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 \(\mathrm{NH}_{2}\) 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 \(\mathrm{NH}_{4}\) HS decomposition at this temperature is [2005] (a) \(0.30\) (b) \(0.18\) (c) \(0.17\) (d) \(0.11\)

Step-by-Step Solution

Answer

The equilibrium constant is approximately 0.11, corresponding to option (d).

Step 1: Write the Balanced Reaction Equation

The decomposition reaction of ammonium hydrogen sulfide \(\mathrm{NH}_4\mathrm{HS}\) can be represented as: \[\mathrm{NH}_4\mathrm{HS} (s) \rightleftharpoons \mathrm{NH}_3 (g) + \mathrm{H}_2\mathrm{S} (g)\]Since the reaction involves only solids and gases, the equilibrium expression will only include the gases.

Step 2: Determine Changes in Pressure

Initially, the pressure of \(\mathrm{NH}_3\) is given as \(0.50\, \mathrm{atm}\). When the system reaches equilibrium, the total pressure in the flask is \(0.84\, \mathrm{atm}\). The pressure increase (\(0.84-0.50=0.34\, \mathrm{atm}\)) is contributed by the gases \(\mathrm{NH}_3\) and \(\mathrm{H}_2\mathrm{S}\) from the decomposition.

Step 3: Establish Relations of Partial Pressures at Equilibrium

Let the pressure contribution of \(\mathrm{H}_2\mathrm{S}\) be \(x\) atm due to decomposition, hence \(\mathrm{NH}_3\) will also have an additional pressure of \(x\) atm.Total increase in pressure due to these gases is \(x + x = 2x = 0.34\, \mathrm{atm}\).Thus, \(x = \frac{0.34}{2} = 0.17\, \mathrm{atm}\).

Step 4: Calculate Equilibrium Partial Pressures

At equilibrium, the total pressure of \(\mathrm{NH}_3\) becomes \(0.50 + 0.17 = 0.67\, \mathrm{atm}\), while \(\mathrm{H}_2\mathrm{S}\) has a pressure of \(0.17\, \mathrm{atm}\).

Step 5: Calculate the Equilibrium Constant Expression

The equilibrium constant \(K_p\) expression for the reaction includes the partial pressures of the gases:\[ K_p = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2\mathrm{S}} \\text{Substitute the pressures: } \K_p = 0.67 \times 0.17 \approx 0.114 \]

Step 6: Identify the Correct Option

The calculated equilibrium constant \(K_p\) is approximately \(0.114\), which when rounded is closest to option (d) \(0.11\).

Key Concepts

Equilibrium Constant

In the field of chemistry, the equilibrium constant, often denoted as \( K_p \) for reactions involving gases, is a fundamental concept. It quantifies the balance between reactants and products in a chemical reaction at equilibrium. This constant is applicable to reactions occurring in a closed system, where the conditions such as temperature and pressure remain consistent. For gaseous reactions, \( K_p \) is calculated using the partial pressures of the gases involved.

In the exercise you studied, the decomposition of ammonium hydrogen sulfide \( (\mathrm{NH}_4\mathrm{HS}) \) into ammonia \( (\mathrm{NH}_3) \) and hydrogen sulfide \( (\mathrm{H}_2\mathrm{S}) \) explains the dependence of \( K_p \) on gas pressures. The balanced equation shows a direct relationship, allowing us to express \( K_p \) as:

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

In this expression, \( P_{\mathrm{NH}_3} \) is the partial pressure of ammonia, and \( P_{\mathrm{H}_2\mathrm{S}} \) is the partial pressure of hydrogen sulfide. The equilibrium constant tells us how far the reaction proceeds towards products at a given temperature, reflecting the extent of the chemical equilibrium. Understanding \( K_p \) helps predict how a system may respond to changes in conditions like temperature, allowing chemists to control and manipulate reactions efficiently.

Pressure Changes

Pressure plays a vital role in chemical reactions involving gases. Changes in pressure can influence the position of equilibrium and the rate of a reaction.

Consider a system at equilibrium, such as the decomposition reaction of ammonium hydrogen sulfide in our exercise. Initially, there's ammonia gas at \(0.50\, \mathrm{atm}\). As the decomposition progresses to equilibrium, additional pressures from new gases formed increase the total pressure of the system to \(0.84\, \mathrm{atm}\).

Let's analyze this step-by-step:

• The increase in pressure results from the production of \( \mathrm{NH}_3 \) and \( \mathrm{H}_2\mathrm{S} \) gases.
• We note the pressure increase of \(0.34\, \mathrm{atm}\), due to the decomposition of \( \mathrm{NH}_4\mathrm{HS}\), which splits into two separate gases.
• Each gas contributes equally to this increase, setting up a situation where \( 2x = 0.34\, \mathrm{atm} \), and therefore, the pressure contribution from each, \( \mathrm{NH}_3 \) and \( \mathrm{H}_2\mathrm{S} \), is \(0.17\, \mathrm{atm}\).

Understanding how pressure affects equilibrium is crucial, as manipulating pressure can shift the equilibrium position according to Le Chatelier's Principle. Increasing pressure generally favors the side of the reaction with fewer gas molecules, an essential tool for controlling reactions in industrial applications.

Decomposition Reaction

Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. These reactions are characterized by the equation:

• \( \text{Reactant} \rightarrow \text{Product}_1 + \text{Product}_2 \)

In our practical example, ammonium hydrogen sulfide \( (\mathrm{NH}_4\mathrm{HS}) \) decomposes into ammonia gas \( (\mathrm{NH}_3) \) and hydrogen sulfide gas \( (\mathrm{H}_2\mathrm{S}) \).

The general processes of decomposition often require energy input to break the chemical bonds in the compound. However, in some situations, as in the case of \( \mathrm{NH}_4\mathrm{HS} \), the decomposition can spontaneously proceed due to favorable conditions. The pressure change observed during the reaction is consistent with forming two gaseous products from a solid reactant.

Decomposition reactions are significant in several fields, including environmental science, where the breakdown of compounds is often needed to recycle materials in nature. They also play a role in various industries, such as the decomposition of carbonates for materials processing. Understanding these reactions helps predict how substances behave and react under different circumstances.