`K_(c)` for `N_(2)O_(4)(g) hArr 2NO_(2)(g)` is `0.00466` at `298 K`. If a `1L` container initially contained `0.8` mol of `N_(2)O_(4)`, what would be the concentrations of `N_(2)O_(4)` and `NO_(2)` at equilibrium? Also calculate the equilibrium concentration of `N_(2)O_(4)` and `NO_(2)` if the volume is halved at the same temperature.

Explanation

To solve the problem, we will go through the following steps:

### Step 1: Write the equilibrium expression
For the reaction:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[NO_2]^2}{[N_2O_4]} \]

### Step 2: Set up the initial concentrations
Initially, we have:
- Moles of \( N_2O_4 = 0.8 \) mol in a 1 L container.
- Therefore, the initial concentration of \( N_2O_4 \) is:
\[ [N_2O_4] = 0.8 \, \text{mol/L} \]
- The initial concentration of \( NO_2 \) is:
\[ [NO_2] = 0 \, \text{mol/L} \]

### Step 3: Define changes at equilibrium
Let \( x \) be the amount of \( N_2O_4 \) that dissociates at equilibrium. Thus:
- At equilibrium, the concentration of \( N_2O_4 \) will be:
\[ [N_2O_4] = 0.8 - x \]
- The concentration of \( NO_2 \) will be:
\[ [NO_2] = 2x \]

### Step 4: Substitute into the equilibrium expression
Substituting these values into the equilibrium expression:
\[ K_c = \frac{(2x)^2}{(0.8 - x)} \]
Given \( K_c = 0.00466 \):
\[ 0.00466 = \frac{4x^2}{0.8 - x} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 0.00466(0.8 - x) = 4x^2 \]
\[ 0.003728 - 0.00466x = 4x^2 \]
Rearranging further:
\[ 4x^2 + 0.00466x - 0.003728 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where:
- \( a = 4 \)
- \( b = 0.00466 \)
- \( c = -0.003728 \)

Calculating the discriminant:
\[ b^2 - 4ac = (0.00466)^2 - 4 \cdot 4 \cdot (-0.003728) \]

Calculating:
\[ = 0.0000217156 + 0.059648 = 0.0596697156 \]

Now, substituting into the quadratic formula:
\[ x = \frac{-0.00466 \pm \sqrt{0.0596697156}}{8} \]

Calculating \( \sqrt{0.0596697156} \):
\[ \approx 0.244 \]

Thus:
\[ x = \frac{-0.00466 \pm 0.244}{8} \]

Taking the positive root:
\[ x \approx \frac{0.23934}{8} \approx 0.02992 \]

### Step 7: Calculate equilibrium concentrations
Now substituting \( x \) back to find concentrations:
- Concentration of \( N_2O_4 \):
\[ [N_2O_4] = 0.8 - 0.02992 \approx 0.77008 \, \text{mol/L} \]
- Concentration of \( NO_2 \):
\[ [NO_2] = 2x = 2 \cdot 0.02992 \approx 0.05984 \, \text{mol/L} \]

### Step 8: Calculate for halved volume
If the volume is halved (0.5 L), the initial concentration of \( N_2O_4 \) becomes:
\[ [N_2O_4] = \frac{0.8}{0.5} = 1.6 \, \text{mol/L} \]

Following similar steps:
Let \( y \) be the amount of \( N_2O_4 \) that dissociates:
- At equilibrium:
\[ [N_2O_4] = 1.6 - y \]
\[ [NO_2] = 2y \]

Using the equilibrium expression:
\[ K_c = \frac{(2y)^2}{(1.6 - y)} = 0.00466 \]
\[ 0.00466(1.6 - y) = 4y^2 \]
\[ 0.007456 - 0.00466y = 4y^2 \]
Rearranging:
\[ 4y^2 + 0.00466y - 0.007456 = 0 \]

Using the quadratic formula again:
Where \( a = 4, b = 0.00466, c = -0.007456 \):
Calculate discriminant:
\[ b^2 - 4ac = (0.00466)^2 - 4 \cdot 4 \cdot (-0.007456) \]
Calculating gives:
\[ 0.0000217156 + 0.119296 = 0.1193177156 \]

Now using the quadratic formula:
\[ y = \frac{-0.00466 \pm \sqrt{0.1193177156}}{8} \]
Calculating \( \sqrt{0.1193177156} \approx 0.345 \):
Taking the positive root:
\[ y \approx \frac{0.34034}{8} \approx 0.04254 \]

### Final Equilibrium Concentrations for Halved Volume
- Concentration of \( N_2O_4 \):
\[ [N_2O_4] = 1.6 - 0.04254 \approx 1.55746 \, \text{mol/L} \]
- Concentration of \( NO_2 \):
\[ [NO_2] = 2y = 2 \cdot 0.04254 \approx 0.08508 \, \text{mol/L} \]

### Summary of Results
1. For the original volume (1 L):
- \( [N_2O_4] \approx 0.77008 \, \text{mol/L} \)
- \( [NO_2] \approx 0.05984 \, \text{mol/L} \)

2. For the halved volume (0.5 L):
- \( [N_2O_4] \approx 1.55746 \, \text{mol/L} \)
- \( [NO_2] \approx 0.08508 \, \text{mol/L} \)