Question

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\(^{-1}\)]

Hint:

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.

Answer 1

Correct Answer: 6

To find the value of \( x \) in the expression for the equilibrium constant \( K_p \), follow these steps:

Reaction: NH₄HS(s) ↔ NH₃(g) + H₂S(g)

Initial Context:

1. Given mass of NH₄HS = 5.1 g.

2. Molecular weight of NH₄HS = 51 g/mol (calculated as 14 + 4 + 1 + 32 + 16).

3. Initial moles of NH₄HS = \( \frac{5.1}{51} = 0.1 \, \text{mol} \).

Decomposition:

Given 20% decomposes, moles decomposed = 0.1 mol × 0.2 = 0.02 mol.

At equilibrium, moles of NH₃ = 0.02 mol and moles of H₂S = 0.02 mol.

Pressure Calculation:

Use the ideal gas law: \( PV = nRT \), where \( V = 2 \, \text{L} \), \( n = 0.02 \), and \( R = 0.082 \).

Pressure of NH₃, \( P_{\text{NH}_3} = \frac{0.02 \times 0.082 \times 300}{2} = 0.246 \, \text{atm} \).

Similarly, pressure of H₂S, \( P_{\text{H}_2\text{S}} = 0.246 \, \text{atm} \).

Equilibrium Constant:

\( K_p = P_{\text{NH}_3} \times P_{\text{H}_2\text{S}} = 0.246 \times 0.246 = 0.060516 \, \text{atm}^2 \).

Given \( K_p = x \times 10^{-2} \).

So, \( 0.060516 = x \times 10^{-2} \) implies \( x = \frac{0.060516}{10^{-2}} = 6.0516 \approx 6 \).

Conclusion: The computed value of \( x \) is 6, which fits within the given range of 6 to 6.