Problem 195 - A Complete Resource Book in Chemistry for JEE Main

Question

An amount of solid \(\mathrm{NH}_{4} \mathrm{HS}\) in placed in a flask already containing ammonia gas at a certain temperature and \(0.50\) atm pressure. Ammonium hydrogen sulphide decomposes to yield \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure is the flask rises to \(0.84 \mathrm{~atm}\), the equilibrium constant for NH HS decomposition at this temperature is (a) \(0.30\) (b) \(0.18\) (c) \(0.17\) (d) \(0.11\)

Step-by-Step Solution

Answer

The equilibrium constant for the decomposition is approximately 0.11 (option d).

Step 1: Identify the Initial and Final Conditions

Initially, the flask contains ammonia gas \(\mathrm{NH}_3\) at a pressure of \(0.50\ \text{atm}\). After the decomposition of \(\mathrm{NH}_4\mathrm{HS}\), the total pressure in the flask is \(0.84\ \text{atm}\). This increase is due to the production of additional gases \(\mathrm{NH}_3\) and \(\mathrm{H}_2\mathrm{S}\).

Step 2: Establish a Relationship between Pressures

We denote \(x\) as the change in pressure due to the decomposition of \(\mathrm{NH}_4\mathrm{HS}\). Since \(\mathrm{NH}_4\mathrm{HS}\) decomposes into one molecule each of \(\mathrm{NH}_3\) and \(\mathrm{H}_2\mathrm{S}\), the pressure change for both is \(x\). Therefore, the equilibrium pressure of \(\mathrm{NH}_3\) is \((0.50 + x)\ \text{atm}\) and for \(\mathrm{H}_2\mathrm{S}\) it is \(x\). The total pressure increase, \(0.84 - 0.50 = 0.34\ \text{atm}\), hence \(x = 0.34\ \text{atm}/2 = 0.17\ \text{atm}\).

Step 3: Calculate the Equilibrium Partial Pressures

The partial pressure of \(\mathrm{NH}_3\) at equilibrium is \(0.50 + 0.17 = 0.67\ \text{atm}\). The partial pressure of \(\mathrm{H}_2\mathrm{S}\) is \(0.17\ \text{atm}\).

Step 4: Determine the Equilibrium Constant Expression

The equilibrium expression for the decomposition \(\mathrm{NH}_4\mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_3(g) + \mathrm{H}_2\mathrm{S}(g)\) is given by \(K_p = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2\mathrm{S}}\), where \(P\) denotes the partial pressures of the gases.

Step 5: Calculate the Equilibrium Constant \(K_p\)

Substitute the equilibrium partial pressures into the expression: \(K_p = 0.67 \times 0.17 = 0.1139\). Given the options, \(K_p \approx 0.11\).

Key Concepts

Ammonium Hydrogen Sulphide Decomposition

Ammonium hydrogen sulphide (\(\text{NH}_4\text{HS}\)) is a solid compound that undergoes decomposition when heated. This process is a key example of a chemical equilibrium reaction.
When \(\text{NH}_4\text{HS}\) decomposes, it forms ammonia (\(\text{NH}_3\)) and hydrogen sulfide (\(\text{H}_2\text{S}\)) gases. In this reaction, the solid decomposes into two separate gaseous products, and this transformation can reach a state called equilibrium.
During equilibrium, the rate of decomposition of \(\text{NH}_4\text{HS}\) to form gases is equal to the rate of recombination of \(\text{NH}_3\) and \(\text{H}_2\text{S}\) back into the solid form. This means that, although the process continues, the concentrations—or in this case, pressures—of the reactants and products remain constant over time.

• The balanced equation for this reaction is: \[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

This setup creates a dynamic balance where no observable changes are detected in the system, although reactions are ongoing.

Equilibrium Constant Calculation

The equilibrium constant, denoted as \(K_p\), provides insight into the relative concentrations of products to reactants at equilibrium in a gas-phase reaction. In the decomposition of ammonium hydrogen sulphide, \(K_p\) is calculated based on the partial pressures of the gaseous products.
For the reaction \(\text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g)\), the equilibrium constant expression is given by:
\[ K_p = P_{\text{NH}_3} \times P_{\text{H}_2\text{S}} \]
Here, \(P_{\text{NH}_3}\) and \(P_{\text{H}_2\text{S}}\) represent the equilibrium partial pressures of ammonia and hydrogen sulfide, respectively.

• Identifying equilibrium pressures involves determining how pressures change as the reaction proceeds to equilibrium.
• In the analyzed case, the initial pressure of \(\text{NH}_3\) was given as \(0.50\ \text{atm}\).
• The increase in total pressure after the reaction reaches equilibrium allows us to calculate the individual pressures of the gases using an algebraic approach.

By substituting these pressures into the \(K_p\) expression, we can easily calculate the equilibrium constant, illustrating the balance between reactants and products in this chemical process.

Gas Pressure Relationships

Understanding gas pressure relationships is essential in analyzing reactions involving gaseous reactants or products. In an equilibrium scenario, the pressures of gases are directly connected to reaction dynamics.
When ammonium hydrogen sulphide decomposes, it creates more gas molecules, increasing the pressure within a closed system like a flask.

• The initial presence of ammonia at \(0.50\ \text{atm}\) and the ultimate rise to \(0.84\ \text{atm}\) imposes that changes occur due to gas production.
• Each particle formed adds pressure to the system, signifying a relationship between the number of gas molecules and total pressure.

This scenario allows us to dissect changes in pressure into components: the original ammonia pressure and the contributions from newly formed gases.
Calculating the equilibrium constant requires understanding these pressure shifts as they directly establish partial pressures of each gaseous component at equilibrium.
Thus, mastering gas pressure relationships offers valuable insight into evaluating chemical equilibrium states, allowing us to predict how pressure changes influence equilibrium reactions in closed systems.