14 8 Finding Equilibrium Concentrations Finding Equilibrium Concentrations
14. 8 Finding Equilibrium Concentrations
Finding Equilibrium Concentrations • Prior to this point, we have looked at how to calculate Kc if given equilibrium concentrations of reactants and products. • Now we will: – 1. ) Find the equilibrium concentrations when we know Kc and all but one of the equilibrium concentrations. – 2. ) Find equilibrium concentrations when we know Kc and only one initial concentration. (This is more difficult!)
Let’s Try a Practice Problem! •
Finding Equilibrium Concentrations from the Equilibrium Constant and Initial Concentrations or Pressure • It is more likely to be tasked with a problem when you only know K and only initial concentrations of reactants. • In this case, we must once again set up an ICE table. However, this time, we don’t know the changes in concentration. So here, they are represented with the variable x. Remember to check Q at the beginning of the reaction to identify which direction the reaction proceeds. If it proceeds right, the change in reaction concentration decreases…but if it proceeds left, the change in product concentration decreases. Straight from the college board: (above)
Let’s Try a Practice Problem! Consider the reaction: N 2(g) + O 2(g) 2 NO(g) Kc = 0. 10 (at 2000 o. C) A reaction mixture at 2000 o. C initially contains [N 2] = 0. 200 M and [O 2] = 0. 200 M. Find the equilibrium concentrations of the reactants and product at this temperature. Set up the ICE table: [N 2] M [O 2] M [NO] M Initial 0. 200 0 Change -x -x +2 x Equilibrium . 200 -x 0. 200 -x 2 x [NO]2 Kc = ------ Continued on next slide [N 2][O 2]
[N 2] M [O 2] M [NO] M Initial 0. 200 0 Change -x -x +2 x Equilibrium . 200 -x 0. 200 -x 2 x X = 0. 027 so… [N 2] =. 200 – 0. 027 = 0. 173 M [O 2] =. 200 – 0. 027 = 0. 173 M [NO] = 2(0. 027) = 0. 054 M Now, if you wanted to check your work, you could plug these equilibrium concentrations back into the original equilibrium expression for this reaction and solve for Kc
Let’s Try Another!!! Consider the reaction: I 2(g) + Cl 2(g) 2 ICl(g) Kp = 81. 9 A reaction mixture at 25 o. C initially contains PI 2 = 0. 100 atm, PCl 2 = 0. 100 atm, and PICl = 0. 100 atm. Find the equilibrium partial pressures of I 2, Cl 2, and ICl at this temperature. Set up an ICE table: PI 2 (atm) PCl 2 (atm) PICl (atm) 0. 100 Change -x -x +2 x Equilibrium 0. 100 -x 0. 100+2 x Initial 0. 100 (PICl)2 Kp = ------- continued on next slide (PI 2)(PCl 2)
PI 2 (atm) PCl 2 (atm) PICl (atm) 0. 100 Change -x -x +2 x Equilibrium 0. 100 -x 0. 100+2 x Initial 0. 100 x = 7. 29 x 10 -2 so…. (PI 2) = 0. 100 – 7. 29 X 10 -2 = 2. 7 X 10 -2 atm (PCl 2) = 0. 100 – 7. 29 X 10 -2 = 2. 7 X 10 -2 atm (PICl) = 0. 100 + 2(7. 29 X 10 -2) = 2. 46 X 10 -1 atm
14. 8 pg. 691 #’s 54 and 60 Read 14. 9 pgs. 677 -683
- Slides: 11