When the reaction, `2NO_(2)(g) hArr N_(2)O_(4)(g)` reaches equilibrium at `298 K`. The partial pressure of `NO_(2)` and `N_(2)O_(4)` are `0.2 Kpa` and `0.4 Kpa`, respectively. What is the equilibrium constant `K_(p)` of the above reaction at `298 K`?

Options

A. `0.1`
B. `0.5`
C. `1.0`
D. `10`

Explanation

To find the equilibrium constant \( K_p \) for the reaction

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

at 298 K, we will follow these steps:

### Step 1: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) is defined in terms of the partial pressures of the products and reactants. For the given reaction, the expression for \( K_p \) is:

\[ K_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2} \]

where \( P_{N_2O_4} \) is the partial pressure of \( N_2O_4 \) and \( P_{NO_2} \) is the partial pressure of \( NO_2 \).

### Step 2: Substitute the given values

From the problem, we know the partial pressures at equilibrium:

- \( P_{NO_2} = 0.2 \, \text{kPa} \)
- \( P_{N_2O_4} = 0.4 \, \text{kPa} \)

Substituting these values into the \( K_p \) expression:

\[ K_p = \frac{0.4 \, \text{kPa}}{(0.2 \, \text{kPa})^2} \]

### Step 3: Calculate \( (P_{NO_2})^2 \)

Calculating \( (0.2 \, \text{kPa})^2 \):

\[ (0.2 \, \text{kPa})^2 = 0.04 \, \text{kPa}^2 \]

### Step 4: Calculate \( K_p \)

Now substituting back into the equation for \( K_p \):

\[ K_p = \frac{0.4 \, \text{kPa}}{0.04 \, \text{kPa}^2} \]

Calculating this gives:

\[ K_p = 10 \]

### Conclusion

Thus, the equilibrium constant \( K_p \) for the reaction at 298 K is:

\[ K_p = 10 \]