In one experiment, certain amount of `NH_(4)I(s)` was heated rapidly in a closed container at `357^(@)C`. The following equilibrium was established:
`NH_(4)I(s) hArrNH_(3)(g)+HI(g)`
but the pressure gradually increases further (when the excess solid residue remains in the vessel) owing to the dissociation of HI. Calculate the final pressure developed at equilibrium. `.
`2HI(g)rarrH_(2)(g)+l_(2)(g) K_(C)=0.065 at 357 ∘ C.

Options

A. `331mm "of" Hg`
B. `335mm "of" Hg`
C. `369 mm "of" Hg`
D. `151 mm "of" Hg`

Answer

To solve the problem, we need to analyze the equilibrium established in the closed container and the subsequent dissociation of HI. Let's break it down step by step.

### Step 1: Establish the Initial Conditions
We start with the equilibrium reaction:
\[ \text{NH}_4\text{I}(s) \rightleftharpoons \text{NH}_3(g) + \text{HI}(g) \]

At equilibrium, let's denote the partial pressure of NH₃ and HI as \( P_{\text{NH}_3} \) and \( P_{\text{HI}} \) respectively. Given that the solid NH₄I does not contribute to the pressure, we can express the total pressure in terms of the gases produced.

### Step 2: Calculate the Initial Pressure
From the problem, we know that at equilibrium, the partial pressures of NH₃ and HI are equal. Let’s denote this common pressure as \( P \):
\[ P_{\text{NH}_3} = P_{\text{HI}} = P \]

Thus, the total pressure \( P_{\text{total}} \) at equilibrium is:
\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{HI}} = P + P = 2P \]

### Step 3: Use the Equilibrium Constant
The equilibrium constant \( K_p \) for the reaction \( \text{NH}_4\text{I}(s) \rightleftharpoons \text{NH}_3(g) + \text{HI}(g) \) can be expressed as:
\[ K_p = \frac{P_{\text{NH}_3} \cdot P_{\text{HI}}}{1} = P^2 \]

Given that \( K_p = 0.065 \):
\[ P^2 = 0.065 \]
\[ P = \sqrt{0.065} \approx 0.255 \text{ atm} \]

### Step 4: Convert to mmHg
To convert the pressure from atm to mmHg, we use the conversion factor \( 1 \text{ atm} = 760 \text{ mmHg} \):
\[ P \approx 0.255 \text{ atm} \times 760 \text{ mmHg/atm} \approx 193.8 \text{ mmHg} \]

### Step 5: Calculate the Total Pressure from HI Dissociation
Next, we consider the dissociation of HI:
\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]
Given \( K_c = 0.065 \) at \( 357^\circ C \), we can express this in terms of partial pressures. Let \( x \) be the change in pressure due to the dissociation of HI.

The initial pressure of HI is \( P \) and after dissociation, the pressures will be:
- \( P_{\text{HI}} = P - 2x \)
- \( P_{\text{H}_2} = x \)
- \( P_{\text{I}_2} = x \)

The equilibrium expression for the dissociation of HI is:
\[ K_p = \frac{P_{\text{H}_2} \cdot P_{\text{I}_2}}{P_{\text{HI}}^2} = \frac{x \cdot x}{(P - 2x)^2} = 0.065 \]

### Step 6: Solve for x
We can rearrange this to find \( x \):
\[ 0.065 = \frac{x^2}{(P - 2x)^2} \]

Substituting \( P \approx 193.8 \text{ mmHg} \):
\[ 0.065 = \frac{x^2}{(193.8 - 2x)^2} \]

This equation can be solved for \( x \) using algebraic methods.

### Step 7: Calculate Final Pressure
Once we find \( x \), we can calculate the final pressures:
- Final pressure of HI: \( P_{\text{HI}} = P - 2x \)
- Final pressure of H₂ and I₂: \( P_{\text{H}_2} = x \) and \( P_{\text{I}_2} = x \)

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

### Step 8: Conclusion
After calculating all values, we arrive at the final pressure at equilibrium.