For the equilibrium
`2SO_(3) (g) Leftrightarrow 2SO_(2) (g)+O_(2)(g)`, the partial pressure of `SO_(y) SO_(2) and O_(2)` gases at 650 K are respectively 0.2, 0.6 bar and 0.4 bar. If the moles of both the oxides of sulphur are so adjusted as equal, what will be the partial pressure of `O_(2)`.

Explanation

To solve the problem step by step, we will follow the equilibrium expression and the conditions given in the question.

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

The equilibrium constant \( K_p \) for this reaction can be expressed as:
\[ K_p = \frac{(P_{SO_2})^2 \cdot (P_{O_2})}{(P_{SO_3})^2} \]
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