One mole of `N_(2)O_(4)(g)` at 300 K is kept in a closed container under one atmosphere. It is heated to 600K when 20% by mass of `N_(2)O_(4)`(g) decomposes to `NO_(2)`(g). The resultant pressure is:

Options

A. 1.2 atm
B. 2.4 atm
C. 2 atm
D. 1 atm

Explanation

To solve the problem step-by-step, we will follow the decomposition reaction of \( N_2O_4 \) and use the ideal gas law to find the resultant pressure after heating.

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

### Step 2: Determine the initial conditions
Initially, we have:
- 1 mole of \( N_2O_4 \)
- Temperature \( T_1 = 300 \, K \)
- Pressure \( P_1 = 1 \, atm \)

### Step 3: Calculate the mass of \( N_2O_4 \) that decomposes
Given that 20% by mass of \( N_2O_4 \) decomposes:
- Molar mass of \( N_2O_4 \) = 92 g/mol
- Mass of 1 mole of \( N_2O_4 \) = 92 g
- Mass decomposed = 20% of 92 g = \( 0.2 \times 92 = 18.4 \, g \)

### Step 4: Calculate the moles of \( N_2O_4 \) that decomposed
Using the molar mass:
\[ \text{Moles of } N_2O_4 \text{ decomposed} = \frac{18.4 \, g}{92 \, g/mol} = 0.2 \, mol \]

### Step 5: Determine the moles of \( N_2O_4 \) and \( NO_2 \) at equilibrium
- Moles of \( N_2O_4 \) remaining = \( 1 - 0.2 = 0.8 \, mol \)
- Moles of \( NO_2 \) produced = \( 2 \times 0.2 = 0.4 \, mol \)

### Step 6: Calculate the total moles at equilibrium
Total moles at equilibrium:
\[ N_2 = 0.8 \, mol + 0.4 \, mol = 1.2 \, mol \]

### Step 7: Use the ideal gas law to find the final pressure
Using the relation \( \frac{P_1}{N_1 T_1} = \frac{P_2}{N_2 T_2} \):
- \( P_1 = 1 \, atm \)
- \( N_1 = 1 \, mol \)
- \( T_1 = 300 \, K \)
- \( N_2 = 1.2 \, mol \)
- \( T_2 = 600 \, K \)

Substituting the values:
\[ \frac{1 \, atm}{1 \, mol \times 300 \, K} = \frac{P_2}{1.2 \, mol \times 600 \, K} \]

### Step 8: Solve for \( P_2 \)
Cross-multiplying gives:
\[ P_2 = \frac{1 \, atm \times 1.2 \, mol \times 600 \, K}{1 \, mol \times 300 \, K} \]
\[ P_2 = \frac{1.2 \times 600}{300} = \frac{720}{300} = 2.4 \, atm \]

### Final Answer
The resultant pressure is \( P_2 = 2.4 \, atm \).

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