1. AP Chem Turn in Equilibrium Lab Challenge
Today: Equilibrium Calculations

2. K vs. Q Equilibrium Constant, K K tells where the equilibrium lies
How likely (to what extent) the reaction is to occur
Reaction Quotient, Q
Snapshot of reaction at some point (where you are right NOW)
Same calculation as K ( [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠] )
Given K, compare Q and K to determine which direction the reaction will shift.

3. Reaction Quotient, Q 3 possible situations
If Q = K, the reaction is already at equilibrium.
If Q < K, the reaction will shift to the right in order to reach equilibrium
(more reactants than should be present at equilibrium)
If Q > K, the reaction will shift to the left in order to reach equilibrium.
(More products than should be present at equilibrium)

4. Reaction Quotient (Q) Example
2 NO2 (g)  N2O4 (g) Kp = 100°C
0.200 moles of NO2 are mixed with moles of N2O4 in a L flask at 100°C. Is the system at equilibrium? If not, determine which direction the reaction must shift to reach equilibrium.
Need to find partial pressures first using PV= nRT
P(4.00 L)= (0.200 mol)(0.0821)(373 K)
Both pressures = 1.53 atm

5. Reaction Quotient (Q) Example
2 NO2 (g)  N2O4 (g) Kp = 100°C Q =
PN2O4
(PNO2)2 =
= 0.654
Q < K, so right now there are more reactants than products. Reaction will shift to the right to reach equilibrium

6. Manipulating Equilibrium Constants
The equilibrium constant of a reaction in the reverse direction is the inverse (reciprocal) of the equilibrium constant of the reaction in the forward direction.
A + B ⇌ C + D Keq = K1
C + D ⇌ A + B Keq = 𝟏 𝑲𝟏

7. Manipulating Equilibrium Constants
The equilibrium constant of a reaction that has been multiplied by a number is equal to the original equilibrium constant raised to the power equal to the number
A + B ⇌ C + D Keq = K1
nA + nB ⇌ nC + nD Keq = K1n

8. Manipulating Equilibrium Constants
The equilibrium constant for a net reaction made up of two or more reactions is the product of the equilibrium constants for the individual reactions.
A + B ⇌ C + D Keq = K1
C + F ⇌ G + A Keq = K2
B + F ⇌ D + G Keq = K1 x K2

9. Example
1. The equilibrium constant of nitrogen and oxygen reacting to form nitrogen monoxide is Kc = 1x10-30 at 25o C. Using this information write the equilibrium expression and calculate Kc for the reverse reaction
Forward Reaction: N2 + O2  2 NO K=1x10-30
Reverse Reaction: 2 NO  N2 + O2
Kc for Reverse Rxn: 𝑵𝟐 [𝑶𝟐] 𝑵𝑶 𝟐 = 𝟏 𝟏𝒙 𝟏𝟎 −𝟑𝟎
K for reverse = 1 x 1030

10. I – initial concentrations/pressures
When solving equilibrium problems, an important tool is a “(R)ICE diagram”.
R –reaction
I – initial concentrations/pressures
C – change in concentration/pressure (using molar ratios)
E – equilibrium concentrations/pressures (I - C)

11. Example
1. A closed system initially contains 1.00 x10-3 M H2 and 2.00 x10-3 M I2 at constant temperature. Calculate Kc for the reaction if 1.87x10-3 M HI is present at equilibrium H2 + I2 2 HI I C E
1.00 x 10-3
2.00 x 10-3
1.87 x 10-3

12. Example
1. A closed system initially contains 1.000x10-3 M H2 and 2.000x10-3 M I2 at constant temperature. Calculate Kc for the reaction if 1.87x10-3 M HI is present at equilibrium H2 + I2 2 HI I C E
1.00 x 10-3
2.00 x 10-3 -x -x
+2x
1.87 x 10-3

13. Example Solve for X 2x =1.87 x 10-3 x = 9.35 x 10-4 I H2 + I2 2 HI C E-x -x
+2x
1.00x x
2.00x10-3 -x
1.87 x 10-3
Solve for X
2x =1.87 x 10-3
x = 9.35 x 10-4

14. Example I 1. H2 + I2 2 HI C E 1.000 x 10-3 2.000 x 10-3 -x -x +2x-x -x
+2x
6.5x10-5
1.065x10-3
1.87 x 10-3
Plug into Kc Expression

