For the reaction `N_2 + 2O_2 hArr 2NO_2`
the equilibrium concentrations in a 1.5L reaction vessel are 5.0 mol of `N_2`, 7.0 mol of `O_2`, and 0.10 mol of `NO_2` Calculate equilibrium constant.

Explanation

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

\[ N_2 + 2O_2 \rightleftharpoons 2NO_2 \]

given the equilibrium concentrations in a 1.5 L reaction vessel, we will follow these steps:

### Step 1: Write the equilibrium expression

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

\[ K_c = \frac{[NO_2]^2}{[N_2][O_2]^2} \]

### Step 2: Calculate the equilibrium concentrations

We have the following moles at equilibrium:

- \( N_2 = 5.0 \, \text{mol} \)
- \( O_2 = 7.0 \, \text{mol} \)
- \( NO_2 = 0.10 \, \text{mol} \)

To find the equilibrium concentrations, we divide the number of moles by the volume of the reaction vessel (1.5 L):

\[ [N_2] = \frac{5.0 \, \text{mol}}{1.5 \, \text{L}} = \frac{5.0}{1.5} \approx 3.33 \, \text{mol/L} \]

\[ [O_2] = \frac{7.0 \, \text{mol}}{1.5 \, \text{L}} = \frac{7.0}{1.5} \approx 4.67 \, \text{mol/L} \]

\[ [NO_2] = \frac{0.10 \, \text{mol}}{1.5 \, \text{L}} = \frac{0.10}{1.5} \approx 0.067 \, \text{mol/L} \]

### Step 3: Substitute the equilibrium concentrations into the \( K_c \) expression

Now we can substitute these concentrations into the equilibrium expression:

\[ K_c = \frac{(0.067)^2}{(3.33)(4.67)^2} \]

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

Calculating the numerator:

\[ (0.067)^2 \approx 0.004489 \]

Calculating the denominator:

\[ (4.67)^2 \approx 21.8089 \]
\[ K_c = \frac{0.004489}{3.33 \times 21.8089} \approx \frac{0.004489}{72.688} \approx 6.17 \times 10^{-5} \]

### Final Result

Thus, the equilibrium constant \( K_c \) for the reaction is approximately:

\[ K_c \approx 6.17 \times 10^{-5} \]

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