Solid `NH_4I` on rapid heating in a closed vessel at `357^@C` develops a constant pressure of `275 mm Hg` owing to partial decomposition of `NH_4I` to `NH_3` and `HI` but the pressure gradually increases further (when the excess of solid residue remains the vessel ) owing to dissociation of `HI` . Calculate the final pressure developed at equilibrium `Kp` for dissociation of `HI` is `0.015` at `357^@C`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Reaction
The initial reaction occurring in the closed vessel is the decomposition of solid `NH4I` into gaseous products:
\[ \text{NH}_4\text{I (s)} \rightleftharpoons \text{NH}_3 (g) + \text{HI} (g) \]

At the initial state, the pressure in the vessel is given as 275 mmHg.

### Step 2: Calculate Partial Pressures
Since the decomposition of `NH4I` produces one mole of `NH3` and one mole of `HI`, the partial pressures of `NH3` and `HI` at equilibrium can be calculated as follows:
- Let the partial pressure of `NH3` be \( P_{\text{NH}_3} \) and that of `HI` be \( P_{\text{HI}} \).
- Since both gases are produced in equal amounts, we have:
\[ P_{\text{NH}_3} = P_{\text{HI}} = \frac{275 \, \text{mmHg}}{2} = 137.5 \, \text{mmHg} \]

### Step 3: Consider the Dissociation of HI
Next, we consider the dissociation of `HI`:
\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]
The equilibrium constant \( K_p \) for this reaction is given as 0.015.

### Step 4: Set Up the Equilibrium Expression
The equilibrium expression for the dissociation of `HI` is:
\[ K_p = \frac{P_{\text{H}_2} \cdot P_{\text{I}_2}}{(P_{\text{HI}})^2} \]
Let \( x \) be the change in pressure due to the dissociation of `HI`. At equilibrium:
- \( P_{\text{HI}} = 137.5 - 2x \)
- \( P_{\text{H}_2} = x \)
- \( P_{\text{I}_2} = x \)

Substituting these into the equilibrium expression gives:
\[ 0.015 = \frac{x \cdot x}{(137.5 - 2x)^2} \]
\[ 0.015 = \frac{x^2}{(137.5 - 2x)^2} \]

### Step 5: Solve for x
Cross-multiplying gives:
\[ 0.015(137.5 - 2x)^2 = x^2 \]
Expanding and rearranging leads to a quadratic equation in terms of \( x \):
\[ 0.015(18906.25 - 550x + 4x^2) = x^2 \]
\[ 283.59375 - 8.25x + 0.015x^2 = x^2 \]
Rearranging gives:
\[ 0.985x^2 - 8.25x + 283.59375 = 0 \]

### Step 6: Use the Quadratic Formula
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 0.985 \), \( b = -8.25 \), and \( c = 283.59375 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-8.25)^2 - 4(0.985)(283.59375) \]
Calculate this value and then find \( x \).

### Step 7: Calculate Final Pressures
Once \( x \) is determined, substitute back to find:
- \( P_{\text{HI}} = 137.5 - 2x \)
- \( P_{\text{H}_2} = x \)
- \( P_{\text{I}_2} = x \)

Finally, the total pressure at equilibrium is:
\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{HI}} + P_{\text{H}_2} + P_{\text{I}_2} \]

### Step 8: Final Calculation
Add all the partial pressures to find the final pressure in the vessel.