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Catalyst 1. Complete the exam reflection.

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1 Catalyst 1. Complete the exam reflection

2 What Our Data Looks Like

3 Multiple Choice Scores Go Up!

4 Woah! Free Responses!

5 PbI2 (s) Pb2+ (aq) + 2I- (aq)
Catalyst 1. Write the equilibrium constant for the reaction: PbI2 (s) Pb2+ (aq) + 2I- (aq) 2. You allow the above reaction to equilibrate. At equilibrium you have 5.2 g of HF, 0.2 M Pb2+ and 0.8 M 2I-. What is the value of the equilibrium constant 3. Does the value of K indicate that the equilibrium favors the products or the reactants? Justify your response.

6 Period 5

7 Lecture 7.2 – Adding Equilibrium Constants and Calculating Equilibrium Concentrations

8 Today’s Learning Targets
LT 7.5 – I can calculate the new equilibrium constant for a chemical reaction after adding constituent chemical reactions and equilibrium constants. LT 7.6 – I can calculate the equilibrium concentration for a chemical of interest using initial concentrations, equilibrium concentrations, and/or the equilibrium constant.

9 Changing K Like H, when we manipulate a chemical reaction, the value of the equilibrium constant changes. Three possible changes: Running Reaction in Reverse Multiplying Reaction by Common Factor Adding Equilibriums

10 Option 1: Running Reactions in Reverse
If we have the equilibrium N2O 2NO2 Then the value of the equilibrium constant is: If we reverse the equilibrium, then it is now: 2NO N2O Therefore, the new equilibrium constant is When a reaction is run in reverse, the equilibrium constant is inversed

11 Option 2: Multiplying by a Constant
Like Hess’s Law, we can multiply equilibriums in order to add them. If we multiply the previous equilibrium by two, then we get: 2N2O 4NO2 Therefore, we need to raise K to the value we multiplied by:

12 Option 3: Adding Equilibriums
Like Hess’s Law, we can add equilibriums to obtain the desired reaction When we add equilibriums, the equilibrium constants are multiplied.

13 Summary Reaction Run in Reverse Multiplying by a Constant
Three possible changes: Reaction Run in Reverse Multiplying by a Constant Adding Reactions

14 Class Example ⇌ ⇌ ⇌ Given the reaction:
HF (aq) H+ (aq) + F- (aq) Kc = 6.8 x 10-4 H2C2O4 (aq) H+ (aq) + C2O42- (aq) Kc = 3.8 x 10-6 Determine the value of Kc for the reaction: 2 HF (aq) + C2O42- (aq) 2 F- (aq) + H2C2O4 (aq)

15 2 NH3 (g) + 3 I2 (g) 6 HI (g) + N2 (g)
Table Talk Given the reaction: H2 (g) + I2 (g) 2HI (g) Kp = 54.0 N2 (g) + 3 H2 (g) 2NH3 (g) Kc = 1.04 x 10-4 Determine the value of Kc for the reaction: 2 NH3 (g) + 3 I2 (g) HI (g) + N2 (g)

16 Stop and Jot ⇌ ⇌ ⇌ The following equilibrium was obtained at 823 K:
CoO (s) + H2 (g) Co (s) + H2O (g) Kc = 67 CoO (s) + CO (g) Co (s) + CO2 (g) Kc = 490 Based on these equilibrium, calculate the equilibrium constant for: H2 (g) + CO2 (g) CO (g) + H2O Assume it is also at 823 K.

17 White Board Problems

18 White Board Problems ⇌ ⇌ ⇌
1. The following two reactions have equilibrium constants listed: SO2 (g) + ½ O2 (g) SO3 (g) K = 2.2 NO2 (g) NO (g) + ½ O2 K = 4.0 What is the equilibrium constant for the reaction: SO2 (g) + NO2 (g) SO3 + NO (g) 2.

19 White Board Problems

20 Cobalt Chloride

21 Calculating Equilibrium Concentrations
Oftentimes, we know the amount of substance that we begin with, but the equilibrium amounts are unknown. We use stoichiometry to help solve for the equilibrium concentrations using the initial conditions and the value of Kc.

22 ICE, ICE, Baby N2 (g) + 3H2 (g)  2NH3 (g)
Use an ICE Table to solve for equilibrium concentrations. N2 (g) H2 (g)  2NH3 (g) Initial Change Equilibrium Initial concentrations go here Result of adding and subtracting I/C Coefficient times x goes here

23 Class Example A L flask if filled with mol of H2 and 2.00 mole of I2 at 448 oC. The value of the equilibrium constant Kc, for the reaction: H2 (g) + I2 (g) HI (g) At 448 oC is What are the equilibrium concentrations of H2, I2, and HI in moles per liter?

24 Table Talk ⇌ For the equilibrium: PCl5 (g) PCl3 (g) + Cl2 (g)
The equilibrium constant, Kp, is at 500 K. A gas cylinder at 500 K charged with PCl5 (g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3, Cl2

25 Class Example A closed system initially containing 1.00 x 10-3 M H2 and x 10-3 M I2 at 448 oC is allowed to reach equilibrium and at equilibrium the HI concentration is 1.87 x 10-3 M. Calculate Kc for the reaction taking place, which is: H2 (g) + I2 (g) 2HI (g)

26 Table Talk Sulfur trioxide decomposes at high temperature in a sealed container: 2 SO2 (g) SO2 (g) + O2 (g) Initially the vessel is charged at 1000 K with So3 (g) at a partial pressure of atm. At equilibrium the SO3 partial pressure is atm. Calculate the value of Kp at 1000 K

27 Around the World There are 8 problems around the room. Complete the problems and check your answers on the backside

28 Exit Ticket 1. The following two reactions have equilibrium constants listed: SO2 (g) + ½ O2 (g) SO3 (g) K = 2.2 NO2 (g) NO (g) + ½ O2 K = 4.0 What is the equilibrium constant for the reaction: SO2 (g) + NO2 (g) SO3 + NO (g) 2. Phosphorus pentachloride decomposes to phosphorus trichloride and chlorine gas. A 1.10 mol/L sample of PCl5 was placed into a vessel. At equilibrium it was determined that the concentration was 0.33 mol/L. Calculate the Keq for this reaction. PCl5(g) ⇌ PCl3(g) + Cl2(g)

29 Rate Yourself Rate yourself on LT 7.1 to LT 7.6

30

31 Closing Time Read: 15.3, 15.4, 15.5, and 15.6 Homework: 15.27, 15.28, 15.51, 15.52, 15.54, 15.57, ,


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