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

Options

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

Explanation

To solve the problem step by step, we will analyze the equilibrium reaction and calculate the average molar mass of the equilibrium mixture.

### Step 1: Write the balanced reaction and define variables
The reaction is:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

Let:
- Initial moles of \( N_2O_4 \) = 1 (assuming 1 mole for simplicity)
- Change in moles of \( N_2O_4 \) = \( -\alpha \)
- Change in moles of \( NO_2 \) = \( +2\alpha \)

At equilibrium:
- Moles of \( N_2O_4 \) = \( 1 - \alpha \)
- Moles of \( NO_2 \) = \( 2\alpha \)

### Step 2: Write the expression for the equilibrium constant \( K_P \)
The equilibrium constant \( K_P \) is given by:
\[ K_P = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]

Using the ideal gas law, we can express the partial pressures in terms of total pressure \( P \):
- Total pressure \( P = P_{N_2O_4} + P_{NO_2} \)
- \( P_{N_2O_4} = \frac{(1 - \alpha)}{(1 + \alpha)} P \)
- \( P_{NO_2} = \frac{2\alpha}{(1 + \alpha)} P \)

Substituting these into the \( K_P \) expression:
\[ K_P = \frac{\left(\frac{2\alpha}{1 + \alpha} P\right)^2}{\frac{(1 - \alpha)}{1 + \alpha} P} \]

### Step 3: Simplify the expression for \( K_P \)
This simplifies to:
\[ K_P = \frac{4\alpha^2 P}{(1 - \alpha)(1 + \alpha)} \]

Given that \( K_P = 4.5 \) atm, we can set up the equation:
\[ 4.5 = \frac{4\alpha^2 \cdot 2}{(1 - \alpha)(1 + \alpha)} \]

### Step 4: Solve for \( \alpha \)
Substituting \( P = 2 \) atm into the equation:
\[ 4.5 = \frac{8\alpha^2}{(1 - \alpha)(1 + \alpha)} \]

Cross-multiplying gives:
\[ 4.5(1 - \alpha)(1 + \alpha) = 8\alpha^2 \]

Expanding and rearranging:
\[ 4.5 - 4.5\alpha^2 = 8\alpha^2 \]
\[ 12.5\alpha^2 = 4.5 \]
\[ \alpha^2 = \frac{4.5}{12.5} = 0.36 \]
\[ \alpha = 0.6 \]

### Step 5: Calculate the average molar mass of the mixture
The average molar mass \( M_{avg} \) can be calculated using:
\[ M_{avg} = \frac{(1 - \alpha)M_{N_2O_4} + (2\alpha)M_{NO_2}}{(1 - \alpha + 2\alpha)} \]

Using molar masses:
- \( M_{N_2O_4} = 92 \, g/mol \)
- \( M_{NO_2} = 46 \, g/mol \)

Substituting values:
\[ M_{avg} = \frac{(1 - 0.6) \cdot 92 + (2 \cdot 0.6) \cdot 46}{1 - 0.6 + 2 \cdot 0.6} \]
\[ M_{avg} = \frac{0.4 \cdot 92 + 1.2 \cdot 46}{1.6} \]
\[ M_{avg} = \frac{36.8 + 55.2}{1.6} \]
\[ M_{avg} = \frac{92}{1.6} = 57.5 \, g/mol \]

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

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