An equilibrium mixture in a vessel of capacity 100 litre contain 1 mol `N_(2)`, 2 mol `O_(2)` and 3 mol NO. Find number of moles of `O_(2)` to be added, so that at new equilibrium the concentration of No is 0.04 mol/lit.

Options

A. `(101)/(18)`
B. `(101)/(9)`
C. `(202)/(9)`
D. None of these

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the balanced chemical equation
The reaction is:
\[ N_2 + O_2 \rightleftharpoons 2 NO \]

### Step 2: Determine the initial concentrations
Given the vessel's capacity is 100 liters, we can calculate the initial concentrations of each species:
- Moles of \( N_2 = 1 \)
- Moles of \( O_2 = 2 \)
- Moles of \( NO = 3 \)

The concentrations are:
\[ [C_{N_2}] = \frac{1 \text{ mol}}{100 \text{ L}} = 0.01 \text{ mol/L} \]
\[ [C_{O_2}] = \frac{2 \text{ mol}}{100 \text{ L}} = 0.02 \text{ mol/L} \]
\[ [C_{NO}] = \frac{3 \text{ mol}}{100 \text{ L}} = 0.03 \text{ mol/L} \]

### Step 3: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant expression for the reaction is:
\[ K_c = \frac{[NO]^2}{[N_2][O_2]} \]
Substituting the initial concentrations:
\[ K_c = \frac{(0.03)^2}{(0.01)(0.02)} = \frac{0.0009}{0.0002} = 4.5 \]

### Step 4: Determine the new equilibrium condition
We want the concentration of \( NO \) to be \( 0.04 \text{ mol/L} \). Therefore, the total moles of \( NO \) at equilibrium will be:
\[ \text{Moles of } NO = 0.04 \text{ mol/L} \times 100 \text{ L} = 4 \text{ mol} \]

### Step 5: Calculate the change in moles
Let \( x \) be the number of moles of \( O_2 \) added. At the new equilibrium:
- Moles of \( NO \) will be \( 3 + 2Y \) (where \( Y \) is the amount of \( N_2 \) reacted).
- Moles of \( N_2 \) will be \( 1 - Y \).
- Moles of \( O_2 \) will be \( 2 + x - Y \).

Setting the equation for \( NO \):
\[ 3 + 2Y = 4 \implies 2Y = 1 \implies Y = 0.5 \]

### Step 6: Substitute back to find moles of \( O_2 \)
Now substituting \( Y \) back:
- Moles of \( N_2 = 1 - 0.5 = 0.5 \)
- Moles of \( O_2 = 2 + x - 0.5 = 1.5 + x \)

### Step 7: Write the new equilibrium constant expression
At the new equilibrium, the concentrations are:
\[ [C_{N_2}] = \frac{0.5}{100}, \quad [C_{O_2}] = \frac{1.5 + x}{100}, \quad [C_{NO}] = \frac{4}{100} \]
Substituting into the \( K_c \) expression:
\[ 9 = \frac{(0.04)^2}{\left(\frac{0.5}{100}\right)\left(\frac{1.5 + x}{100}\right)} \]
\[ 9 = \frac{0.0016}{\frac{0.5(1.5 + x)}{10000}} \]
\[ 9 = \frac{0.0016 \times 10000}{0.5(1.5 + x)} \]
\[ 9 = \frac{16}{0.5(1.5 + x)} \implies 9 \times 0.5(1.5 + x) = 16 \]
\[ 4.5(1.5 + x) = 16 \implies 1.5 + x = \frac{16}{4.5} \approx 3.56 \]
\[ x \approx 3.56 - 1.5 = 2.06 \]

### Final Answer
The number of moles of \( O_2 \) to be added is approximately \( 2.06 \) moles.

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