The Kp of the reaction is `NH_4HS(s)⇌NH_3(g)+H_2S( g)`. If the total pressure at equilibrium is 30 atm.

Answer

To solve the problem regarding the equilibrium constant \( K_p \) for the reaction:

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

given that the total pressure at equilibrium is 30 atm, we can follow these steps:

### Step 1: Write the expression for total pressure at equilibrium

At equilibrium, let the partial pressures of \( NH_3 \) and \( H_2S \) be \( p \) atm each. Since both gases are produced in a 1:1 ratio from the dissociation of \( NH_4HS \), the total pressure \( P_{total} \) can be expressed as:

\[ P_{total} = P_{NH_3} + P_{H_2S} = p + p = 2p \]

### Step 2: Substitute the given total pressure

We know from the problem that the total pressure at equilibrium is 30 atm. Therefore, we can set up the equation:

\[ 2p = 30 \text{ atm} \]

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

To find \( p \), divide both sides of the equation by 2:

\[ p = \frac{30}{2} = 15 \text{ atm} \]

### Step 4: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) for the reaction is given by the product of the partial pressures of the products:

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

Substituting the value of \( p \):

\[ K_p = p \times p = p^2 \]

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

Now substitute \( p = 15 \) atm into the equation for \( K_p \):

\[ K_p = 15 \times 15 = 225 \text{ atm}^2 \]

### Final Answer

Thus, the value of \( K_p \) for the reaction is:

\[ K_p = 225 \text{ atm}^2 \]

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