Respuesta :
Answer: The pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.
Explanation:
We are given:
Equilibrium partial pressure of [tex]O_2[/tex] = 0.29 atm
For the given chemical equation:
[tex]2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)[/tex]
Initial: a
At eqllm: a-2x 2x x
Calculating for the value of 'x'
[tex]\Rightarrow x=0.29[/tex]
Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm
Equilibrium partial pressure of [tex]NO_2[/tex] = a - 2x = a - 2(0.29) = a - 0.58
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{O_2}\times (p_{NO})^2}{(p_{NO_2})^2}[/tex]
We are given:
[tex]K_p=158[/tex]
Putting values in above expression, we get:
[tex]158=\frac{0.29\times (0.58)^2}{(a-0.58)^2}\\\\a=0.555,0.604[/tex]
Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.
So, [tex]a=0.604[/tex]
Equilibrium partial pressure of [tex]NO_2[/tex] = (a - 0.58) = (0.604 - 0.58) = 0.024 atm
Hence, the pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.
