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.

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.

more infor would

a lot