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 `K_p` for dissociation of HI is 0.015 at `357^@C`.

Options

A. 309.75 atm
B. 30.975 atm
C. 3.0975 atm
D. 0.30975 atm

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the decomposition reaction of NH₄I
The decomposition of ammonium iodide (NH₄I) can be represented as:
\[ \text{NH}_4\text{I (s)} \rightleftharpoons \text{NH}_3 (g) + \text{HI} (g) \]

### Step 2: Determine the initial pressure
The problem states that the initial pressure developed due to the partial decomposition of NH₄I is 275 mm Hg. This pressure is due to the formation of NH₃ and HI.

### Step 3: Calculate the partial pressures of NH₃ and HI
Since 1 mole of NH₄I produces 1 mole of NH₃ and 1 mole of HI, we can denote the partial pressures of NH₃ and HI as \( P \). Therefore, we have:
\[ P + P = 275 \, \text{mm Hg} \]
\[ 2P = 275 \]
\[ P = \frac{275}{2} = 137.5 \, \text{mm Hg} \]

### Step 4: Write the expression for Kp
The equilibrium constant \( K_p \) for the reaction can be expressed as:
\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{HI}}}{P_{\text{NH}_4\text{I}}} \]
Since NH₄I is a solid, its activity is considered to be 1. Thus:
\[ K_p = P_{\text{NH}_3} \cdot P_{\text{HI}} \]
Substituting the values:
\[ K_p = 137.5 \cdot 137.5 = 18906.25 \, \text{mm Hg}^2 \]

### Step 5: Consider the dissociation of HI
The dissociation of HI can be represented as:
\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]
Let \( x \) be the amount of HI that dissociates. The equilibrium expression for this reaction is:
\[ K_p = \frac{(P_H2)(P_I2)}{(P_{HI})^2} \]

### Step 6: Set up the equilibrium expression
At equilibrium, the pressures will be:
- Pressure of HI = \( P - x \)
- Pressure of H₂ = \( \frac{x}{2} \)
- Pressure of I₂ = \( \frac{x}{2} \)

Thus, we can write:
\[ K_p = \frac{\left(\frac{x}{2}\right) \left(\frac{x}{2}\right)}{(P - x)^2} = 0.015 \]

### Step 7: Substitute the known values into the Kp equation
Substituting \( K_p = 0.015 \) and \( P = 137.5 \):
\[ 0.015 = \frac{\left(\frac{x}{2}\right)^2}{(137.5 - x)^2} \]

### Step 8: Solve for x
Cross-multiplying gives:
\[ 0.015(137.5 - x)^2 = \left(\frac{x}{2}\right)^2 \]
Expanding and simplifying will yield a quadratic equation in terms of \( x \).

### Step 9: Calculate the final pressure at equilibrium
The total pressure at equilibrium will be:
\[ P_{\text{total}} = P + A + (P - A) + \frac{A}{2} + \frac{A}{2} \]
Where \( A \) is the pressure change due to the dissociation of HI.

### Step 10: Final calculation
After calculating \( A \) from the previous steps, substitute back into the total pressure equation to find the final pressure at equilibrium.