When 5.1 g of solid NH\(_4\)HS is introduced into a two litre evacuated flask at 27\(^\circ\)C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K\(_p\) for the reaction at 27\(^\circ\)C is \(x \times 10^{-2}\). The value of x is _________. (Integer answer) [Given R = 0.082 L atm K\(^{-1}\) mol\(^{-1}\)]
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For equilibria involving pure solids or liquids, remember that their activities are taken as 1 and they do not appear in the equilibrium constant expression. Here, K\(_p\) only depends on the partial pressures of the gaseous products.
Step 1: Understanding the Question:
We are given the initial amount of a solid reactant and the extent of its decomposition at equilibrium. We need to calculate the equilibrium constant K\(_p\). Step 2: The Equilibrium Reaction:
The decomposition of solid ammonium hydrosulfide is:
\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]
Step 3: Calculate Moles of Gaseous Products at Equilibrium:
- Molar mass of NH\(_4\)HS = 14.0 + 4(1.0) + 32.1 = 51.1 g/mol.
- Initial moles of NH\(_4\)HS = \( \frac{5.1 \, \text{g}}{51.1 \, \text{g/mol}} \approx 0.1 \) mol.
- The problem states that 20% of the solid decomposes. Moles decomposed = \( 0.20 \times 0.1 = 0.02 \) mol.
- From the stoichiometry, moles of NH\(_3\) formed = 0.02 mol, and moles of H\(_2\)S formed = 0.02 mol. Step 4: Calculate Partial Pressures at Equilibrium:
We can use the Ideal Gas Law, \(P = \frac{nRT}{V}\), for each gas.
- T = 27\(^\circ\)C = 300 K.
- V = 2 L.
- R = 0.082 L atm K\(^{-1}\) mol\(^{-1}\).
Since the moles of NH\(_3\) and H\(_2\)S are equal, their partial pressures will be equal.
\[ P_{\text{NH}_3} = P_{\text{H}_2\text{S}} = \frac{(0.02 \, \text{mol}) \times (0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (300 \, \text{K})}{2 \, \text{L}} = 0.246 \, \text{atm} \]
Step 5: Calculate K\(_p\):
The equilibrium constant expression for this reaction is:
\[ K_p = P_{\text{NH}_3} \times P_{\text{H}_2\text{S}} \]
Since the partial pressures are equal:
\[ K_p = (0.246) \times (0.246) = (0.246)^2 \approx 0.0605 \]
Step 6: Find the value of x:
We are given that \(K_p = x \times 10^{-2}\).
\[ 0.0605 = x \times 10^{-2} \]
\[ x = 6.05 \]
The value of x as an integer is 6.
