At 273 K and 1atm , 10 litre of `N_(2)O_(4)` decompose to `NO_(4)` decompoes to ` NO_(2)` according to equation
`N_(2)O_(4)(g)hArr2NO_(@)(G)`
What is degree of dissociation `(alpha)` when the original volume is 25% less then that os existing volume?

Options

A. `0.25`
B. `0.33`
C. `0.66`
D. `0.5`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]
This means that 1 mole of \( N_2O_4 \) decomposes into 2 moles of \( NO_2 \).

### Step 2: Initial Moles Calculation
At the start, we have 10 liters of \( N_2O_4 \) at 273 K and 1 atm. Using the ideal gas law, we can find the number of moles of \( N_2O_4 \):
\[ PV = nRT \]
Where:
- \( P = 1 \, \text{atm} \)
- \( V = 10 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 273 \, \text{K} \)

Rearranging for \( n \):
\[ n = \frac{PV}{RT} = \frac{1 \times 10}{0.0821 \times 273} \]
Calculating this gives:
\[ n \approx 0.446 \, \text{moles of } N_2O_4 \]

### Step 3: Define Degree of Dissociation
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). At equilibrium:
- Moles of \( N_2O_4 \) remaining = \( n(1 - \alpha) \)
- Moles of \( NO_2 \) produced = \( 2n\alpha \)

Total moles at equilibrium:
\[ n(1 - \alpha) + 2n\alpha = n(1 + \alpha) \]

### Step 4: Volume Consideration
It is given that the original volume is 25% less than the existing volume. Let the equilibrium volume be \( V \):
- Original volume = \( V - 0.25V = 0.75V \)

Using the ideal gas law again, we know that the number of moles is proportional to the volume:
\[ 0.75n(1 + \alpha) = n \]
Cancelling \( n \):
\[ 0.75(1 + \alpha) = 1 \]

### Step 5: Solve for \( \alpha \)
Rearranging the equation:
\[ 1 + \alpha = \frac{1}{0.75} \]
\[ 1 + \alpha = \frac{4}{3} \]
\[ \alpha = \frac{4}{3} - 1 = \frac{1}{3} \]

### Step 6: Conclusion
Thus, the degree of dissociation \( \alpha \) is:
\[ \alpha = 0.33 \]

### Final Answer
The degree of dissociation \( \alpha \) is \( 0.33 \).

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