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 need to find the equilibrium constant \( K_c \) for the reaction:

\[ NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g) \]

### Step 1: Understand the equilibrium expression
The equilibrium constant \( K_p \) in terms of partial pressures for the reaction can be expressed as:

\[ K_p = \frac{P_{NH_3} \cdot P_{H_2S}}{P_{NH_4HS}} \]

Since \( NH_4HS \) is a solid, its activity is 1 and does not appear in the expression. Thus, we have:

\[ K_p = P_{NH_3} \cdot P_{H_2S} \]

### Step 2: Determine the partial pressures
Given that the total pressure at equilibrium is 20 atm, we can let the partial pressures of \( NH_3 \) and \( H_2S \) be \( P \) each. Therefore, we have:

\[ P_{NH_3} + P_{H_2S} = 20 \text{ atm} \]

Since both gases are produced in equal amounts, we can say:

\[ P + P = 20 \text{ atm} \]

This simplifies to:

\[ 2P = 20 \text{ atm} \]

Thus,

\[ P = 10 \text{ atm} \]

### Step 3: Calculate \( K_p \)
Now substituting the values of the partial pressures into the expression for \( K_p \):

\[ K_p = P_{NH_3} \cdot P_{H_2S} = 10 \text{ atm} \cdot 10 \text{ atm} = 100 \text{ atm}^2 \]

### Step 4: Relate \( K_p \) to \( K_c \)
We can relate \( K_p \) and \( K_c \) using the formula:

\[ K_p = K_c \cdot R^n \cdot T^{\Delta n} \]

Where:
- \( R \) is the gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in moles of gas (moles of products - moles of reactants).

### Step 5: Calculate \( \Delta n \)
For the reaction:

- Moles of gaseous products = 2 (1 mole of \( NH_3 \) + 1 mole of \( H_2S \))
- Moles of gaseous reactants = 0 (since \( NH_4HS \) is a solid)

Thus,

\[ \Delta n = 2 - 0 = 2 \]

### Step 6: Convert temperature to Kelvin
The temperature given is \( 127^\circ C \). To convert to Kelvin:

\[ T = 127 + 273 = 400 \text{ K} \]

### Step 7: Substitute values into the equation
Now we can substitute the values into the equation to find \( K_c \):

\[ 100 = K_c \cdot (0.0821)^{2} \cdot (400)^{2} \]

Calculating \( (0.0821)^{2} \) and \( (400)^{2} \):

\[ (0.0821)^{2} \approx 0.00673641 \]
\[ (400)^{2} = 160000 \]

Thus,

\[ 100 = K_c \cdot 0.00673641 \cdot 160000 \]

Calculating \( 0.00673641 \cdot 160000 \):

\[ 0.00673641 \cdot 160000 \approx 1077.82 \]

### Step 8: Solve for \( K_c \)
Now we can solve for \( K_c \):

\[ K_c = \frac{100}{1077.82} \approx 0.0927 \]

Thus, the value of \( K_c \) is approximately \( 0.092 \).

### Final Answer
The equilibrium constant \( K_c \) for the reaction is approximately \( 0.092 \).

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