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Understanding Equilibrium: Kp vs. Kc Constants in Chemistry

Explore the relationship between the equilibrium constants Kp and Kc, Dalton’s Law of Partial Pressures, and derive the formula for finding Kp using Ideal Gas Law. Solve equilibrium problems with step-by-step examples.

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Understanding Equilibrium: Kp vs. Kc Constants in Chemistry

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  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

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