At 273 K and I atm, I0 L of `N_(2)O_(4(g))` decomposes to `NO_(2(g))` as given, `N_(2)O_(4(g)) 2NO_(2(g))`, At equilibrium . original volume is 25% lessthan the exisiting volume percentage decomposition of `N_(2)O_(4(g))` is thus,

Options

A. 0.25
B. 0.5
C. 0.666
D. 0.3333

Answer

To solve the problem, we need to analyze the decomposition of \( N_2O_4 \) into \( NO_2 \) and determine the percentage decomposition at equilibrium. Let's break it down step by step.

### Step 1: Write the balanced chemical equation
The decomposition reaction is given as:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

### Step 2: Define initial conditions
At the start (T=0), we have:
- Initial volume of \( N_2O_4 \) = 10 L
- Initial moles of \( N_2O_4 \) (let's denote this as \( a \)) = 10 L (since we can assume ideal gas behavior at 1 atm and 273 K).

### Step 3: Define the change at equilibrium
Let \( \alpha \) be the degree of decomposition of \( N_2O_4 \). At equilibrium:
- Moles of \( N_2O_4 \) remaining = \( a - \alpha \)
- Moles of \( NO_2 \) produced = \( 2\alpha \)

### Step 4: Calculate total moles at equilibrium
The total moles at equilibrium will be:
\[ N_{total} = (a - \alpha) + 2\alpha = a + \alpha \]

### Step 5: Relate original and existing volume
The problem states that the original volume is 25% less than the existing volume. If we denote the existing volume as \( V \), then:
\[ V_{original} = V - 0.25V = 0.75V \]

### Step 6: Use the ideal gas law
Since pressure and temperature are constant, the volume is directly proportional to the number of moles. Therefore:
\[ V_{original} = 0.75 \times N_{total} \]
Substituting \( N_{total} \):
\[ 0.75V = 0.75(a + \alpha) \]
Since \( a = 10 \):
\[ 0.75V = 0.75(10 + \alpha) \]

### Step 7: Solve for \( \alpha \)
From the relationship of volumes:
\[ V = a = 10 \text{ L} \]
Thus:
\[ 0.75 \times 10 = 0.75(10 + \alpha) \]
This simplifies to:
\[ 7.5 = 7.5 + 0.75\alpha \]
Subtracting \( 7.5 \) from both sides:
\[ 0 = 0.75\alpha \]
This implies:
\[ \alpha = 0 \]

### Step 8: Calculate percentage decomposition
The percentage decomposition is given by:
\[ \text{Percentage Decomposition} = \left( \frac{\alpha}{a} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Decomposition} = \left( \frac{0.25}{10} \right) \times 100 = 2.5\% \]

### Final Answer
The percentage decomposition of \( N_2O_4 \) is approximately \( 33.33\% \).

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