Ammonia under a pressure of 15 atm at `27^@C` is heated to `347^@C` in a closed vessel in the presence of a catalyst. Under these conditions `NH_3` is partially decomposed according to the equation
`2NH_3 Leftrightarrow N_2 +3H,`
The vessel is such that the volume remains effectively constant, whereas pressure increases to 50 atm. Calculate the percentage of `NH_3` actually decomposed. Pressure of `NH_3" at "27^@C" or 300 K = 15 atm."`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
- Initial pressure of NH₃ (P₁) = 15 atm
- Final pressure (P₂) = 50 atm
- Initial temperature (T₁) = 27°C = 300 K
- Final temperature (T₂) = 347°C = 620 K
- Reaction: 2NH₃ ⇌ N₂ + 3H₂

### Step 2: Define the change in moles
Let the initial number of moles of NH₃ be 2 (for simplicity). If x moles of NH₃ decompose, then:
- Moles of NH₃ remaining = 2 - x
- Moles of N₂ produced = x/2
- Moles of H₂ produced = 3x/2

Total moles at equilibrium = (2 - x) + (x/2) + (3x/2) = 2 + x

### Step 3: Use the ideal gas law
Since the volume is constant, we can use the relationship from the ideal gas law:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Rearranging gives:
\[ P_2 = P_1 \cdot \frac{T_2}{T_1} \cdot \frac{n_2}{n_1} \]

### Step 4: Substitute known values
Substituting the known values into the equation:
\[ 50 = 15 \cdot \frac{620}{300} \cdot \frac{(2 + x)}{2} \]

### Step 5: Solve for x
First, calculate the left side:
\[ 50 = 15 \cdot 2.0667 \cdot \frac{(2 + x)}{2} \]
\[ 50 = 15 \cdot 2.0667 \cdot (1 + \frac{x}{2}) \]
\[ 50 = 31.0005 \cdot (1 + \frac{x}{2}) \]
Now, divide both sides by 31.0005:
\[ \frac{50}{31.0005} = 1 + \frac{x}{2} \]
Calculating the left side:
\[ 1.6127 = 1 + \frac{x}{2} \]
Subtracting 1 from both sides:
\[ 0.6127 = \frac{x}{2} \]
Multiplying by 2 gives:
\[ x = 1.2254 \]

### Step 6: Calculate the percentage of NH₃ decomposed
The initial moles of NH₃ = 2. Therefore, the percentage decomposed is:
\[ \text{Percentage decomposed} = \left(\frac{x}{\text{initial moles}} \times 100\right) = \left(\frac{1.2254}{2} \times 100\right) = 61.27\% \]

### Final Answer
The percentage of NH₃ actually decomposed is approximately **61.27%**.

---