Question

N2(g) O2(g) <------> 2 NO(g) At a certain temperature, the above reaction has a Kp of 0.050. If nitrogen and oxygen gas are present in an equilibrium mixture at partial pressures of 0.20 atm and 0.60 atm, respectively, what must be the partial pressure of NO(gas)

Answer 1

The partial pressure of NO(g) at equilibrium is approximately 0.077 atm, calculated using the given equilibrium constant (Kp = 0.050) and the partial pressures of reactants N₂ and O₂.

Step-by-step explanation:

The student is asking about the calculation of an unknown equilibrium partial pressure of a product in a chemical reaction based on given reactant partial pressures and the equilibrium constant (Kp).

Given the reaction N2(g) + O2(g) ⇌ 2 NO(g) with a Kp of 0.050 and partial pressures for N2 and O2 as 0.20 atm and 0.60 atm respectively, we can set up the expression for Kp:

Kp = [NO]^2 / ([N2][O2])

From this, we can solve for the partial pressure of NO given that Kp = 0.050:

0.050 = [NO]^2 / (0.20 * 0.60)

Through rearrangement and solving for [NO], we deduce the equilibrium partial pressure of NO. If [NO] is the partial pressure of NO, then:

[NO]^2 = 0.050 * 0.20 * 0.60

[NO]^2 = 0.006

[NO] = √0.006

[NO] = 0.077 atm (approximately)

Therefore, the partial pressure of NO(gas) at equilibrium is about 0.077 atm.