The equilibrium pressure of
`NH_(4)CN(s) hArr NH_(3)(g)+HCN(g)` is `2.98` atm. Calculate `K_(p)`

Explanation

To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction:

\[ \text{NH}_4\text{CN}(s) \rightleftharpoons \text{NH}_3(g) + \text{HCN}(g) \]

Given that the equilibrium pressure is \( 2.98 \) atm, we will follow these steps:

### Step 1: Understand the Reaction
The reaction involves the decomposition of solid ammonium cyanide (\( \text{NH}_4\text{CN} \)) into ammonia gas (\( \text{NH}_3 \)) and hydrogen cyanide gas (\( \text{HCN} \)). Since \( \text{NH}_4\text{CN} \) is a solid, it does not contribute to the pressure.

### Step 2: Define the Variables
At equilibrium, let the partial pressure of \( \text{NH}_3 \) be \( P \) and the partial pressure of \( \text{HCN} \) also be \( P \). Therefore, the total pressure at equilibrium is:

\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{HCN}} = P + P = 2P \]

### Step 3: Set Up the Equation
We know from the problem statement that the total pressure at equilibrium is \( 2.98 \) atm. Thus, we can write:

\[ 2P = 2.98 \text{ atm} \]

### Step 4: Solve for \( P \)
To find \( P \), we divide the total pressure by 2:

\[ P = \frac{2.98}{2} = 1.49 \text{ atm} \]

### Step 5: Calculate \( K_p \)
The equilibrium constant \( K_p \) is given by the product of the partial pressures of the gaseous products:

\[ K_p = P_{\text{NH}_3} \times P_{\text{HCN}} = P \times P = P^2 \]

Substituting the value of \( P \):

\[ K_p = (1.49)^2 \]

### Step 6: Perform the Calculation
Calculating \( K_p \):

\[ K_p = 1.49 \times 1.49 = 2.2201 \text{ atm}^2 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) is approximately:

\[ K_p \approx 2.22 \text{ atm}^2 \]

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