Ammonium hydrogen sulphide dissociated according to the equation,
`NH_4HS(s) hArr NH_3(g) +H_2S(g)`. If the observed pressure of the mixture is 1.12 atm at `106^@C`, what is the equilibrium constant `K_p` of the reactions ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium expression
The dissociation of ammonium hydrogen sulfide is given by the equation:
\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

Since solids do not contribute to the equilibrium expression, we can write the equilibrium constant \( K_p \) as:
\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}}}{1} = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} \]

### Step 2: Define the pressures at equilibrium
Let \( P_1 \) be the partial pressure of both ammonia \( (\text{NH}_3) \) and hydrogen sulfide \( (\text{H}_2\text{S}) \) at equilibrium. Therefore, we have:
- \( P_{\text{NH}_3} = P_1 \)
- \( P_{\text{H}_2\text{S}} = P_1 \)

### Step 3: Relate the total pressure to partial pressures
The total observed pressure of the gas mixture at equilibrium is given as 1.12 atm. Since both gases contribute to the total pressure, we can write:
\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{H}_2\text{S}} = P_1 + P_1 = 2P_1 \]

### Step 4: Solve for \( P_1 \)
Using the total pressure:
\[ 2P_1 = 1.12 \]
\[ P_1 = \frac{1.12}{2} = 0.56 \, \text{atm} \]

### Step 5: Calculate \( K_p \)
Now we can substitute \( P_1 \) back into the expression for \( K_p \):
\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} = P_1 \cdot P_1 = 0.56 \cdot 0.56 \]

Calculating this gives:
\[ K_p = 0.3136 \, \text{atm}^2 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) for the reaction is:
\[ K_p = 0.3136 \, \text{atm}^2 \]

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