For a reaction equilibrium, `N_2O_4` ​ (g)⇌2`NO_2` ​ (g), the concentrations of N2O4 and NO2 ​ at equilibrium are 3.6 and 1.2 mol/L respectively. The value of K c ​ , for the reaction is:

Answer

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

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

given the concentrations of \( N_2O_4 \) and \( NO_2 \) at equilibrium, follow these steps:

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

The equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]

For the reaction \( N_2O_4 \rightleftharpoons 2NO_2 \), the expression becomes:

\[ K_c = \frac{[NO_2]^2}{[N_2O_4]} \]

### Step 2: Substitute the equilibrium concentrations

From the problem, we know:

- Concentration of \( N_2O_4 = 3.6 \, \text{mol/L} \)
- Concentration of \( NO_2 = 1.2 \, \text{mol/L} \)

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

\[ K_c = \frac{(1.2)^2}{3.6} \]

### Step 3: Calculate \( K_c \)

Now, calculate \( (1.2)^2 \):

\[ (1.2)^2 = 1.44 \]

Now substitute this value back into the equation for \( K_c \):

\[ K_c = \frac{1.44}{3.6} \]

### Step 4: Perform the division

Now, divide \( 1.44 \) by \( 3.6 \):

\[ K_c = 0.4 \]

### Final Answer

Thus, the value of \( K_c \) for the reaction is:

\[ \boxed{0.4} \]