For the reaction equilibrium,
` N_2O_4 (g) hArr 2NO_2 (g)`
the concentrations of ` N_2O_4 and NO_2 ` at equilibrium are ` 4.8 xx 10^(-2) and 1.2 xx 10 ^(-2) " mol " L^(-1)` respectively . The value of `K_c ` for the reaction is

Options

A. ` 3.3 xx 10 ^(-2) " mol " L^(-1)`
B. ` 3 xx 10 ^(-1) " mol " L^(-1)`
C. ` 3 xx 10 ^(-3) " mol " L^(-1)`
D. ` 3 xx 10 ^(3) " mol " L^(-1)`

Explanation

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

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

we will follow these steps:

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

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

\[ K_c = \frac{[\text{Products}]^{\text{stoichiometric coefficient}}}{[\text{Reactants}]^{\text{stoichiometric coefficient}}} \]

For the given reaction, the expression becomes:

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

### Step 2: Substitute the equilibrium concentrations

From the problem, we know the equilibrium concentrations:

- \( [N_2O_4] = 4.8 \times 10^{-2} \, \text{mol/L} \)
- \( [NO_2] = 1.2 \times 10^{-2} \, \text{mol/L} \)

Now, substitute these values into the \( K_c \) expression:

\[ K_c = \frac{(1.2 \times 10^{-2})^2}{(4.8 \times 10^{-2})} \]

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

First, calculate \( (1.2 \times 10^{-2})^2 \):

\[ (1.2 \times 10^{-2})^2 = 1.44 \times 10^{-4} \]

Now substitute this value into the \( K_c \) expression:

\[ K_c = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}} \]

Next, perform the division:

\[ K_c = \frac{1.44 \times 10^{-4}}{4.8 \times 10^{-2}} = \frac{1.44}{4.8} \times 10^{-4 + 2} = \frac{1.44}{4.8} \times 10^{-2} \]

Calculating \( \frac{1.44}{4.8} \):

\[ \frac{1.44}{4.8} = 0.3 \]

Thus,

\[ K_c = 0.3 \times 10^{-2} = 3.0 \times 10^{-3} \, \text{mol/L} \]

### Final Answer

The value of \( K_c \) for the reaction is:

\[ K_c = 3.0 \times 10^{-3} \, \text{mol/L} \]

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