15. Example 𝐾= 𝐻𝐼 2 𝐻2 [𝐼2] 𝐾= 1.87 𝑋 10 −3 2 6.5 𝑋 10 −5 [1.065 𝑋 10 −3 ]
𝐾= 𝐻𝐼 2 𝐻2 [𝐼2]
𝐾= 𝑋 10 − 𝑋 10 −5 [1.065 𝑋 10 −3 ]
K = 51

16. Example 2 SO3(g) ⇌ 2 SO2(g) + O2(g) I C E
2. Sulfur trioxide decomposes at high temperatures. Initially the vessel is charged at 1000K with SO3(g) at a partial pressure of atm. At equilibrium the SO3 (g) partial pressure is atm. Calculate the value of Kp at 1000K
2 SO3(g) ⇌ 2 SO2(g) + O2(g) I C E
0.500 atm
-2x
+2x +x
0.200 atm 2x x

17. Example 2 SO3(g) ⇌ 2 SO2(g) + O2(g) Solve for X 0.5 – 2x = 0.2I C E
0.500 atm
-2x
+2x +x
0.200 atm 2x x
Solve for X
0.5 – 2x = 0.2
2x = 0.3
x = 0.15 atm

18. Example 2 SO3(g) ⇌ 2 SO2(g) + O2(g) 0.30 atm 0.15 atm x = 0.15 atmI C E
0.500 atm
-2x
+2x +x
0.200 atm 2x x
0.30 atm
0.15 atm
x = 0.15 atm
Kp = 𝑷𝑺𝑶𝟐 𝟐 𝑷𝑶𝟐 𝑷𝑺𝑶𝟑 𝟐
= 𝟎.𝟑 𝟐(𝟎.𝟏𝟓) 𝟎.𝟐 𝟐
=0.338 atm

19. Example 2 NOCl(g) ⇌ 2 NO(g) + Cl2 (g)
3. Gaseous NOCl decomposes to form the gases NO and Cl2. At 35°C, the equilibrium constant is 1.6 x In an experiment where 1.0 mol NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations? Kc = 1.6 x 10-5
2 NOCl(g) ⇌ 2 NO(g) + Cl2 (g) I C E
0.5 M
0 M
0 M
+ x
- 2x
+ 2x
0.5 – 2x 2x x

20. Example 𝐾= 𝑁𝑂 2[𝐶𝑙2] 𝑁𝑂𝐶𝑙 2 = 1.6 x 10-5
Technically, you’d need to plug into the quadratic equation to solve for this…
HOWEVER!
Since K is relatively small (1.6x10-5), you can assume that the changes (x) is negligibly small. This will make the math a lot less tedious.

21. Example Assume “-x” is negligible: x = 0.010

22. Example 2 NOCl(g) ⇌ 2 NO(g) + Cl2 (g) I C E 0.5 M 0 M 0 M + x - 2x
You can plug the ‘E’ values back in to the K expression to check your answer

23. Example 2 NO(g) + O2(g) ⇌ 2 NO2(g)
4) moles of NO and moles of O2 are placed in an empty 2.00 L flask. The system is allowed to establish equilibrium. What will be the equilibrium concentration of each species in the flask? Kc = 5.00 x 10-6
2 NO(g) O2(g) ⇌ 2 NO2(g) I C E
0.300 M
0.375 M
-2x
- x
+ 2x
0.300 – 2x x 2x

24. Example
𝐾= 𝑁𝑂2 2 𝑁𝑂 2[𝑂2] = 5.0 x 10-6
5.0 x 10-6 = 2𝑥 −2𝑥 2[0.375 −𝑥]
Technically, you’d need to plug into the quadratic equation to solve for this…
HOWEVER!
Since K is relatively small (5.0x10-6), you can assume that the changes (x) is negligibly small. This will make the math a lot less tedious.

25. Example Assume “-x” is negligible: x = 2.05 x 10-4
Note: you can only make this assumption if “x” is less than 5% of K. Otherwise, you have to rearrange and plug into the quadratic formula.

26. Example I 2) 2 NO(g) + O2(g) ⇌ 2 NO2(g) C E 0.300 M 0.375 M -2x - x
-2x
- x
+ 2x
0.300 – 2x x 2x M M
4.1 x 10-4 M
X = 2.05 x 10-4
You can plug the ‘E’ values back in to the K expression to check your answer