For the reaction, `N_(2)O_(4)(g) hArr 2NO_(2)(g)`, the concentration of an equilibrium mixture at `298 K` is `N_(2)O_(4)=4.50xx10^(-2) mol L^(-1)` and `NO_(2)=1.61xx10^(-2) mol L^(-1)`. What is the value of equilibrium constant?

Explanation

To find the equilibrium constant (K_eq) for the reaction \( N_2O_4(g) \rightleftharpoons 2NO_2(g) \), we will follow these steps:

### Step 1: Write the expression for the equilibrium constant
The equilibrium constant expression for the reaction is given by:
\[ K_{eq} = \frac{[NO_2]^2}{[N_2O_4]} \]
where \([NO_2]\) is the concentration of nitrogen dioxide and \([N_2O_4]\) is the concentration of dinitrogen tetroxide at equilibrium.

### Step 2: Substitute the given concentrations into the expression
From the problem, we know:
- \([N_2O_4] = 4.50 \times 10^{-2} \, \text{mol L}^{-1}\)
- \([NO_2] = 1.61 \times 10^{-2} \, \text{mol L}^{-1}\)

Now, substituting these values into the equilibrium expression:
\[ K_{eq} = \frac{(1.61 \times 10^{-2})^2}{4.50 \times 10^{-2}} \]

### Step 3: Calculate the square of the concentration of \(NO_2\)
Calculating \((1.61 \times 10^{-2})^2\):
\[ (1.61 \times 10^{-2})^2 = 2.5921 \times 10^{-4} \]

### Step 4: Divide by the concentration of \(N_2O_4\)
Now, substituting back into the expression:
\[ K_{eq} = \frac{2.5921 \times 10^{-4}}{4.50 \times 10^{-2}} \]

### Step 5: Perform the division
Calculating the division:
\[ K_{eq} = \frac{2.5921 \times 10^{-4}}{4.50 \times 10^{-2}} = 0.0576 \]

### Step 6: Adjust the scientific notation
To express this in scientific notation:
\[ K_{eq} = 5.76 \times 10^{-2} \, \text{mol L}^{-1} \]

### Final Answer
Thus, the value of the equilibrium constant \( K_{eq} \) is:
\[ K_{eq} = 5.76 \times 10^{-2} \, \text{mol L}^{-1} \]

---