For the following equilibrium, `2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g)` the total equilibrium pressure is `P_(1)`. If volume of the system is reduced to `1//2` of this initial volume then equilibrium is restablished. The new equilibrium total pressure will be `:`

Options

A. `2P_(1)`
B. `3P_(1)`
C. `3.5P_(1)`
D. less than `2P_(1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium reaction
The equilibrium reaction given is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

### Step 2: Define initial conditions
Let the initial partial pressures at equilibrium be:
- \( P_{SO_2} = x \)
- \( P_{O_2} = y \)
- \( P_{SO_3} = z \)

The total initial equilibrium pressure \( P_1 \) can be expressed as:
\[ P_1 = x + y + z \]

### Step 3: Analyze the effect of volume change
When the volume of the system is reduced to half, according to the ideal gas law, the pressure will double (since pressure is inversely proportional to volume). Therefore, the new pressures before re-establishing equilibrium will be:
- \( P_{SO_2} = 2x \)
- \( P_{O_2} = 2y \)
- \( P_{SO_3} = 2z \)

### Step 4: Set up the new equilibrium expression
At the new equilibrium, let’s denote the change in the amount of \( SO_2 \) and \( O_2 \) as they react to form \( SO_3 \). Let \( \alpha \) be the degree of dissociation of \( O_2 \). The new pressures will be:
- For \( SO_2 \): \( 2x - 2\alpha \)
- For \( O_2 \): \( 2y - \alpha \)
- For \( SO_3 \): \( 2z + 2\alpha \)

### Step 5: Write the total pressure at new equilibrium
The total pressure \( P_2 \) at the new equilibrium can be expressed as:
\[ P_2 = (2x - 2\alpha) + (2y - \alpha) + (2z + 2\alpha) \]

### Step 6: Simplify the expression
Combining the terms, we get:
\[ P_2 = 2x + 2y + 2z - 2\alpha - \alpha + 2\alpha \]
This simplifies to:
\[ P_2 = 2(x + y + z) - \alpha \]
Since \( x + y + z = P_1 \), we can substitute:
\[ P_2 = 2P_1 - \alpha \]

### Step 7: Conclusion about the new equilibrium pressure
Since \( \alpha \) is a positive value (the degree of dissociation), we conclude that:
\[ P_2 < 2P_1 \]
Thus, the new equilibrium total pressure will be less than \( 2P_1 \).

### Final Answer
The new equilibrium total pressure \( P_2 \) is less than \( 2P_1 \).

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