The equilibrium `NH_(4)HS(s)hArr NH_(3)(g)+H_(2)S(g)`, is followed to set - up at `127^(@)C` in a closed vessel. The total pressure at equillibrium was 20 atm. The `K_(C)` for the reaction is

Options

A. `0.092 M^(2)`
B. `0.085M^(2)`
C. `3.045M^(2)`
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium expression
The equilibrium reaction is:
\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

Since NH4HS is a solid, it does not appear in the equilibrium expression. The equilibrium constant \( K_p \) can be expressed as:
\[ K_p = P_{\text{NH}_3} \times P_{\text{H}_2\text{S}} \]

### Step 2: Define the total pressure at equilibrium
Given that the total pressure at equilibrium is 20 atm, we can denote the partial pressures of NH3 and H2S as \( P_{\text{NH}_3} \) and \( P_{\text{H}_2\text{S}} \) respectively. Since both gases are produced in a 1:1 ratio, we can let:
\[ P_{\text{NH}_3} = P \]
\[ P_{\text{H}_2\text{S}} = P \]

Thus, the total pressure can be expressed as:
\[ P + P = 20 \text{ atm} \]
\[ 2P = 20 \text{ atm} \]
\[ P = 10 \text{ atm} \]

### Step 3: Calculate \( K_p \)
Now substituting the values of the partial pressures into the \( K_p \) expression:
\[ K_p = P_{\text{NH}_3} \times P_{\text{H}_2\text{S}} = P \times P = 10 \text{ atm} \times 10 \text{ atm} = 100 \text{ atm}^2 \]

### Step 4: Relate \( K_p \) to \( K_c \)
We know that:
\[ K_p = K_c (RT)^{\Delta n} \]
Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 127 + 273 = 400 \, \text{K} \)
- \( \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 2 - 0 = 2 \)

### Step 5: Substitute values into the equation
Now we can substitute the values into the equation:
\[ 100 = K_c \times (0.0821 \times 400)^2 \]

Calculating \( (0.0821 \times 400)^2 \):
\[ 0.0821 \times 400 = 32.84 \]
\[ (32.84)^2 = 1075.1856 \]

Thus, we have:
\[ 100 = K_c \times 1075.1856 \]

### Step 6: Solve for \( K_c \)
Now, solving for \( K_c \):
\[ K_c = \frac{100}{1075.1856} \approx 0.0929 \, \text{molarity}^2 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is approximately:
\[ K_c \approx 0.0929 \, \text{molarity}^2 \]

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