Problem 59 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

When \(3.06 \mathrm{~g}\) of solid \(\mathrm{NH}_{4} \mathrm{HS}\) is introduced into a two litre evacuated flask at \(27^{\circ} \mathrm{C}, 30 \%\) of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate \(K_{c}\) and \(K_{p}\) for the reaction at \(27^{\circ} \mathrm{C}\). (ii) What would happen to the equilibrium when more solid \(\mathrm{NH}_{4} \mathrm{HS}\) is introduced into the flask?

Step-by-Step Solution

Answer

(i) \(K_c = 8.1 \times 10^{-5}\), \(K_p = 6.05 \times 10^{-3}\). (ii) Equilibrium stays the same.

Step 1: Determine Initial Molality

The molar mass of NH₄HS is approximately 51 g/mol. Given 3.06 g, the number of moles is \(\frac{3.06}{51}\approx 0.06\) moles. Placed in a 2 L flask, the initial molality is \(\frac{0.06}{2} = 0.03\) mol/L.

Step 2: Calculate Change in Moles

30% of the NH₄HS decomposes. Therefore, \(0.03 \times 0.3 = 0.009\) mol/L decomposes to NH₃ and H₂S.

Step 3: Write Balanced Equation and Express Equilibrium Concentrations

The decomposition reaction is: \(\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{H}_{2}\mathrm{S}(g)\). \[ [\text{NH}_3] = [\text{H}_2\text{S}] = 0 + 0.009 = 0.009 \text{ mol/L}. \]

Step 4: Calculate Kc

The expression for \(K_{c}\) is \(K_{c} = [\text{NH}_3][\text{H}_2\text{S}]\). Therefore, \[ K_{c} = (0.009)(0.009) = 8.1 \times 10^{-5}. \]

Step 5: Relate Kc to Kp Using Temperature

The relationship between \(K_{c}\) and \(K_{p}\) for a gaseous reaction is given by \[ Kp = Kc \times (RT)^{\Delta n}. \] Here, \(\Delta n = 1 + 1 - 0 = 2\). At 27°C (or 300 K), and using \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \), \[ Kp = 8.1 \times 10^{-5} \times (0.0821 \times 300)^2 \approx 6.05 \times 10^{-3}. \]

Step 6: Analyze the Effect of Adding More Solid

Adding more solid NH₄HS does not affect the equilibrium concentrations of gaseous products. The equilibrium will remain unchanged as the equilibrium expression does not include solids.

Key Concepts

Equilibrium Constant

The equilibrium constant is a measure of the extent to which a chemical reaction will proceed to form products from reactants at equilibrium. It indicates the ratio of the concentrations of products to the concentrations of reactants, each raised to the power of their coefficients in the balanced equation. The equilibrium constant comes in two forms:

• \( K_c \) for reactions in solutions, where concentrations are measured in ext{mol/L}.
• \( K_p \) for gaseous reactions, where partial pressures are used instead of concentrations.

For the decomposition of \( \text{NH}_4\text{HS} \), calculating \( K_c \) involves determining the molarity of the gaseous products just like we did with \( \text{NH}_3 \) and \( \text{H}_2\text{S} \). With \( [\text{NH}_3] = [\text{H}_2\text{S}] = 0.009 \text{ mol/L} \), \( K_c \) is \( 8.1 \times 10^{-5} \). \( K_p \) can be calculated using the equation \( K_p = K_c \times (RT)^{\Delta n} \), where \( R \) is the gas constant and \( \Delta n \) is the change in the number of moles of gas.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. In our case, \( \text{NH}_4\text{HS} \) is the compound that decomposes. Generally, this can be represented by the formula: \[ AB \rightarrow A + B. \]The decomposition we are discussing is \( \text{NH}_4\text{HS (s)} \rightarrow \text{NH}_3 (g) + \text{H}_2\text{S} (g) \). 30% decomposition implies a partial reaction where only a portion of the compound changes, making it a dynamic process that eventually reaches equilibrium when the forward and backward reaction rates equalize.

Gaseous Products

Gaseous products, like \( \text{NH}_3 \) and \( \text{H}_2\text{S} \) in this reaction, play a crucial role in determining the equilibrium constant in gaseous reactions. When decomposition of \( \text{NH}_4\text{HS} \) occurs, the formation of these gaseous products impacts how the reaction progresses towards equilibrium.

• At equilibrium, the concentrations of all species involved remain constant.
• The concentration of \( \text{NH}_3 \) and \( \text{H}_2\text{S} \) determine \( K_c \) for the reaction.

Their presence means that calculations involving gaseous products will relate directly to changes in pressure, making \( K_p \) relevant alongside \( K_c \) for a full understanding of the equilibrium.

Effect of Solid in Equilibrium

An important aspect of equilibrium reactions involving solids is recognizing their non-contribution to the equilibrium expression. Solids, such as \( \text{NH}_4\text{HS} \) in the given reaction, have constant concentrations that do not appear in the equilibrium constant expressions, \( K_c \) or \( K_p \). When additional \( \text{NH}_4\text{HS} \) is added:

• The concentration of gaseous products remains unaffected at equilibrium.
• The result is no change in the equilibrium position, as the additional solid does not alter the partial pressures or concentrations of the gases.

This is because the equilibrium constants are only affected by changes involving species in the gaseous or aqueous state.