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

The value of \(\mathrm{K}_{\mathrm{p}}\) is 0.109, which corresponds to option (a) 0.109.

Step 1: Understand the Reaction

The given reaction is \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{~g})\). Since \(\mathrm{NH}_{4} \mathrm{HS}\) is a solid, it does not appear in the equilibrium expression for \(\mathrm{K}_{\mathrm{p}}\). Therefore, \(\mathrm{K}_{\mathrm{p}}\) will be expressed in terms of the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\).

Step 2: Write the Expression for \(\mathrm{K}_{\mathrm{p}}\)

For the reaction \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\), the equilibrium constant \(\mathrm{K}_{\mathrm{p}}\) can be expressed as: \[ \mathrm{K}_{\mathrm{p}} = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2 \mathrm{S}} \] Here, \(P_{\mathrm{NH}_3}\) and \(P_{\mathrm{H}_2 \mathrm{S}}\) are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\), respectively.

Step 3: Determine the Partial Pressures

Since both \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) are produced from the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\), and the reaction is in equilibrium, their partial pressures are equal. Let the partial pressure of each be \(x\). Also, the total equilibrium pressure is given as \(0.660 \mathrm{~atm}\). Therefore, \[ 2x = 0.660 \quad \Rightarrow \quad x = 0.330 \mathrm{~atm} \] This means \(P_{\mathrm{NH}_3} = 0.330 \mathrm{~atm}\) and \(P_{\mathrm{H}_2 \mathrm{S}} = 0.330 \mathrm{~atm}\).

Step 4: Calculate \(\mathrm{K}_{\mathrm{p}}\)

Substitute the values of the partial pressures into the expression for \(\mathrm{K}_{\mathrm{p}}\): \[ \mathrm{K}_{\mathrm{p}} = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2 \mathrm{S}} = 0.330 \times 0.330 = 0.1089 \approx 0.109 \] The closest answer to this value in the options given is (a) 0.109.

Key Concepts

Understanding Partial Pressure in Chemical Equilibrium

In chemical reactions that occur in the gas phase, understanding and calculating partial pressure is crucial. Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. Imagine a container filled with different gases; each gas contributes to the total pressure of the system. This contribution is known as its partial pressure. For chemical reactions at equilibrium, especially when dealing with equilibrium constants like the equilibrium constant for pressure (\(K_p\)), partial pressure is a key concept. In the reaction of ammonium hydrogen sulfide (\(\mathrm{NH}_{4} \mathrm{HS}\)) dissociating into ammonia and hydrogen sulfide, each of these gases has a partial pressure that contributes to the total pressure observed at equilibrium.

• The total pressure is the sum of the partial pressures of all gases present.
• If the total pressure is given and the gases are produced in a 1:1 ratio, then their partial pressures are equal.

Understanding partial pressure helps calculate \(K_p\), as it's determined by the product of the partial pressures of the gases involved.

The Role of Ammonium Hydrogen Sulfide in the Reaction

Ammonium hydrogen sulfide (\(\mathrm{NH}_{4} \mathrm{HS}\)) plays a critical role in the chemical equilibrium of the reaction provided. It serves as the initial reactant that decomposes into ammonia (\(\mathrm{NH}_3\)) and hydrogen sulfide (\(\mathrm{H}_2 \mathrm{S}\)) gases.

• In the given equation: \(\mathrm{NH}_{4} \mathrm{HS}\,(\mathrm{s}) \rightleftharpoons \mathrm{NH}_3\,(\mathrm{~g}) + \mathrm{H}_2 \mathrm{S}\,(\mathrm{~g})\), ammonium hydrogen sulfide is in its solid form and doesn't directly contribute to the equilibrium constant expression.
• This is because \(\mathrm{K}_p\) expressions involve only gases and aqueous solutions; solids don't appear in the equilibrium expression.
• Despite its exclusion from the \(K_p\) calculation, \(\mathrm{NH}_{4} \mathrm{HS}\) is essential as it's the compound that produces the gases whose partial pressures define the equilibrium state.

This solid compound sets the stage for the equilibrium by dictating the initial conditions under which the gases are formed and how they contribute to the pressure.

Basics of Chemical Equilibrium and \(K_p\)

Chemical equilibrium is achieved when the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products over time. In the context of gas reactions, \(K_p\) represents the equilibrium constant expressed in terms of partial pressure.

• The equilibrium expression for a reaction like \(\mathrm{NH}_{4} \mathrm{HS}\,(\mathrm{s}) \rightleftharpoons \mathrm{NH}_3\,(\mathrm{~g}) + \mathrm{H}_2 \mathrm{S}\,(\mathrm{~g})\) is: \(K_p = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2 \mathrm{S}}\).
• Because the partial pressures are equivalent in this reaction's equilibrium, the equation simplifies to \(K_p = (P_{\mathrm{one~gas}})^2\).
• This helps in finding \(K_p\) by using the total pressure of the system, given the 1:1 formation ratio of the products.

Understanding these equilibrium basics and how \(K_p\) is derived and calculated is crucial in mastering chemical reactions involving gases.