The equilibrium pressure of
`NH_(4)CN(s) hArr NH_(3)(g)+HCN(g)` is `2.98` atm. Calculate `K_(p)`. If `NH_(4)CN(s)` is allowed to decompose in presence of `NH_(3)` at `0.25` atm, calculate partial pressure of `HCN` at equilibrium.

Explanation

To solve the problem step by step, we will first calculate the equilibrium constant \( K_p \) and then find the partial pressure of \( HCN \) at equilibrium when \( NH_3 \) is present at \( 0.25 \) atm.

### Step 1: Write the Reaction
The decomposition reaction is given as:
\[ NH_4CN(s) \rightleftharpoons NH_3(g) + HCN(g) \]

### Step 2: Understand the Given Information
We know that the equilibrium pressure of the system is \( 2.98 \) atm. This total pressure is the sum of the partial pressures of the gaseous products, \( NH_3 \) and \( HCN \).

### Step 3: Define Partial Pressures at Equilibrium
Let the partial pressure of \( NH_3 \) at equilibrium be \( P \) and the partial pressure of \( HCN \) at equilibrium also be \( P \) (since they are produced in a 1:1 ratio).

Thus, the total pressure at equilibrium can be expressed as:
\[ P_{total} = P_{NH_3} + P_{HCN} = P + P = 2P \]

### Step 4: Calculate Partial Pressures
From the total pressure given:
\[ 2P = 2.98 \, \text{atm} \]
Solving for \( P \):
\[ P = \frac{2.98}{2} = 1.49 \, \text{atm} \]
Thus, the partial pressures are:
\[ P_{NH_3} = 1.49 \, \text{atm} \quad \text{and} \quad P_{HCN} = 1.49 \, \text{atm} \]

### Step 5: Calculate \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{NH_3} \cdot P_{HCN}}{P_{NH_4CN}} \]
Since \( NH_4CN \) is a solid, its activity is taken as 1:
\[ K_p = P_{NH_3} \cdot P_{HCN} = 1.49 \cdot 1.49 = 2.2201 \]

### Step 6: Set Up the New Equilibrium Condition
Now, we need to find the partial pressure of \( HCN \) when \( NH_3 \) is present at \( 0.25 \) atm. Let the partial pressure of \( HCN \) at equilibrium be \( P' \).

At equilibrium, the total pressure can be expressed as:
\[ P_{total} = P' + 0.25 \]

### Step 7: Use the Equilibrium Constant
Using the previously calculated \( K_p \):
\[ K_p = P_{NH_3} \cdot P_{HCN} = 0.25 \cdot P' = 2.2201 \]
Rearranging gives:
\[ P' = \frac{2.2201}{0.25} = 8.8804 \]

### Step 8: Formulate the Equation
Now, we need to set up the equation based on the new equilibrium:
\[ P' \cdot (P' + 0.25) = 2.2201 \]
This leads to:
\[ P'^2 + 0.25P' - 2.2201 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula \( P' = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 0.25, c = -2.2201 \):
\[ P' = \frac{-0.25 \pm \sqrt{(0.25)^2 - 4 \cdot 1 \cdot (-2.2201)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \sqrt{0.0625 + 8.8804} = \sqrt{8.9429} \approx 2.99 \]
Then:
\[ P' = \frac{-0.25 \pm 2.99}{2} \]
Taking the positive root:
\[ P' = \frac{2.74}{2} = 1.37 \, \text{atm} \]

### Final Answer
The partial pressure of \( HCN \) at equilibrium is \( 1.37 \, \text{atm} \).

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