Consider the following reaction (assume an ideal gas mixture) : 2NOBr(g) 2NO(g) + Br₂(g) A 1.0-liter vessel was initially filled with pure NOBr,at a pressure of 4.0 atm,at 300 K.
-After equilibrium was established,the partial pressure of NOBr was 3.1 atm.What is K_p for the reaction?

Question

Consider the following reaction (assume an ideal gas mixture) : 2NOBr(g)

2NO(g) + Br₂(g) A 1.0-liter vessel was initially filled with pure NOBr,at a pressure of 4.0 atm,at 300 K.
-After equilibrium was established,the partial pressure of NOBr was 3.1 atm.What is K_p for the reaction?

Quizplus · Accepted Answer

The equilibrium constant for the reaction is given by the expression:

Kp = (PNO)^2(PBr2)/PNOBr^2

where PNO, PBr2, and PNOBr are the partial pressures of NO, Br2, and NOBr, respectively, at equilibrium.

We are given that the initial pressure of NOBr is 4.0 atm and the partial pressure of NOBr at equilibrium is 3.1 atm. We can calculate the partial pressures of NO and Br2 at equilibrium using the stoichiometry of the reaction:

2NOBr(g) ⇌ 2NO(g) + Br2(g)

At equilibrium, the change in the partial pressure of NOBr is -0.9 atm, which means that the partial pressures of NO and Br2 both increase by 0.45 atm. Thus, at equilibrium:

PNO = 0.45 atm 
PBr2 = 0.45 atm

Substituting these values and the value for PNOBr into the expression for Kp gives:

Kp = (0.45)^2(0.45)/3.1^2 = 0.038

Therefore, the answer is B.

Consider the Following Reaction (Assume an Ideal Gas Mixture): 2NOBr(g)