Complete Question
The complete question is shown on the first uploaded image
Answer:
The pressure is [tex][O_2] = 4.8 *10^{-5} \ atm[/tex]
Explanation:
From the question we are told that
The pressure of [tex]SO_3[/tex] is [tex][SO_3 ] = 0.63 \ atm[/tex]
The pressure of [tex]SO_2[/tex] is [tex][SO_ 2] = 0.30 \ atm[/tex]
The equilibrium constant is [tex]K_p = 1.2 *10^{-5}[/tex]
The reaction is
[tex]2SO_3 _{(g)}[/tex] ⇔ [tex]2SO_2_{(g)} + O_2 _{(g)}[/tex]
Generally the equilibrium constant is mathematically represented as
[tex]K_p = \frac{(SO_2)^2 * (O_2)}{(SO_3)^2}[/tex]
=> [tex][O_2] = \frac{k_p * [SO_3] ^2 }{[SO_2]^2}[/tex]
substituting values
[tex][O_2] = \frac{1.2 *10^{-5} * 0.60 ^2 }{0.30^2}[/tex]
[tex][O_2] = 4.8 *10^{-5} \ atm[/tex]
