For the following equilibrium reaction `N_(2)O_(4)(g)hArr 2NO_(2)(g)`, `NO_(2)` is 50% of the total volume at a given temperature. Hence, vapour density of the equilibrium mixture is `:`

Options

A. 34.5
B. 25
C. 23
D. 20

Explanation

To solve the problem of finding the vapor density of the equilibrium mixture for the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \), where \( NO_2 \) constitutes 50% of the total volume at a given temperature, we can follow these steps:

### Step 1: Understanding the Reaction
The equilibrium reaction is:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]
Let’s denote the initial moles of \( N_2O_4 \) as 1 mole. At equilibrium, let \( x \) be the degree of dissociation of \( N_2O_4 \).

### Step 2: Establishing Moles at Equilibrium
At equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - x \)
- Moles of \( NO_2 \) = \( 2x \)

Thus, the total moles at equilibrium will be:
\[ \text{Total moles} = (1 - x) + 2x = 1 + x \]

### Step 3: Using the Given Information
We know that \( NO_2 \) constitutes 50% of the total volume. Therefore, the moles of \( NO_2 \) at equilibrium can be expressed as:
\[ \frac{2x}{1 + x} = 0.5 \]

### Step 4: Solving for \( x \)
Cross-multiplying gives:
\[ 2x = 0.5(1 + x) \]
Expanding this, we get:
\[ 2x = 0.5 + 0.5x \]
Rearranging terms:
\[ 2x - 0.5x = 0.5 \implies 1.5x = 0.5 \implies x = \frac{1}{3} \]

### Step 5: Finding the Molar Masses
The molar mass of \( N_2O_4 \) is:
\[ M(N_2O_4) = 2(14) + 4(16) = 28 + 64 = 92 \, \text{g/mol} \]
The molar mass of \( NO_2 \) is:
\[ M(NO_2) = 14 + 2(16) = 14 + 32 = 46 \, \text{g/mol} \]

### Step 6: Calculating Vapor Density
Vapor density (\( D \)) is defined as:
\[ D = \frac{\text{Molar mass}}{\text{Volume}} \]
Using the formula for vapor density in terms of the degree of dissociation:
\[ D = \frac{M(N_2O_4)(1 - x) + M(NO_2)(2x)}{1 + x} \]
Substituting the values:
\[ D = \frac{92(1 - \frac{1}{3}) + 46(2 \cdot \frac{1}{3})}{1 + \frac{1}{3}} \]
Calculating the numerator:
\[ D = \frac{92 \cdot \frac{2}{3} + 46 \cdot \frac{2}{3}}{\frac{4}{3}} = \frac{\frac{184 + 92}{3}}{\frac{4}{3}} = \frac{276/3}{4/3} = \frac{276}{4} = 69 \]

### Step 7: Final Calculation for Vapor Density
Since we need to find the vapor density in terms of the average molar mass:
\[ D = \frac{3 \times 46}{4} = 34.5 \]

### Final Answer
Thus, the vapor density of the equilibrium mixture is:
\[ \boxed{34.5} \]