrogen sulfide, the reactants and products are more reactive and respond swiftly to changes in the system. These reactions reach an equilibrium framework where the rate of the forward reaction matches the rate of the reverse reaction, explaining the constant state of the products and reactants involved.

• Ammonium hydrogen sulfide breaks into ammonia and hydrogen sulfide in gas form.
• This reaction demonstrates how gases can shift between reactants and products, balancing themselves to maintain equilibrium.

Understanding gaseous reactions provides insight into how pressure and concentration affect the pace and direction of chemical processes in real-world scenarios.

Equilibrium Pressure

Equilibrium pressure refers to the total pressure in the system when a reaction has reached dynamic equilibrium, meaning the formation and decomposition happen at the same rate. In the context of the given reaction, the total pressure includes the sum of all gaseous products.

For the reaction \( \text{NH}_4 \text{HS(s)} \rightleftharpoons \text{NH}_3(\text{g}) + \text{H}_2\text{S}(\text{g}) \), the equilibrium pressure is provided as 0.660 atm. This is a crucial piece of information because:

• It allows us to calculate the partial pressure of each gaseous product since they are formed in a 1:1 ratio.
• The pressure of each gas is half of the total equilibrium pressure, hence \( P_{\text{NH}_3} = P_{\text{H}_2\text{S}} = 0.330 \text{ atm} \).

From here, understanding each gas's contribution helps in determining \( K_p \) for the reaction and ensures comprehensive insight into how individual pressures contribute to the overall reaction equilibrium.