`NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)`
`The equilibrium pressure at `25 degree Celsius ` is 0.660 atm . What is Kp for the reaction ?

Options

A. 0.109
B. `0.218`
C. `1.89`
D. `2.18`

Answer

To find the equilibrium constant \( K_p \) for the reaction

\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]

given that the equilibrium pressure at 25°C is 0.660 atm, we can follow these steps:

### Step 1: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) is defined as:

\[ K_p = \frac{(P_{\text{NH}_3})^{\text{n}} \cdot (P_{\text{H}_2\text{S}})^{\text{m}}}{(P_{\text{NH}_4\text{HS}})^{\text{k}}} \]

Where:
- \( P_{\text{NH}_3} \) is the partial pressure of ammonia,
- \( P_{\text{H}_2\text{S}} \) is the partial pressure of hydrogen sulfide,
- \( P_{\text{NH}_4\text{HS}} \) is the partial pressure of the solid, which is not included in the expression for \( K_p \) since solids do not contribute to the pressure.

### Step 2: Determine the total pressure and the partial pressures

At equilibrium, we know that the total pressure \( P_{total} \) is given as 0.660 atm. Since both products (NH₃ and H₂S) are produced in equal amounts, we can denote the partial pressures as follows:

Let:
- \( P_{\text{NH}_3} = x \)
- \( P_{\text{H}_2\text{S}} = x \)

Thus, the total pressure can be expressed as:

\[ P_{total} = P_{\text{NH}_3} + P_{\text{H}_2\text{S}} = x + x = 2x \]

Setting this equal to the total pressure:

\[ 2x = 0.660 \, \text{atm} \]

### Step 3: Solve for \( x \)

To find \( x \):

\[ x = \frac{0.660}{2} = 0.330 \, \text{atm} \]

So, we have:

\[ P_{\text{NH}_3} = 0.330 \, \text{atm} \]
\[ P_{\text{H}_2\text{S}} = 0.330 \, \text{atm} \]

### Step 4: Substitute the partial pressures into the \( K_p \) expression

Now we can substitute the values into the \( K_p \) expression:

\[ K_p = P_{\text{NH}_3} \cdot P_{\text{H}_2\text{S}} = (0.330) \cdot (0.330) \]

### Step 5: Calculate \( K_p \)

Calculating \( K_p \):

\[ K_p = 0.330 \times 0.330 = 0.1089 \, \text{atm}^2 \]

### Final Result

Thus, rounding to three significant figures, we find:

\[ K_p \approx 0.109 \, \text{atm}^2 \]