A reaction system in equilibrium according to reaction `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` in one litre vessel at a given temperature was found to be `0.12` mole each of `SO_(2)` and `SO_(3)` and `5` mole of `O_(2)` In another vessel of one litre contains `32` g of `SO_(2)` at the same temperature. What mass of `O_(2)` must be added to this vessel in order that at equilibrium `20%` of `SO_(2)` is oxidized to `SO_(3)`?

Options

A. `0.4125`g
B. `11.6` g
C. `1.6` g
D. None of these

Answer

To solve the problem step by step, we will follow the outlined process to determine the mass of \( O_2 \) that must be added to the vessel.

### Step 1: Understand the Reaction and Initial Conditions
The reaction is given as:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
In the first vessel, we have:
- \( [SO_2] = 0.12 \) moles
- \( [SO_3] = 0.12 \) moles
- \( [O_2] = 5 \) moles

### Step 2: Calculate the Equilibrium Constant \( K_c \)
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[SO_3]^2}{[SO_2]^2 \cdot [O_2]} \]
Substituting the values:
\[ K_c = \frac{(0.12)^2}{(0.12)^2 \cdot 5} = \frac{0.0144}{0.0144 \cdot 5} = \frac{0.0144}{0.072} = 0.2 \]

### Step 3: Analyze the Second Vessel
In the second vessel, we have \( 32 \) g of \( SO_2 \). First, we need to convert this mass into moles:
\[ \text{Moles of } SO_2 = \frac{32 \text{ g}}{64 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 4: Determine the Amount of \( SO_2 \) Oxidized
We need to find out how much \( SO_2 \) is oxidized. If \( 20\% \) of \( SO_2 \) is oxidized:
\[ \text{Moles of } SO_2 \text{ oxidized} = 0.5 \times \frac{20}{100} = 0.1 \text{ moles} \]
Thus, the remaining \( SO_2 \) will be:
\[ \text{Remaining } SO_2 = 0.5 - 0.1 = 0.4 \text{ moles} \]

### Step 5: Set Up the Stoichiometry for \( O_2 \)
From the balanced equation, for every \( 2 \) moles of \( SO_2 \) oxidized, \( 1 \) mole of \( O_2 \) is consumed. Therefore, for \( 0.1 \) moles of \( SO_2 \) oxidized:
\[ \text{Moles of } O_2 \text{ consumed} = \frac{0.1}{2} = 0.05 \text{ moles} \]

### Step 6: Calculate the Moles of \( O_2 \) at Equilibrium
Let \( X \) be the moles of \( O_2 \) added. The moles of \( O_2 \) at equilibrium will be:
\[ \text{Equilibrium } O_2 = X - 0.05 \]

### Step 7: Write the Equilibrium Expression for the Second Vessel
Using the same \( K_c \) value:
\[ K_c = \frac{[SO_3]^2}{[SO_2]^2 \cdot [O_2]} \]
At equilibrium:
- \( [SO_3] = 0.1 \) moles
- \( [SO_2] = 0.4 \) moles
- \( [O_2] = X - 0.05 \) moles

Substituting into the \( K_c \) expression:
\[ 0.2 = \frac{(0.1)^2}{(0.4)^2 \cdot (X - 0.05)} \]
This simplifies to:
\[ 0.2 = \frac{0.01}{0.16 \cdot (X - 0.05)} \]

### Step 8: Solve for \( X \)
Cross-multiplying gives:
\[ 0.2 \cdot 0.16 \cdot (X - 0.05) = 0.01 \]
\[ 0.032(X - 0.05) = 0.01 \]
\[ X - 0.05 = \frac{0.01}{0.032} \approx 0.3125 \]
\[ X \approx 0.3125 + 0.05 = 0.3625 \text{ moles} \]

### Step 9: Convert Moles of \( O_2 \) to Mass
To find the mass of \( O_2 \):
\[ \text{Mass of } O_2 = \text{Moles} \times \text{Molar Mass} = 0.3625 \text{ moles} \times 32 \text{ g/mol} = 11.6 \text{ g} \]

### Final Answer
The mass of \( O_2 \) that must be added is **11.6 g**.