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

Answer

To find the degree of dissociation of ammonia gas (\(NH_3\)) under the given conditions, we can follow these steps:

### Step 1: Understand the Reaction
The dissociation of ammonia can be represented as:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \]

### Step 2: Initial Conditions
- Initial pressure of \(NH_3\) = 15 atm
- Temperature = 300 K (which we will convert to Celsius later)

### Step 3: Convert Temperature
The equilibrium pressure is given at 300 °C. We need to convert this to Kelvin for calculations:
\[ 300 °C = 300 + 273 = 573 K \]

### Step 4: Use the Ideal Gas Law
Using the ideal gas law, we can relate the pressures at different temperatures:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Where:
- \(P_1 = 15 \, \text{atm}\) (initial pressure)
- \(T_1 = 300 \, \text{K}\)
- \(T_2 = 573 \, \text{K}\)

Rearranging gives:
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]
Substituting the values:
\[ P_2 = 15 \, \text{atm} \times \frac{573 \, \text{K}}{300 \, \text{K}} = 28.65 \, \text{atm} \]

### Step 5: Set Up the Equilibrium Expression
At equilibrium, let \(x\) be the degree of dissociation of \(NH_3\). The changes in pressure can be expressed as:
- Pressure of \(NH_3\) at equilibrium = \(28.65 - 2x\)
- Pressure of \(N_2\) at equilibrium = \(x\)
- Pressure of \(H_2\) at equilibrium = \(3x\)

The total pressure at equilibrium is given as:
\[ P_{total} = P_{NH_3} + P_{N_2} + P_{H_2} \]
Substituting the pressures:
\[ 40.11 = (28.65 - 2x) + x + 3x \]
Simplifying this gives:
\[ 40.11 = 28.65 + 2x \]

### Step 6: Solve for \(x\)
Rearranging the equation:
\[ 2x = 40.11 - 28.65 \]
\[ 2x = 11.46 \]
\[ x = \frac{11.46}{2} = 5.73 \]

### Step 7: Calculate Degree of Dissociation
The degree of dissociation \(\alpha\) can be calculated as:
\[ \alpha = \frac{2x}{P_{initial}} \]
Where \(P_{initial} = 28.65 \, \text{atm}\):
\[ \alpha = \frac{2 \times 5.73}{28.65} \]
Calculating this gives:
\[ \alpha = \frac{11.46}{28.65} \approx 0.4 \]

### Final Answer
The degree of dissociation of \(NH_3\) is approximately \(0.4\).

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