1 / 7

Equilibrium Pressure

Equilibrium Pressure. If the values at equilibrium are given in partial pressure, then solving for the constant is the same, but use K p instead of K c . What makes them different: Kc = equilibrium constant based on molarity and concentration.

whardison
Download Presentation

Equilibrium Pressure

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Equilibrium Pressure • If the values at equilibrium are given in partial pressure, then solving for the constant is the same, but use Kp instead of Kc. • What makes them different:Kc = equilibrium constant based on molarity and concentration. Kp = equilibrium constant based on partial pressures in atm.

  2. The mole ratio in a mixture gas is proportional to the partial pressures of each of the gases. • Dalton’s Law of Partial Pressures – the total pressure in a system is equal to the sum of the individual pressures. • Total = Pressure of + Pressure of ….Pressure Gas #1 Gas #2

  3. 2 NO2 (g) 2 NO (g) + O2 (g) • Determine the Kp for the above reaction when at equilibrium, the pressure of nitrogen dioxide is 0.425 atm, nitrogen monoxide is 0.270 atm, and oxygen is 0.100 atm. • Kp = [NO]2 [O2] [NO2]2 • Kp = [0.270]2 [0.100] [0.425]2 • Kp= 0.0404

  4. Relationship between Kc and Kp We use the Ideal Gas Law, and derive the relationship between Kc and Kp. Kp= Kc(RT)n Kp = Equilibrium constant in partial pressure Kc = Equilibrium constant in molar concentrations R = Ideal Gas constant (always 0.082) T = Temperature (in Kelvin) n = Change in moles of GASES ONLY (moles of gas products) – (moles of gas reactants)

  5. 2 NO2 (g) 2 NO (g) + O2 (g) • At equilibrium, the pressure of nitrogen dioxide is 0.425 atm, nitrogen monoxide is 0.270 atm, and oxygen is 0.100 atm. Determine the Kp for the above reaction.

  6. CO(g) + Cl2(g)  COCl2(g) • At equilibrium the concentrations at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. • Start with the equilibrium expression for Kc: Kc = [COCl2] [CO][Cl2] Kc = [0.14 M] = 216 [0.012][0.054]

  7. CO(g) + Cl2(g)  COCl2(g) • At equilibrium the concentrations at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. • To solve for Kp we use the Ideal gas law eq.Kp= Kc(RT)n Kp= 216 x (.082 x 347)(1-2) Kp = 7.6

More Related