Exhibit 14-6 Consider the reaction at equilibrium below and its corresponding equilibrium constant to answer the following question(s).
2 NO₂ (g) 2 NO (g) + O₂(g) K p = 4.48*10 - 13
A pressure of 0.45 atm of NO₂ is introduced into a container and allowed to come to equilibrium. Complete the ICE equilibrium table shown below to get started. Partial pressure S 2 NO₂ 2 NO O₂Initial 0.45 M Change At equilibrium
-Refer to Exhibit 14-6. What relationship listed below can be used to solve for x in this table and ultimately the equilibrium partial pressures of each substance?

Question

Exhibit 14-6 Consider the reaction at equilibrium below and its corresponding equilibrium constant to answer the following question(s).
2 NO₂ (g)  2 NO (g) + O₂(g) K p = 4.48*10 - 13 
A pressure of 0.45 atm of NO₂ is introduced into a container and allowed to come to equilibrium. Complete the ICE equilibrium table shown below to get started. Partial pressure S 2 NO₂ 2 NO O₂Initial 0.45 M Change At equilibrium
-Refer to Exhibit 14-6. What relationship listed below can be used to solve for x in this table and ultimately the equilibrium partial pressures of each substance?

Quizplus · Accepted Answer

The equation of the reaction given is 2 NO₂ (g) ⇌ 2 NO (g) + O₂(g) with the equilibrium constant, Kp = 4.48×10^-13. Using the ICE table, the initial partial pressure of NO₂ is 0.45 atm, the change is -2x atm for NO₂ and +2x atm for NO and +x atm for O₂, and the equilibrium partial pressure of NO₂ is (0.45-2x) atm, NO is (2x) atm, and O₂ is (x) atm. The expression for Kp is (PNO)^2 * (PO₂) / (PNO₂)^2, or (2x)^2 * (x) / (0.45-2x)^2. Setting this expression equal to the given Kp value and solving for x will lead to the equilibrium partial pressures. Simplifying this equation gives D) 4x*(x) / (0.45-2x)^2 = 4.48×10^-13.

Exhibit 14-6 Consider the Reaction at Equilibrium Below and Its