0.1 mole of `N_(2)O_(4)(g)` was sealed in a tude under one atmospheric conditions at `25^(@)C` Calculate the number of moles of `NO_(2)(g)` present , if the equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)(K_(P)=0.14)` is reached after some time :

Options

A. `1.8xx10^(2)`
B. `2.8xx10^(2)`
C. `0.036`
D. `2.8xx10^(-2)`

Answer

To solve the problem, we need to determine the number of moles of \( NO_2(g) \) present at equilibrium when \( 0.1 \) mole of \( N_2O_4(g) \) is sealed in a tube and the equilibrium constant \( K_p \) for the reaction is given as \( 0.14 \).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction is:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

2. **Set Up Initial Conditions:**
Initially, we have:
- Moles of \( N_2O_4 = 0.1 \) moles
- Moles of \( NO_2 = 0 \) moles

3. **Define Change in Moles:**
Let \( x \) be the number of moles of \( N_2O_4 \) that dissociate at equilibrium. Thus:
- Moles of \( N_2O_4 \) at equilibrium = \( 0.1 - x \)
- Moles of \( NO_2 \) at equilibrium = \( 2x \)

4. **Total Moles at Equilibrium:**
The total moles at equilibrium will be:
\[ \text{Total moles} = (0.1 - x) + 2x = 0.1 + x \]

5. **Write the Expression for \( K_p \):**
The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
where \( P \) represents the partial pressures.

6. **Calculate Partial Pressures:**
The partial pressures can be expressed in terms of mole fractions and total pressure (1 atm):
- \( P_{NO_2} = \frac{2x}{0.1 + x} \times 1 \)
- \( P_{N_2O_4} = \frac{0.1 - x}{0.1 + x} \times 1 \)

7. **Substitute into the \( K_p \) Expression:**
Substituting the expressions for partial pressures into the \( K_p \) equation:
\[ 0.14 = \frac{\left(\frac{2x}{0.1 + x}\right)^2}{\frac{0.1 - x}{0.1 + x}} \]
This simplifies to:
\[ 0.14 = \frac{4x^2}{(0.1 + x)^2} \cdot \frac{0.1 + x}{0.1 - x} \]
Which further simplifies to:
\[ 0.14 = \frac{4x^2}{(0.1 + x)(0.1 - x)} \]

8. **Cross Multiply:**
Cross multiplying gives:
\[ 0.14(0.1 + x)(0.1 - x) = 4x^2 \]
Expanding the left side:
\[ 0.14(0.01 - x^2) = 4x^2 \]
This leads to:
\[ 0.0014 - 0.14x^2 = 4x^2 \]
Combining like terms:
\[ 0.0014 = 4.14x^2 \]

9. **Solve for \( x^2 \):**
\[ x^2 = \frac{0.0014}{4.14} \approx 0.000338 \]
Taking the square root:
\[ x \approx 0.0184 \]

10. **Calculate Moles of \( NO_2 \):**
The number of moles of \( NO_2 \) at equilibrium is:
\[ \text{Moles of } NO_2 = 2x = 2 \times 0.0184 \approx 0.0368 \text{ moles} \]

### Final Answer:
The number of moles of \( NO_2(g) \) present at equilibrium is approximately \( 0.0368 \) moles.