One mole of `N_2O_4(g)` at 300 K I kept in a closed container under one atmosphere. It is heated to 600 k when `N_2O_4(g)` decomposes to `NO_2(g)` . If the resultant pressure is 2.4 atm, the percentage dissociation by mass of `N_2O_4(g)` is

Options

A. 0.1
B. 0.2
C. 0.3
D. 0.4

Explanation

To solve the problem of finding the percentage dissociation by mass of \( N_2O_4(g) \) when it is heated, we will follow these steps:

### Step 1: Write the decomposition reaction
The decomposition of \( N_2O_4(g) \) can be represented as:
\[ N_2O_4(g) \rightleftharpoons 2 NO_2(g) \]

### Step 2: Set up initial conditions
Initially, we have:
- Moles of \( N_2O_4 = 1 \)
- Moles of \( NO_2 = 0 \)

### Step 3: Define the change in moles
Let \( \alpha \) be the degree of dissociation of \( N_2O_4 \). At equilibrium:
- Moles of \( N_2O_4 = 1 - \alpha \)
- Moles of \( NO_2 = 2\alpha \)

### Step 4: Calculate total moles at equilibrium
The total number of moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + 2\alpha = 1 + \alpha \]

### Step 5: Use the ideal gas law to find the relationship between pressure and moles
Using the ideal gas law \( PV = nRT \), we can express the pressure at equilibrium:
Given that the total pressure at equilibrium is 2.4 atm, we can set up the equation:
\[ P = \frac{nRT}{V} \]
Where:
- \( P = 2.4 \, \text{atm} \)
- \( n = 1 + \alpha \)
- \( R \) is the gas constant
- \( T = 600 \, \text{K} \)

### Step 6: Calculate the volume using initial conditions
Initially, at 300 K and 1 atm, we can find the volume:
\[ PV = nRT \implies 1 \times V = 1 \times R \times 300 \implies V = 300R \]

### Step 7: Substitute volume back into the equilibrium pressure equation
Now substituting back into the equilibrium pressure equation:
\[ 2.4 = \frac{(1 + \alpha)R \times 600}{300R} \]
Cancelling \( R \) from both sides:
\[ 2.4 = \frac{(1 + \alpha) \times 600}{300} \]

### Step 8: Simplify and solve for \( \alpha \)
Simplifying gives:
\[ 2.4 = 2(1 + \alpha) \implies 1 + \alpha = 1.2 \implies \alpha = 0.2 \]

### Step 9: Calculate the percentage dissociation
The percentage dissociation is given by:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.2 \times 100 = 20\% \]

### Final Answer
The percentage dissociation by mass of \( N_2O_4(g) \) is **20%**.

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