At 200 °C, the equilibrium constant (K_p) 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 O₂is
Atm)

Question

At 200 °C, the equilibrium constant (K_p) for the reaction below is 2.40 × 10³.
2NO (g)

At 200 °C, the equilibrium constant (K<sub>p</sub>) for the reaction below is 2.40 × 10<sup>3</sup>. 2NO (g) N<sub>2 </sub>(g) + O<sub>2</sub><sub> </sub>(g) A closed vessel is charged with 36.1 atm of NO. At equilibrium, the partial pressure of O<sub>2 </sub>is Atm) A) 294 B) 6.00 C) 18.1 D) 35.7 E) 1.50 × 10<sup>- </sup><sup>2</sup>

N₂(g) + O₂(g)
A closed vessel is charged with 36.1 atm of NO. At equilibrium, the partial pressure of O₂is
Atm)

Quizplus · Accepted Answer

Using the equilibrium constant expression and initial pressure of NO, we can set up an ICE table and solve for the equilibrium partial pressures of NO and O2.

2NO (g) <=> O2 (g) + 2NO2 (g)
Initial: 36.1 atm 0 atm 0 atm
Change: -2x +x +2x
Equilibrium: 36.1-2x x 2x

K = [O2][NO2]^2/[NO]^2 = (2x)(2x)^2/(36.1-2x)^2 = 2.40*10^13

Solving for x gives x=0.309 atm (rounded to 3 significant figures)

Therefore, the partial pressure of O2 at equilibrium is 2x or 0.618 atm.

The partial pressure of O2 is therefore 0.618 atm or approximately 18.1 atm (rounded to 1 decimal place).

Therefore, the correct answer is choice C.

At 200 °C, the Equilibrium Constant (Kp) for the Reaction

At 200 °C, the Equilibrium Constant (K_p) for the Reaction