1. 14.8 Finding Equilibrium Concentrations

2. 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!)

3. Let’s Try a Practice Problem!
Diatomic iodine [I2] decomposes at high a temperature to form I atoms according to the reaction: I2(g) 2I(g) Kc = at 1200oC In an equilibrium mixture, the concentration of I2 is 0.10 M. What is the equilibrium concentration of I? [I]2 Kc = [I2] = [I] = 0.011(0.10 𝑀) = 3.3X10-2 M 0.10 M

4. Straight from the college board: (above)
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)

5. Let’s Try a Practice Problem!
Consider the reaction: N2(g) + O2(g) 2NO(g) Kc = 0.10 (at 2000oC) A reaction mixture at 2000oC initially contains [N2] = M and [O2] = M. Find the equilibrium concentrations of the reactants and product at this temperature. Set up the ICE table: [NO]2 Kc = Continued on next slide  [N2][O2]
[N2] M
[O2] M
[NO] M
Initial
0.200
Change -x
+2x
Equilibrium
.200-x
0.200-x 2x

6. [NO]2 Kc = ------------ [N2][O2] (2x)2 0. 10 = ------------- (0
[NO]2 Kc = [N2][O2] (2x) = (0.200-x)2 2x 0.10 = x 0.10 (0.200-x) = 2x = − 0.10 x = 2x – 0.32x =2x = 2.32 x = x = Now, substitute in for x to determine the equilibrium concentrations: Continued on next slide 

7. [N2] M
[O2] M
[NO] M
Initial
0.200
Change -x
+2x
Equilibrium
.200-x
0.200-x 2x
X = so…
[N2] = .200 – = M
[O2] = .200 – = M
[NO] = 2(0.027) = 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

8. Let’s Try Another!!!
Consider the reaction: I2(g) + Cl2(g) 2ICl(g) Kp = 81.9 A reaction mixture at 25oC initially contains PI2 = atm, PCl2 = atm, and PICl = atm. Find the equilibrium partial pressures of I2, Cl2, and ICl at this temperature. Set up an ICE table: (PICl)2 Kp = continued on next slide  (PI2)(PCl2)
PI2 (atm)
PCl2 (atm)
PICl (atm)
Initial
0.100
Change -x
+2x
Equilibrium
0.100-x x

9. (PICl)2 Kp = ------------- (PI2)(PCl2) (0. 100 +2x)2 81
(PICl)2 Kp = (PI2)(PCl2) ( x) = (0.100-x) X 81.9 = x − 81.9 x = x – 9.05x = x = 11.05x so… x = 7.29x10-2 Now, plug this value in to find the partial pressure of the gases in this reaction at equilibrium. Next slide 

10. PI2 (atm)
PCl2 (atm)
PICl (atm)
Initial
0.100
Change -x
+2x
Equilibrium
0.100-x x
x = 7.29x10-2 so…. (PI2) = – 7.29X10-2 = 2.7X10-2 atm (PCl2) = – 7.29X10-2 = 2.7X10-2 atm (PICl) = (7.29X10-2) = 2.46X10-1 atm

11. 14.8 pg. 691 #’s 54 and 60 Read 14.9 pgs