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 \(\mathrm{NH}_{4} \mathrm{HS}\) 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 for the decomposition is 0.11, which corresponds to option (d).

Step 1: Write the Balanced Equation

The decomposition reaction of ammonium hydrogen sulfide (\(\mathrm{NH}_4\mathrm{HS}\)) is given by the equation: \[\mathrm{NH}_4\mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_3(g) + \mathrm{H}_2\mathrm{S}(g)\] This equation shows that one mole of \(\mathrm{NH}_4\mathrm{HS}\) decomposes to form one mole each of ammonia (\(\mathrm{NH}_3\)) and hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)).

Step 2: Determine Changes in Pressure

Initially, the pressure due to \(\mathrm{NH}_3\) is \(0.50\, \text{atm}\). Let the increase in pressure due to decomposition be \(x\) atm. Therefore, at equilibrium, the pressure of \(\mathrm{NH}_3\) becomes \(0.50 + x\) atm, and the pressure of \(\mathrm{H}_2\mathrm{S}\) is \(x\) atm.

Step 3: Use Total Pressure at Equilibrium

The total pressure at equilibrium is given as \(0.84\, \text{atm}\). Thus, we have \((0.50 + x) + x = 0.84\). Solve this equation to find the value of \(x\).

Step 4: Solve for x

Simplify the equation from Step 3: \[0.50 + 2x = 0.84\] \[2x = 0.34\] \[x = 0.17\] Hence, the equilibrium pressure of \(\mathrm{H}_2\mathrm{S}\) is \(0.17\) atm, and the added pressure of \(\mathrm{NH}_3\) due to decomposition is also \(0.17\) atm.

Step 5: Calculate the Equilibrium Constant

The equilibrium constant \(K_p\) for the decomposition reaction is given by: \[K_p = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2\mathrm{S}} = (0.50 + 0.17) \times 0.17\] \[= 0.67 \times 0.17 = 0.1139\] Which, when considering significant figures, rounds to \(0.11\).

Step 6: Select the Correct Option

Comparing the calculated \(K_p = 0.11\) with the provided options, the correct answer is option (d).

Key Concepts

Equilibrium Constant

The equilibrium constant (\(K_p\)) in a chemical reaction measures the ratio of the concentrations of products to reactants at equilibrium, each raised to their respective stoichiometric coefficients. It varies depending on the reaction and specific conditions like temperature. For gaseous reactions, it's common to use partial pressures, resulting in \(K_p\).

• The equilibrium constant is a unique value for a given reaction at a specific temperature.
• A larger \(K_p\) indicates a reaction strongly favoring product formation, while a smaller value suggests limited product formation.
• It reflects the position of equilibrium — either leaning towards products or reactants.

In this case, \(K_p = 0.11\) means the reaction doesn't favor the production of \(\mathrm{NH}_3\) and \(\mathrm{H}_2\mathrm{S}\) significantly, but achieves a balance between the reactants and products. The temperature at which this equilibrium constant is calculated is crucial, as any change in temperature will likely affect \(K_p\).

Gas Laws

Gas laws describe how the pressure, volume, and temperature of a gas are interrelated. In this exercise, we primarily utilize Dalton's Law of Partial Pressures. This law states that the total pressure exerted by a gaseous mixture is the sum of the partial pressures of each gas in the mixture.

• Initially, the partial pressure of ammonia, \(\mathrm{NH}_3\), was \(0.50\, \text{atm}\).
• The reaction resulted in additional gases being produced, resulting in an increase in total pressure to \(0.84\, \text{atm}\).
• By subtracting the initial pressures from the total equilibrium pressure, we determined the change in pressure due to decomposition.

Hence, understanding how gas laws relate leads to solving for the equilibrium pressures of the gases involved, contributing significantly to calculating the equilibrium constant.

Decomposition Reaction

A decomposition reaction involves the breakdown of a single compound into two or more simpler substances. These reactions are typically identified by their reactants, which break apart to form products. In this exercise, the decomposition of ammonium hydrogen sulfide (\(\mathrm{NH}_4\mathrm{HS}\)) is illustrated.

• The balanced equation is: \[\mathrm{NH}_4\mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_3(g) + \mathrm{H}_2\mathrm{S}(g)\]
• The solid \(\mathrm{NH}_4\mathrm{HS}\) decomposes to produce gaseous ammonia (\(\mathrm{NH}_3\)) and hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\)).
• One mole of \(\mathrm{NH}_4\mathrm{HS}\) decomposes to form one mole of each product gas.

This single-compound decomposition is crucial in understanding how the equilibrium pressures of the generated gases are established. Recognizing the stoichiometry of the reaction aids in determining the specific pressures and consequently \(K_p\)\.