The equilibrium constant `K_(p)` for the following reaction is `4.5`
`N_(2)O_(4)(g)hArr2NO_(2)(g)` What would be the average molar mass `("in"g//mol)` of an equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` formed by the dissociation of pure `N_(2)O_(4)` at a total pressure of `2` atm ?

Options

A. `69`
B. `57.5`
C. `80.5`
D. `85.5`

Answer

To find the average molar mass of an equilibrium mixture of \( N_2O_4 \) and \( NO_2 \) formed by the dissociation of pure \( N_2O_4 \) at a total pressure of 2 atm, we can follow these steps:

### Step 1: Write the Reaction and Define Variables
The reaction is:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]
Let:
- \( \alpha \) = degree of dissociation of \( N_2O_4 \)
- Initially, we have 1 mole of \( N_2O_4 \) and 0 moles of \( NO_2 \).

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

Total moles at equilibrium:
\[ n_{total} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 3: Calculate Partial Pressures
Using the total pressure \( P = 2 \) atm, we can find the partial pressures:
- Partial pressure of \( NO_2 \):
\[ P_{NO_2} = \left(\frac{2\alpha}{1 + \alpha}\right) \cdot 2 = \frac{4\alpha}{1 + \alpha} \]
- Partial pressure of \( N_2O_4 \):
\[ P_{N_2O_4} = \left(\frac{1 - \alpha}{1 + \alpha}\right) \cdot 2 = \frac{2(1 - \alpha)}{1 + \alpha} \]

### Step 4: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{\left(\frac{4\alpha}{1 + \alpha}\right)^2}{\frac{2(1 - \alpha)}{1 + \alpha}} \]
Substituting \( K_p = 4.5 \):
\[ 4.5 = \frac{16\alpha^2}{2(1 - \alpha)(1 + \alpha)} = \frac{8\alpha^2}{(1 - \alpha)(1 + \alpha)} \]

### Step 5: Cross-Multiply and Rearrange
Cross-multiplying gives:
\[ 4.5(1 - \alpha)(1 + \alpha) = 8\alpha^2 \]
Expanding and rearranging:
\[ 4.5 - 4.5\alpha^2 = 8\alpha^2 \implies 4.5 = 12.5\alpha^2 \implies \alpha^2 = \frac{4.5}{12.5} = 0.36 \]
Taking the square root:
\[ \alpha = 0.6 \]

### Step 6: Calculate Molar Mass of the Mixture
Now, we can calculate the average molar mass of the mixture:
\[ \text{Average Molar Mass} = \left(\frac{1 - \alpha}{1 + \alpha} \cdot M_{N_2O_4}\right) + \left(\frac{2\alpha}{1 + \alpha} \cdot M_{NO_2}\right) \]
Where:
- \( M_{N_2O_4} = 92 \, \text{g/mol} \)
- \( M_{NO_2} = 46 \, \text{g/mol} \)

Calculating the mole fractions:
- Mole fraction of \( N_2O_4 \):
\[ \frac{1 - 0.6}{1 + 0.6} = \frac{0.4}{1.6} = 0.25 \]
- Mole fraction of \( NO_2 \):
\[ \frac{2 \cdot 0.6}{1 + 0.6} = \frac{1.2}{1.6} = 0.75 \]

Now substituting into the average molar mass formula:
\[ \text{Average Molar Mass} = (0.25 \cdot 92) + (0.75 \cdot 46) \]
Calculating:
\[ = 23 + 34.5 = 57.5 \, \text{g/mol} \]

### Final Answer
The average molar mass of the equilibrium mixture is approximately \( 57.5 \, \text{g/mol} \).

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