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 \(K_{p}\) for the reaction? (a) \(0.109\) (b) \(0.218\) (c) \(1.89\) (d) \(2.18\)

Step-by-Step Solution

Answer

The value of \( K_p \) for the reaction is \( 0.109 \), corresponding to option (a).

Step 1: Understanding the Reaction

The reaction given is \( \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \). This is a decomposition reaction where solid ammonium hydrogen sulfide dissociates into gaseous ammonia and hydrogen sulfide.

Step 2: Recognizing the State of Matter Impact on \( K_p \)

Since \( \mathrm{NH}_{4} \mathrm{HS} \) is a solid, it does not appear in the expression for the equilibrium constant \( K_p \). Only gases are included in \( K_p \) expressions.

Step 3: Writing the \( K_p \) Expression

The equilibrium expression for \( K_p \) is given by the formula \( K_p = \frac{P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2\mathrm{S}}}{1} \). Because the solid does not appear, only the pressures of the gases are considered.

Step 4: Setting up Equality of Gaseous Pressures

At equilibrium, the pressure of \( \mathrm{NH}_3 \) equals the pressure of \( \mathrm{H}_2\mathrm{S} \), since both are produced in a 1:1 ratio. Given the total equilibrium pressure \( P_{\text{total}} = 0.660 \text{ atm} \), each gas has a partial pressure of \( 0.330 \text{ atm} \).

Step 5: Calculating \( K_p \)

Substitute the partial pressures into the \( K_p \) expression:\[K_p = (0.330)(0.330)\]Calculate to find:\[K_p = 0.1089\]

Step 6: Rounding and Choosing the Nearest Value

Round the calculated \( K_p \) value of \( 0.1089 \) to three significant figures, which is \( 0.109 \). Ensure this matches one of the multiple choice answers.

Key Concepts

Decomposition Reaction

In chemistry, a decomposition reaction involves breaking down a single compound into two or more different substances. In our exercise, the decomposition reaction is shown as \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\). Here, solid ammonium hydrogen sulfide \(\mathrm{NH}_{4} \mathrm{HS}\) dissociates into gaseous ammonia \(\mathrm{NH}_{3}\) and hydrogen sulfide \(\mathrm{H}_{2}\mathrm{S}\).

It's important to recognize that such reactions often involve a solid breaking down into gases or liquids. These reactions help in studying various equilibrium concepts because they clearly illustrate how different states of matter behave during a reaction. Understanding that solids do not participate directly in the equilibrium expression is key.

Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. In a closed container where different gases coexist, each gas contributes to the total pressure, irrespective of its type. The total pressure is the sum of these partial pressures.

In the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\), both gases formed at equilibrium - ammonia \(\mathrm{NH}_3\) and hydrogen sulfide \(\mathrm{H}_2\mathrm{S}\) - have equal partial pressures due to their 1:1 molar relationship in the reaction. If the total pressure is \(0.660\,\mathrm{atm}\), each gas contributes \(0.330\,\mathrm{atm}\) to this total, assuming they are the only gases present. Calculating partial pressures accurately is essential for subsequent calculations, such as determining equilibrium constants.

Equilibrium Expression

An equilibrium expression is a mathematical representation that describes the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients, in a reversible reaction at equilibrium. For the given reaction, the equilibrium expression for gas pressures is represented by \(K_p\), which only includes the pressures of gases, not solids or pure liquids.

Our reaction \(\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{H}_{2} \mathrm{S}(g)\) has the equilibrium constant \(K_p\) calculated as:

• \[K_p = \frac{P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2\mathrm{S}}}{1}\]

Solids are omitted since their concentrations don't change. Remember, equilibrium expressions differ for \(K_p\) and \(K_c\), where \(K_p\) uses pressures and \(K_c\) uses molar concentrations.

Gaseous Equilibrium

When a chemical reaction involving gases reaches a state where the rates of the forward and reverse reactions are equal, it is said to be in gaseous equilibrium. This means the concentrations of reactants and products remain constant over time.

In the case of \(\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{H}_{2} \mathrm{S}(g)\), gaseous equilibrium is established when the pressure of the products remains unchanged despite ongoing forward and reverse reactions. The equilibrium state is dynamic, meaning that reactions are continuously occurring, but no net change is observed in the system. This is a central concept in understanding how reactions proceed and stabilize in closed systems.

Gaseous equilibrium also entails how factors like temperature and pressure can shift the position of equilibrium, thereby changing the equilibrium constant for a reaction.