1.0 mole of `AB_(5)(g)` is placed in a closed container under one atmosphere and at 300K. It is heated to 600K, when 20% by mass of it dissociates as
`AB_(5)(g)toAB(g)+2B_(2)(g)`. The resultant pressure is

Options

A. 1.2 atm
B. 2.4 atm
C. 1.4 atm
D. 2.8 atm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial conditions
We have 1 mole of \( AB_5(g) \) in a closed container at 1 atmosphere and 300 K.

### Step 2: Calculate the amount dissociated
Given that 20% by mass of \( AB_5 \) dissociates, we first need to find the mass of \( AB_5 \). Since we have 1 mole, we can denote the molar mass of \( AB_5 \) as \( M \). The mass of 1 mole of \( AB_5 \) is \( M \) grams.

The mass that dissociates is:
\[ \text{Mass dissociated} = 0.2 \times M = 0.2M \text{ grams} \]

### Step 3: Calculate the moles of \( AB_5 \) that dissociate
The number of moles that dissociate can be calculated using the molar mass:
\[ \text{Moles dissociated} = \frac{0.2M}{M} = 0.2 \text{ moles} \]

### Step 4: Determine the moles of products formed
From the reaction:
\[ AB_5(g) \rightarrow AB(g) + 2B_2(g) \]
For every mole of \( AB_5 \) that dissociates, 1 mole of \( AB \) and 2 moles of \( B_2 \) are produced. Therefore, if 0.2 moles of \( AB_5 \) dissociate:
- Moles of \( AB \) produced = 0.2 moles
- Moles of \( B_2 \) produced = \( 2 \times 0.2 = 0.4 \) moles

### Step 5: Calculate the total moles at equilibrium
Initially, we had 1 mole of \( AB_5 \). After dissociation:
- Remaining moles of \( AB_5 \) = \( 1 - 0.2 = 0.8 \) moles
- Total moles at equilibrium = moles of \( AB_5 \) + moles of \( AB \) + moles of \( B_2 \)
\[ \text{Total moles} = 0.8 + 0.2 + 0.4 = 1.4 \text{ moles} \]

### Step 6: Use the ideal gas law to find the new pressure
We can use the ideal gas law, \( PV = nRT \). Since the volume \( V \) and the gas constant \( R \) are constant, we can set up a ratio of the pressures and moles at the two different temperatures:
\[ \frac{P_1}{P_2} = \frac{n_1 T_1}{n_2 T_2} \]
Where:
- \( P_1 = 1 \) atm (initial pressure)
- \( n_1 = 1 \) mole (initial moles)
- \( T_1 = 300 \) K (initial temperature)
- \( n_2 = 1.4 \) moles (moles at equilibrium)
- \( T_2 = 600 \) K (final temperature)

Rearranging gives:
\[ P_2 = P_1 \times \frac{n_2}{n_1} \times \frac{T_2}{T_1} \]
Substituting the values:
\[ P_2 = 1 \times \frac{1.4}{1} \times \frac{600}{300} = 1 \times 1.4 \times 2 = 2.8 \text{ atm} \]

### Final Answer
The resultant pressure at 600 K is **2.8 atm**.