Question

0.60 atm of SO3 and 0.30 atm of SO2are placed in a container and the system is allowed to reach equilibrium. Calculate the pressure of O2(g) at equilibrium.

Answer 1

Complete Question

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Answer:

The pressure is $[O_2] = 4.8 *10^{-5} \ atm$

Explanation:

From the question we are told that

     The  pressure of  $SO_3$ is  $[SO_3 ] = 0.63 \ atm$

     The  pressure of  $SO_2$ is  $[SO_ 2] = 0.30 \ atm$

      The equilibrium constant is $K_p = 1.2 *10^{-5}$

     The  reaction is

           $2SO_3 _{(g)}$ ⇔ $2SO_2_{(g)} + O_2 _{(g)}$

Generally the equilibrium constant is mathematically represented as

           $K_p = \frac{(SO_2)^2 * (O_2)}{(SO_3)^2}$

=>         $[O_2] = \frac{k_p * [SO_3] ^2 }{[SO_2]^2}$

substituting values

            $[O_2] = \frac{1.2 *10^{-5} * 0.60 ^2 }{0.30^2}$

             $[O_2] = 4.8 *10^{-5} \ atm$