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 : - 0.1 [tex]2 \times 0.1[/tex]
Equilibrium : (x - 0.1) 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 N₂O₄ is 0.12 atm.
What is the decomposition reaction of N₂O₄?
The decomposition of N₂O₄ is determined as follows;
N₂O₄ → 2NO₂
ICE table can be created as follows;
N₂O₄ → 2NO₂
I: x 0
C: 0.1 2(0.1)
E: (x - 0.1) (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 N₂O₄ is 0.12 atm.
Learn more about decomposition here: https://brainly.com/question/14539143
#SPJ5
