For the following equilibrium reaction ,
`N_2O_4(g) hArr 2NO_2(g)`
`NO_2` is 50% of total volume at given temperature . Hence vapour density of the equilibrium mixture is

Options

A. 34.5
B. 25
C. `23.0`
D. 20

Answer

To solve the problem, we need to find the vapor density of the equilibrium mixture for the reaction:

\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

Given that NO2 is 50% of the total volume at equilibrium, we can follow these steps:

### Step 1: Define the Initial and Equilibrium Conditions
At the start (initially), we have:
- 1 mole of \( N_2O_4 \)
- 0 moles of \( NO_2 \)

At equilibrium, let \( x \) be the degree of dissociation of \( N_2O_4 \). Therefore, we have:
- Moles of \( N_2O_4 \) at equilibrium = \( 1 - x \)
- Moles of \( NO_2 \) at equilibrium = \( 2x \)

### Step 2: Total Moles at Equilibrium
The total number of moles at equilibrium is:
\[ \text{Total moles} = (1 - x) + 2x = 1 + x \]

### Step 3: Determine the Volume Contribution
Given that \( NO_2 \) is 50% of the total volume, we can express this as:
\[ \frac{2x}{1 + x} = \frac{1}{2} \]

### Step 4: Solve for \( x \)
Cross-multiplying gives:
\[ 2x = \frac{1}{2}(1 + x) \]
\[ 4x = 1 + x \]
\[ 4x - x = 1 \]
\[ 3x = 1 \implies x = \frac{1}{3} \]

### Step 5: Calculate the Moles at Equilibrium
Now we can find the moles of each component at equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - x = 1 - \frac{1}{3} = \frac{2}{3} \)
- Moles of \( NO_2 \) = \( 2x = 2 \times \frac{1}{3} = \frac{2}{3} \)

### Step 6: Calculate the Total Moles
Total moles at equilibrium:
\[ \text{Total moles} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

### Step 7: Calculate the Vapor Density
To find the vapor density, we need the molecular weights:
- Molecular weight of \( N_2O_4 = 2(14) + 4(16) = 28 + 64 = 92 \, g/mol \)
- Molecular weight of \( NO_2 = 14 + 2(16) = 14 + 32 = 46 \, g/mol \)

Now we calculate the average molecular weight of the mixture:
\[ \text{Average molecular weight} = \frac{(\text{moles of } N_2O_4 \times \text{molecular weight of } N_2O_4) + (\text{moles of } NO_2 \times \text{molecular weight of } NO_2)}{\text{Total moles}} \]
\[ = \frac{\left(\frac{2}{3} \times 92\right) + \left(\frac{2}{3} \times 46\right)}{\frac{4}{3}} \]
\[ = \frac{\frac{184}{3} + \frac{92}{3}}{\frac{4}{3}} = \frac{\frac{276}{3}}{\frac{4}{3}} = \frac{276}{4} = 69 \]

### Step 8: Calculate Vapor Density
Vapor density is defined as the average molecular weight divided by 2:
\[ \text{Vapor Density} = \frac{69}{2} = 34.5 \]

### Final Answer
The vapor density of the equilibrium mixture is **34.5**.