Explanation
$$2SO_2 + O_2 \rightarrow2SO_3$$
$$K_p$$ can be calculated as:
$$K_p=\dfrac{[P_{SO_3^{2-}}]}{[P_{SO_2}^2]\times [P_{O_2}]}$$
As given:
$$[P_{SO_3}]=[P_{SO_2}]$$
So,
$$5=\dfrac{1}{[O_2]}$$
$$P_{O_2}=\dfrac{1}{5}$$
$$P_{O_2}=0.2$$
So, partial pressure of oxygen will be 0.2 atm.
0.2 atm will be the partial pressure of $$O_2$$.
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