For the equilibrium:
`LiCl.3NH_(3(s))hArrLiCl.NH_(3(s))+2NH_(3)`, `K_(p)=9 atm^(2)`
at `40^(@)C`. A `5 litre` vessel contains `0.1` mole of `LiCl.NH_(3)`. How many mole of `NH_(3)` should be added to the flask at this temperture to derive the backward reaction for completion?

Options

A. `0.2`
B. `0.59`
C. `0.69`
D. `0.79`

Answer

To solve the problem step by step, we will analyze the equilibrium reaction and apply the necessary calculations to find out how many moles of NH₃ should be added to the flask.

### Step 1: Write the equilibrium reaction and the expression for Kp
The equilibrium reaction is:
\[ \text{LiCl} \cdot 3\text{NH}_3 (s) \rightleftharpoons \text{LiCl} \cdot \text{NH}_3 (s) + 2\text{NH}_3 (g) \]

The equilibrium constant \( K_p \) is given as:
\[ K_p = 9 \, \text{atm}^2 \]

### Step 2: Determine the reverse reaction and its Kp
The reverse reaction is:
\[ \text{LiCl} \cdot \text{NH}_3 (s) + 2\text{NH}_3 (g) \rightleftharpoons \text{LiCl} \cdot 3\text{NH}_3 (s) \]

For the reverse reaction, the equilibrium constant \( K_p' \) is:
\[ K_p' = \frac{1}{K_p} = \frac{1}{9} \, \text{atm}^{-2} \]

### Step 3: Set up the initial conditions
In a 5-liter vessel, we have:
- Initial moles of \( \text{LiCl} \cdot \text{NH}_3 = 0.1 \, \text{moles} \)
- Initial moles of \( \text{NH}_3 = x \) (to be determined)

### Step 4: Define the changes at equilibrium
Let \( y \) be the amount of \( \text{NH}_3 \) added. At equilibrium:
- Moles of \( \text{LiCl} \cdot \text{NH}_3 \) will be \( 0.1 - y \)
- Moles of \( \text{NH}_3 \) will be \( y + 2(0.1 - y) = 0.2 - y \)

### Step 5: Calculate the partial pressures
Since the pressure is only due to the gaseous \( \text{NH}_3 \):
\[ P_{\text{NH}_3} = \frac{(y + 0.2 - y)RT}{V} = \frac{(0.2 - y)RT}{5} \]

### Step 6: Apply the Kp expression
From the equilibrium constant expression:
\[ K_p' = \frac{P_{\text{NH}_3}^2}{1} = \frac{(0.2 - y)^2 \cdot (RT/5)^2}{1} \]

Setting this equal to \( \frac{1}{9} \):
\[ \frac{(0.2 - y)^2 \cdot (RT/5)^2}{1} = \frac{1}{9} \]

### Step 7: Substitute values and solve for y
Using \( R = 0.0821 \, \text{L atm/(K mol)} \) and \( T = 313 \, \text{K} \):
\[ P_{\text{NH}_3} = 3 \, \text{atm} \]
\[ 3 = \frac{(0.2 - y) \cdot (0.0821 \cdot 313)}{5} \]

Solving for \( y \):
1. Calculate \( 0.0821 \cdot 313 \approx 25.7 \)
2. Thus, \( 3 = \frac{(0.2 - y) \cdot 25.7}{5} \)
3. Rearranging gives \( 15 = (0.2 - y) \cdot 25.7 \)
4. Solving for \( 0.2 - y \) gives \( 0.2 - y = \frac{15}{25.7} \approx 0.584 \)
5. Therefore, \( y = 0.2 - 0.584 \approx -0.384 \)

### Step 8: Calculate the total moles of NH₃ needed
Finally, the total moles of \( \text{NH}_3 \) to be added is:
\[ x = y + 0.1 \approx 0.7837 \, \text{moles} \]

### Conclusion
The number of moles of \( \text{NH}_3 \) that should be added to the flask is approximately:
\[ \text{Answer} \approx 0.79 \, \text{moles} \]

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