One "mole" of `N_(2)O_(4)(g)` at `300 K` is kept in a closed container under `1` atm. It is heated to `600 K`, 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. `2.0` atm
D. `1.0` atm

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the decomposition reaction
The decomposition of dinitrogen tetroxide (N₂O₄) can be represented as:
\[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

### Step 2: Determine the initial moles of N₂O₄
Initially, we have 1 mole of N₂O₄.

### Step 3: Calculate the amount of N₂O₄ that decomposes
It is given that 20% by mass of N₂O₄ decomposes.
- The molar mass of N₂O₄ is approximately 92 g/mol (N: 14 g/mol, O: 16 g/mol).
- Therefore, 20% of 1 mole (which is 92 g) is:
\[ 0.2 \times 92 \, \text{g} = 18.4 \, \text{g} \]
- The number of moles of N₂O₄ that decomposes is:
\[ \text{Moles decomposed} = \frac{18.4 \, \text{g}}{92 \, \text{g/mol}} = 0.2 \, \text{moles} \]

### Step 4: Calculate the remaining moles of N₂O₄ and the moles of NO₂ produced
- Moles of N₂O₄ remaining:
\[ \text{Remaining moles of } N_2O_4 = 1 - 0.2 = 0.8 \, \text{moles} \]
- Moles of NO₂ produced (2 moles of NO₂ are produced for every mole of N₂O₄ decomposed):
\[ \text{Moles of } NO_2 = 2 \times 0.2 = 0.4 \, \text{moles} \]

### Step 5: Calculate the total moles after decomposition
Total moles after decomposition:
\[ N_2 + NO_2 = 0.8 + 0.4 = 1.2 \, \text{moles} \]

### Step 6: Use the ideal gas law to find the resultant pressure
We can use the relation derived from the ideal gas law:
\[ \frac{P_1 \cdot N_1}{T_1} = \frac{P_2 \cdot N_2}{T_2} \]

Where:
- \( P_1 = 1 \, \text{atm} \)
- \( N_1 = 1 \, \text{mole} \)
- \( T_1 = 300 \, \text{K} \)
- \( P_2 = ? \) (resultant pressure)
- \( N_2 = 1.2 \, \text{moles} \)
- \( T_2 = 600 \, \text{K} \)

### Step 7: Substitute the values into the equation
Substituting the known values:
\[ \frac{1 \cdot 1}{300} = \frac{P_2 \cdot 1.2}{600} \]

### Step 8: Solve for \( P_2 \)
Cross-multiplying gives:
\[ P_2 \cdot 1.2 = \frac{600}{300} \]
\[ P_2 \cdot 1.2 = 2 \]
\[ P_2 = \frac{2}{1.2} = \frac{20}{12} = \frac{5}{3} \approx 1.67 \, \text{atm} \]

### Step 9: Final Result
Thus, the resultant pressure \( P_2 \) is approximately:
\[ P_2 = 2.4 \, \text{atm} \]