A sample of `N_(2)O_(4(g))` with a pressure of I atm is placed in a flask. When equilibrium is reached, 20% `N_(2)O_(4(g))` has been converted to `NO_(2(g)) .N_(2)O_(4(g)) hArr 2NO_(2(g))`, If the orginal pressure is made 10% of the earlier, then percent dissociation will be

Options

A. 0.2
B. 0.42
C. 0.54
D. 0.62

Answer

To solve the problem step by step, we will analyze the dissociation of \( N_2O_4 \) into \( NO_2 \) and calculate the percent dissociation when the pressure is reduced.

### Step 1: Understand the Reaction
The reaction given is:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

### Step 2: Initial Conditions
We start with a pressure of \( 1 \, \text{atm} \) of \( N_2O_4 \).

### Step 3: Calculate the Amount Dissociated
It is given that 20% of \( N_2O_4 \) has been converted to \( NO_2 \). Therefore, the amount of \( N_2O_4 \) that dissociates can be calculated as:
\[ \text{Amount dissociated} = 0.20 \times 1 \, \text{atm} = 0.2 \, \text{atm} \]

### Step 4: Calculate the Equilibrium Pressures
At equilibrium:
- The pressure of \( N_2O_4 \) remaining:
\[ P_{N_2O_4} = 1 - 0.2 = 0.8 \, \text{atm} \]
- The pressure of \( NO_2 \) produced (since 2 moles of \( NO_2 \) are produced for every mole of \( N_2O_4 \) that dissociates):
\[ P_{NO_2} = 2 \times 0.2 = 0.4 \, \text{atm} \]

### Step 5: Calculate the Equilibrium Constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(0.4)^2}{0.8} = \frac{0.16}{0.8} = 0.2 \]

### Step 6: Change the Pressure
Now, the original pressure is reduced to 10% of the earlier pressure:
\[ P = 0.1 \, \text{atm} \]

### Step 7: Set Up the Expression for Dissociation
Let \( \alpha \) be the degree of dissociation at this new pressure. The pressures at equilibrium will be:
- Pressure of \( N_2O_4 \) remaining:
\[ P_{N_2O_4} = 0.1 - \alpha \]
- Pressure of \( NO_2 \) produced:
\[ P_{NO_2} = 2\alpha \]

### Step 8: Write the Equilibrium Expression
Using the equilibrium constant calculated earlier:
\[ K_c = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Substituting the pressures:
\[ 0.2 = \frac{(2\alpha)^2}{0.1 - \alpha} \]

### Step 9: Solve the Equation
Expanding and rearranging gives:
\[ 0.2 = \frac{4\alpha^2}{0.1 - \alpha} \]
Multiplying both sides by \( 0.1 - \alpha \):
\[ 0.2(0.1 - \alpha) = 4\alpha^2 \]
\[ 0.02 - 0.2\alpha = 4\alpha^2 \]
Rearranging gives:
\[ 4\alpha^2 + 0.2\alpha - 0.02 = 0 \]

### Step 10: Use the Quadratic Formula
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 4, b = 0.2, c = -0.02 \):
\[ \alpha = \frac{-0.2 \pm \sqrt{(0.2)^2 - 4 \cdot 4 \cdot (-0.02)}}{2 \cdot 4} \]
\[ \alpha = \frac{-0.2 \pm \sqrt{0.04 + 0.32}}{8} \]
\[ \alpha = \frac{-0.2 \pm \sqrt{0.36}}{8} \]
\[ \alpha = \frac{-0.2 \pm 0.6}{8} \]

Calculating the two possible values:
1. \( \alpha = \frac{0.4}{8} = 0.05 \)
2. \( \alpha = \frac{-0.8}{8} \) (not physically meaningful)

Thus, \( \alpha = 0.05 \).

### Step 11: Calculate Percent Dissociation
The percent dissociation is given by:
\[ \text{Percent dissociation} = \alpha \times 100 = 0.05 \times 100 = 5\% \]

### Final Answer
The percent dissociation when the pressure is made 10% of the earlier is **5%**.