`K_(p) "is" atm^(2)` for the reaction: `LiCI.3NH_(3)(s)hArrLiCI.NH_(3)(s)+2NH_(3)(g) "at" 40^(@)C`. How many moles of ammonia must be added at this temperature to a `5` litre flask containing `0.1` mole of LiCI. `NH_(3)` in order to completely convert the solid to `LiCI.3NH_(3)`?

Answer

To solve the problem step by step, we will analyze the equilibrium reaction and calculate how many moles of ammonia (NH₃) must be added to convert all the lithium chloride-ammonia complex (LiCl·NH₃) to the triammoniate complex (LiCl·3NH₃).

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

The equilibrium constant \( K_p \) is given as \( 1.8 \, \text{atm}^2 \).

### Step 2: Determine the initial conditions
We start with:
- 0.1 moles of LiCl·NH₃ in a 5-liter flask.
- The initial moles of NH₃ gas are 0.

### Step 3: Set up the equilibrium expression
The equilibrium expression for the reaction is:
\[ K_p = \frac{P_{\text{NH}_3}^2}{1} \]
Where \( P_{\text{NH}_3} \) is the partial pressure of NH₃ gas at equilibrium.

### Step 4: Calculate the moles of NH₃ at equilibrium
Let \( x \) be the moles of NH₃ added to the system. At equilibrium:
- The moles of NH₃ gas will be \( 2 \times 0.1 + x \) (because 0.1 moles of LiCl·NH₃ will produce 0.2 moles of NH₃ gas).
- The total moles of NH₃ gas will be \( 0.2 + x \).

### Step 5: Calculate the pressure of NH₃
Using the ideal gas law:
\[ P = \frac{nRT}{V} \]
Where:
- \( n = 0.2 + x \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 313 \, \text{K} \)
- \( V = 5 \, \text{L} \)

The pressure \( P \) becomes:
\[ P = \frac{(0.2 + x) \times 0.0821 \times 313}{5} \]

### Step 6: Substitute into the equilibrium expression
Substituting \( P \) into the \( K_p \) expression:
\[ K_p = \frac{(P_{\text{NH}_3})^2}{1} = \left(\frac{(0.2 + x) \times 0.0821 \times 313}{5}\right)^2 \]

Setting this equal to \( 1.8 \):
\[ 1.8 = \left(\frac{(0.2 + x) \times 0.0821 \times 313}{5}\right)^2 \]

### Step 7: Solve for \( x \)
1. Calculate \( \frac{0.0821 \times 313}{5} \):
\[ \frac{0.0821 \times 313}{5} \approx 5.134 \]
2. Set up the equation:
\[ 1.8 = (5.134(0.2 + x))^2 \]
3. Take the square root:
\[ \sqrt{1.8} = 5.134(0.2 + x) \]
\[ 1.34 \approx 5.134(0.2 + x) \]
4. Solve for \( 0.2 + x \):
\[ 0.2 + x \approx \frac{1.34}{5.134} \approx 0.261 \]
5. Therefore:
\[ x \approx 0.261 - 0.2 \approx 0.061 \]

### Conclusion
To completely convert the solid LiCl·NH₃ to LiCl·3NH₃, approximately **0.061 moles** of ammonia must be added.

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