An equilibrium mixture in a vessel of capacity `100` litre contain `1 "mol" N_(2).2"mol"O_(2) "and" 3 "mol" NO`. Number of moles of `O_(2)` to be added so that at new equilibrium the conc. Of `NO` is found to be `0.04` mol//lt.

Options

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

Answer

To solve the problem step by step, we will follow the outlined approach to find the number of moles of \( O_2 \) to be added so that the concentration of \( NO \) reaches \( 0.04 \) mol/L.

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

### Step 2: Calculate initial concentrations
Given:
- Volume of the vessel = 100 L
- Moles of \( N_2 = 1 \) mol
- Moles of \( O_2 = 2 \) mol
- Moles of \( NO = 3 \) mol

Using the formula for concentration:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]

Calculating the initial concentrations:
- Concentration of \( N_2 \):
\[ \text{Concentration of } N_2 = \frac{1 \text{ mol}}{100 \text{ L}} = 0.01 \text{ mol/L} \]
- Concentration of \( O_2 \):
\[ \text{Concentration of } O_2 = \frac{2 \text{ mol}}{100 \text{ L}} = 0.02 \text{ mol/L} \]
- Concentration of \( NO \):
\[ \text{Concentration of } NO = \frac{3 \text{ mol}}{100 \text{ L}} = 0.03 \text{ mol/L} \]

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

### Step 4: Establish new equilibrium after adding \( a \) moles of \( O_2 \)
Let \( a \) be the number of moles of \( O_2 \) added. The new concentrations will be:
- \( [N_2] = 0.01 - X \)
- \( [O_2] = 0.02 - \frac{X}{2} + \frac{a}{100} \)
- \( [NO] = 0.03 + \frac{2X}{100} \)

Given that the new concentration of \( NO \) is \( 0.04 \) mol/L:
\[ 0.03 + \frac{2X}{100} = 0.04 \]
This simplifies to:
\[ \frac{2X}{100} = 0.01 \Rightarrow 2X = 1 \Rightarrow X = 0.005 \]

### Step 5: Substitute \( X \) into the equilibrium expressions
Now substituting \( X \) back into the expressions for concentrations:
- New concentration of \( N_2 \):
\[ [N_2] = 0.01 - 0.005 = 0.005 \text{ mol/L} \]
- New concentration of \( O_2 \):
\[ [O_2] = 0.02 - \frac{0.005}{2} + \frac{a}{100} = 0.02 - 0.0025 + \frac{a}{100} = 0.0175 + \frac{a}{100} \]
- New concentration of \( NO \):
\[ [NO] = 0.04 \text{ mol/L} \]

### Step 6: Set up the equilibrium constant equation for the new equilibrium
Using the equilibrium constant \( K_c = 4.5 \):
\[ 4.5 = \frac{(0.04)^2}{(0.005)(0.0175 + \frac{a}{100})} \]
Calculating \( (0.04)^2 = 0.0016 \):
\[ 4.5 = \frac{0.0016}{0.005(0.0175 + \frac{a}{100})} \]
Cross-multiplying gives:
\[ 4.5 \times 0.005(0.0175 + \frac{a}{100}) = 0.0016 \]
\[ 0.0225(0.0175 + \frac{a}{100}) = 0.0016 \]
Dividing both sides by \( 0.0225 \):
\[ 0.0175 + \frac{a}{100} = \frac{0.0016}{0.0225} \approx 0.0711 \]

### Step 7: Solve for \( a \)
Rearranging gives:
\[ \frac{a}{100} = 0.0711 - 0.0175 = 0.0536 \]
Multiplying by \( 100 \):
\[ a = 5.36 \text{ moles} \]

### Conclusion
Thus, the number of moles of \( O_2 \) to be added is approximately \( 5.36 \) moles.

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