Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at `300^@C`.The degree of dissociation of `NH_3` will be :

Options

A. `0.6`
B. `0.4`
C. Unpredictable
D. None of these

Explanation

To find the degree of dissociation of ammonia gas (NH₃) in the given scenario, we will follow these steps:

### Step 1: Understand the Initial Conditions
We know that ammonia gas is introduced into a rigid vessel at a pressure of 15 atm and a temperature of 300 K.

### Step 2: Determine the Equilibrium Conditions
At equilibrium, the total pressure in the vessel is given as 40.11 atm.

### Step 3: Convert Temperature to Kelvin
The temperature is given as 300 °C, which we need to convert to Kelvin:
\[ T = 300 + 273 = 573 \, K \]

### Step 4: Use the Ideal Gas Law to Find Equilibrium Pressure
Using the relationship from the ideal gas law, we can set up the equation:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Where:
- \( P_1 = 15 \, atm \) (initial pressure)
- \( T_1 = 300 \, K \)
- \( T_2 = 573 \, K \)
- \( P_2 \) is the pressure we need to find at equilibrium.

Substituting the known values:
\[ P_2 = P_1 \times \frac{T_2}{T_1} = 15 \times \frac{573}{300} \]

Calculating \( P_2 \):
\[ P_2 = 15 \times 1.91 = 28.65 \, atm \]

### Step 5: Write the Dissociation Reaction
The dissociation of ammonia can be represented as:
\[ 2 \, NH_3 \rightleftharpoons N_2 + 3 \, H_2 \]

### Step 6: Set Up the Pressure Changes
Let \( P \) be the pressure of ammonia that dissociates. Initially, we have:
- Pressure of NH₃ = \( 28.65 \, atm \)
- At equilibrium:
- Pressure of NH₃ = \( 28.65 - 2P \)
- Pressure of N₂ = \( P \)
- Pressure of H₂ = \( 3P \)

### Step 7: Write the Total Pressure Equation
The total pressure at equilibrium is given by:
\[ P_{total} = (28.65 - 2P) + P + 3P = 28.65 + 2P \]

Setting this equal to the total pressure at equilibrium:
\[ 28.65 + 2P = 40.11 \]

### Step 8: Solve for \( P \)
Rearranging the equation:
\[ 2P = 40.11 - 28.65 \]
\[ 2P = 11.46 \]
\[ P = \frac{11.46}{2} = 5.73 \, atm \]

### Step 9: Calculate the Degree of Dissociation
The degree of dissociation \( \alpha \) can be calculated as:
\[ \alpha = \frac{2P}{P_{initial}} \]
Where \( P_{initial} = 40.11 \, atm \).

Substituting the values:
\[ \alpha = \frac{2 \times 5.73}{40.11} \]
\[ \alpha = \frac{11.46}{40.11} \approx 0.286 \]

### Final Answer
The degree of dissociation of NH₃ is approximately **0.286**.

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