Answered

At 200 °C, the equilibrium constant (Kp) for the reaction below is 2.40 × 103. 2NO (g) ↔ N2 (g) + O2 (g) A closed vessel is charged with 36.1 atm of NO. At equilibrium, the partial pressure of O2 is __________ atm.

Answer :

codedmog101

Answer:

17.9 atm.

Explanation:

Okay, in the question above we are given the following information; the equilibrium constant (Kp) for the reaction= 2.40 × 10^3, and the initial partial pressure of NO= 36.1 atm.

The equilibrium constant,kp can be represented mathematically by using the formula below;

kp = P(N2) P(O2) / P^2 (NO).

The equation for the balanced Reaction is given below as;

2 NO <======> N2 + O2.

Therefore, at EQUILIBRIUM the partial pressure of NO = 36.1 - 2x and the partial pressure of N2 and O2 = x respectively.

So, let us go back to the formula for Calculating kp which is;

2.40 × 10^3 = x^2 / 36.1 - 2x.

Therefore, solving for the value of x in the question we have x = 17.9.

The partial pressure of the oxygen in the reaction mixture is 17.9 atm. Thus option C is correct.

From the balanced chemical equation, the partial pressure of NO is twice the partial pressure of Nitrogen and Oxygen.

Given, equilibrium constant = [tex]\rm 2.40\;\times\;10^3[/tex]

The partial pressure of system = 36.1 atm

Partial pressure of Nitrogen = x

Partial pressure of oxygen = x

Partial pressure of NO = 36.1 - 2x

Equilibrium constant = [tex]\rm \dfrac{[pN_2]\;[pO_2]}{[pNO]}[/tex]

[tex]\rm 2.40\;\times\;10^3[/tex] = [tex]\rm \dfrac{[x]\;[x]}{[36.1-2x]}[/tex]

[tex]\rm 2.40\;\times\;10^3[/tex] [tex]\times[/tex] (36.1 - 2x) = [tex]\rm x^2[/tex]

x = 17.9 atm.

Thus, the partial pressure of the oxygen in the reaction mixture is 17.9 atm. Thus option C is correct.

For more information about the partial pressure, refer to the link:

https://brainly.com/question/15075781

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