Reaction between nitrogen and oxygen takes place as following:
`2N_(2(g))+O_(2)hArr2N_(2)O_((g))`
If a mixture of `0.482 "mole" N_(2)` and `0.933 "mole"` of `O_(2)` is placed in a reaction vessel of volume `10 litre` and allowed to form `N_(2)O` at a temperature for which `K_(c)=2.0xx10^(-37)litre mol^(-1)`. Determine the composition of equilibrium mixture.

Explanation

To determine the composition of the equilibrium mixture for the reaction:

\[ 2N_{2(g)} + O_{2(g)} \rightleftharpoons 2N_{2}O_{(g)} \]

we will follow these steps:

### Step 1: Write the balanced equation
The balanced equation for the reaction is already given as:

\[ 2N_{2(g)} + O_{2(g)} \rightleftharpoons 2N_{2}O_{(g)} \]

### Step 2: Determine initial moles and concentrations
We are given:
- Initial moles of \( N_2 = 0.482 \, \text{moles} \)
- Initial moles of \( O_2 = 0.933 \, \text{moles} \)
- Volume of the reaction vessel = 10 L

To find the initial concentrations:
- Concentration of \( N_2 = \frac{0.482 \, \text{moles}}{10 \, \text{L}} = 0.0482 \, \text{mol/L} \)
- Concentration of \( O_2 = \frac{0.933 \, \text{moles}}{10 \, \text{L}} = 0.0933 \, \text{mol/L} \)

### Step 3: Set up the change in concentration at equilibrium
Let \( x \) be the amount of \( N_2 \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- Change in concentration of \( N_2 = -2x \)
- Change in concentration of \( O_2 = -x \)
- Change in concentration of \( N_2O = +2x \)

At equilibrium, the concentrations will be:
- \( [N_2] = 0.0482 - 2x \)
- \( [O_2] = 0.0933 - x \)
- \( [N_2O] = 2x \)

### Step 4: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[N_2O]^2}{[N_2]^2[O_2]} \]

Substituting the equilibrium concentrations:

\[ K_c = \frac{(2x)^2}{(0.0482 - 2x)^2(0.0933 - x)} \]

### Step 5: Substitute the value of \( K_c \)
Given \( K_c = 2.0 \times 10^{-37} \), we can set up the equation:

\[ 2.0 \times 10^{-37} = \frac{(2x)^2}{(0.0482 - 2x)^2(0.0933 - x)} \]

### Step 6: Approximate \( x \)
Since \( K_c \) is very small, we can assume that \( x \) is very small compared to the initial concentrations. Thus, we can approximate:

\[ 0.0482 - 2x \approx 0.0482 \quad \text{and} \quad 0.0933 - x \approx 0.0933 \]

This simplifies our equation to:

\[ 2.0 \times 10^{-37} = \frac{(2x)^2}{(0.0482)^2(0.0933)} \]

### Step 7: Solve for \( x \)
Rearranging gives:

\[ (2x)^2 = 2.0 \times 10^{-37} \times (0.0482)^2 \times (0.0933) \]

Calculating the right-hand side:

\[ (2x)^2 = 2.0 \times 10^{-37} \times 0.002321 \times 0.0933 \]

Calculating this will yield:

\[ (2x)^2 \approx 4.33 \times 10^{-39} \]

Taking the square root:

\[ 2x \approx 6.58 \times 10^{-20} \quad \Rightarrow \quad x \approx 3.29 \times 10^{-20} \]

### Step 8: Calculate equilibrium concentrations
Now we can find the equilibrium concentrations:

- \( [N_2] = 0.0482 - 2(3.29 \times 10^{-20}) \approx 0.0482 \)
- \( [O_2] = 0.0933 - (3.29 \times 10^{-20}) \approx 0.0933 \)
- \( [N_2O] = 2(3.29 \times 10^{-20}) \approx 6.58 \times 10^{-20} \)

### Final Answer
The composition of the equilibrium mixture is approximately:
- \( [N_2] \approx 0.0482 \, \text{mol/L} \)
- \( [O_2] \approx 0.0933 \, \text{mol/L} \)
- \( [N_2O] \approx 6.58 \times 10^{-20} \, \text{mol/L} \)