1. Assign. #13.2 – Adding Equilibrium Constants and Calculating Equilibrium Concentrations

2. 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

3. 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: ⇌

4. 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.

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

6. 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) ⇌ ⇌ ⇌

7. 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) ⇌ ⇌ ⇌

8. 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. ⇌ ⇌ ⇌

9. Example Problems

10. 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.

11. 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

12. 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? ⇌

13. 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 ⇌

14. 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) ⇌

15. 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 ⇌