Respuesta :
Answer: Option (B) is the correct answer.
Explanation:
Expression for the given decomposition reaction is as follows.
      [tex]N_{2}O_{4} \rightarrow 2NO_{2}[/tex]
Let us assume that x concentration of [tex]N_{2}O_{4}[/tex] is present at the initial stage. Therefore, according to the ICE table,
          [tex]N_{2}O_{4} \rightarrow 2NO_{2}[/tex]
Initial :        x          0
Change : Â Â Â - 0.1 Â Â Â Â [tex]2 \times 0.1[/tex]
Equilibrium : (x - 0.1) Â Â Â Â Â Â 0.2
Now, expression for [tex]K_{p}[/tex] of this reaction is as follows.
   [tex]K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}[/tex]
Putting the given values into the above formula as follows.
     [tex]K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}[/tex]
         [tex]2 = \frac{(0.2)^{2}}{(x - 0.1)}[/tex]
        [tex]2 \times (x - 0.1) = (0.2)^{2}[/tex]
              x = 0.12
This means that [tex]P_{N_{2}O_{4}}[/tex] = x = 0.12 atm.
Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.
We can conclude that the initial pressure in the container before the decomposition of Nâ‚‚Oâ‚„ is 0.12 atm.
What is the decomposition reaction of Nâ‚‚Oâ‚„?
The decomposition of Nâ‚‚Oâ‚„ is determined as follows;
N₂O₄ → 2NO₂
ICE table can be created as follows;
   N₂O₄ → 2NO₂
I:    x       0
C: Â Â 0.1 Â Â Â Â 2(0.1)
E: (x - 0.1) Â Â (0.2 - 0)
Expression for equilibrium constant;
[tex]K_p = \frac{P^2NO_2}{PN_2O_4} \\\\2 = \frac{0.2^2}{(x - 0.1)} \\\\2(x - 0.1) = 0.2^2\\\\x-0.1 = 0.02\\\\x = 0.12[/tex]
Thus, we can conclude that the initial pressure in the container before the decomposition of Nâ‚‚Oâ‚„ is 0.12 atm.
Learn more about decomposition here: https://brainly.com/question/14539143
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