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, we will follow these steps:

### Step 1: Determine the mass of N₂O₄ that decomposes
Given that 20% by mass of 1 mole of N₂O₄ decomposes, we first need to find the mass of 1 mole of N₂O₄. The molar mass of N₂O₄ is approximately 92 g/mol.

\[ \text{Mass of N₂O₄} = 92 \text{ g} \]
\[ \text{Mass decomposed} = 20\% \text{ of } 92 \text{ g} = 0.20 \times 92 = 18.4 \text{ g} \]

### Step 2: Calculate the number of moles of N₂O₄ that decomposes
To find the number of moles of N₂O₄ that decomposes, we use the formula:

\[ \text{Moles of N₂O₄ decomposed} = \frac{\text{Mass decomposed}}{\text{Molar mass of N₂O₄}} = \frac{18.4 \text{ g}}{92 \text{ g/mol}} = 0.2 \text{ mol} \]

### Step 3: Determine the moles of products formed
According to the reaction:
\[ \text{N₂O₄} \rightarrow 2 \text{NO₂} \]
For every mole of N₂O₄ that decomposes, 2 moles of NO₂ are produced. Therefore, if 0.2 moles of N₂O₄ decompose, the moles of NO₂ produced will be:

\[ \text{Moles of NO₂} = 2 \times \text{Moles of N₂O₄ decomposed} = 2 \times 0.2 = 0.4 \text{ mol} \]

### Step 4: Calculate the total moles at equilibrium
Initially, we had 1 mole of N₂O₄. After 0.2 moles decompose, the remaining moles of N₂O₄ will be:

\[ \text{Remaining moles of N₂O₄} = 1 - 0.2 = 0.8 \text{ mol} \]

Thus, the total moles at equilibrium (N₂O₄ + NO₂) will be:

\[ \text{Total moles} = \text{Remaining moles of N₂O₄} + \text{Moles of NO₂} = 0.8 + 0.4 = 1.2 \text{ mol} \]

### Step 5: Use the ideal gas law to find the final pressure
We will use the ideal gas law, which states:

\[ PV = nRT \]

Since the volume (V) remains constant, we can relate the initial and final states:

\[ \frac{P_1}{n_1 T_1} = \frac{P_2}{n_2 T_2} \]

Where:
- \( P_1 = 1 \text{ atm} \)
- \( n_1 = 1 \text{ mol} \)
- \( T_1 = 300 \text{ K} \)
- \( n_2 = 1.2 \text{ mol} \)
- \( T_2 = 600 \text{ K} \)

Rearranging gives us:

\[ P_2 = P_1 \times \frac{n_2}{n_1} \times \frac{T_2}{T_1} \]

Substituting in the values:

\[ P_2 = 1 \text{ atm} \times \frac{1.2 \text{ mol}}{1 \text{ mol}} \times \frac{600 \text{ K}}{300 \text{ K}} = 1 \times 1.2 \times 2 = 2.4 \text{ atm} \]

### Final Answer
The resultant pressure is **2.4 atm**.

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