The density of an equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` at 1 atm and 373.5K is 2.0 g/L.
Calculate `K_(C)` for the reaction `N_(2)O_(2)(g) iff 2NO_(2)(g)`

Answer

To calculate the equilibrium constant \( K_c \) for the reaction:

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

given the density of the equilibrium mixture, we can follow these steps:

### Step 1: Use the Ideal Gas Law to find the molar mass of the mixture

The ideal gas equation is:

\[ PV = nRT \]

Where:
- \( P \) = pressure (1 atm)
- \( V \) = volume (1 L for simplicity)
- \( n \) = number of moles
- \( R \) = gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (373.5 K)

We can relate density (\( D \)) to molar mass (\( M \)) using the equation:

\[ PM = DRT \]

Given that the density \( D \) is 2 g/L, we can rearrange the equation to find the molar mass \( M \):

\[ M = \frac{DRT}{P} \]

Substituting the values:

\[ M = \frac{(2 \, \text{g/L}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (373.5 \, \text{K})}{1 \, \text{atm}} \]

Calculating this gives:

\[ M = \frac{(2) \times (0.0821) \times (373.5)}{1} \approx 61.32 \, \text{g/mol} \]

### Step 2: Calculate the vapor density of the mixture

The vapor density (\( VD \)) is given by:

\[ VD = \frac{M}{2} \]

Thus,

\[ VD = \frac{61.32 \, \text{g/mol}}{2} \approx 30.66 \, \text{g/L} \]

### Step 3: Calculate the degree of dissociation (\( \alpha \))

The initial vapor density of \( N_2O_4 \) is:

\[ VD_{initial} = \frac{M_{N_2O_4}}{2} = \frac{92 \, \text{g/mol}}{2} = 46 \, \text{g/L} \]

Using the relationship between the initial and equilibrium vapor densities:

\[ \frac{VD_{initial}}{VD_{equilibrium}} = \frac{C}{C(1+\alpha)} \]

Where \( C \) is the concentration of \( N_2O_4 \) initially. Thus,

\[ \frac{46}{30.66} = \frac{1}{1+\alpha} \]

Solving for \( \alpha \):

\[ 1 + \alpha = \frac{46}{30.66} \]

\[ \alpha = \frac{46}{30.66} - 1 \approx 0.5 \]

### Step 4: Calculate the equilibrium concentrations

Assuming 1 mole of \( N_2O_4 \) initially:

- At equilibrium, \( N_2O_4 \) remaining = \( 1 - \alpha = 0.5 \)
- \( NO_2 \) produced = \( 2\alpha = 1 \)

Thus, the total moles at equilibrium = \( 0.5 + 1 = 1.5 \).

### Step 5: Calculate the partial pressures

The partial pressure of \( N_2O_4 \):

\[ P_{N_2O_4} = \frac{0.5}{1.5} \times 1 \, \text{atm} = \frac{1}{3} \, \text{atm} \]

The partial pressure of \( NO_2 \):

\[ P_{NO_2} = \frac{1}{1.5} \times 1 \, \text{atm} = \frac{2}{3} \, \text{atm} \]

### Step 6: Calculate \( K_p \)

Using the equilibrium expression:

\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(\frac{2}{3})^2}{\frac{1}{3}} = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{3} \]

### Step 7: Convert \( K_p \) to \( K_c \)

Using the relationship:

\[ K_p = K_c (RT)^{\Delta n} \]

Where \( \Delta n = 2 - 1 = 1 \):

\[ K_c = \frac{K_p}{RT} \]

Substituting the values:

\[ K_c = \frac{\frac{4}{3}}{(0.0821)(373.5)} \]

Calculating \( K_c \):

\[ K_c \approx \frac{1.33}{30.66} \approx 0.043 \, \text{mol/L} \]

### Final Answer:

\[ K_c \approx 0.043 \, \text{mol/L} \]

---