1. Weekly Review Week of 9/24-9/28
Geoffrey Geberth 10/1/18

2. Chemical Equilibria

3. Activity (ai)
Measure of how the free energy of a compound changes as a reaction proceeds
Unitless
Gas
𝑎 𝑖 = 𝑃 𝑖 𝑃 °
Solutions
𝑎 𝑖 = 𝑖 𝐶 °
Pure solids and liquids
𝑎 𝑖 =1

4. Mass Action Expression
xX+𝑏𝐵⇄𝑐𝐶+𝑑𝐷
𝑎 𝐶 𝑐 𝑎 𝐷 𝑑 𝑎 𝑋 𝑥 𝑎 𝐵 𝑏
Remember: Products/Reactants!
Can be done with concentrations or pressures
𝐶 𝑐 𝐷 𝑑 𝑋 𝑥 𝐵 𝑏
𝑃 𝐶 𝑐 𝑃 𝐷 𝑑 𝑃 𝑋 𝑥 𝑃 𝐵 𝑏
Remember: The standard state of a pure solid or liquid is 1
These expressions give us the Reaction Quotient (Q)

5. Practice 𝑁 2 𝑔 +3 𝐻 2 𝑔 ⇌2 𝑁𝐻 3 ⇒ 𝑁𝐻 3 2 𝑁 2 𝐻 2 3
𝑁 2 𝑔 +3 𝐻 2 𝑔 ⇌2 𝑁𝐻 3
⇒ 𝑁𝐻 𝑁 𝐻 2 3
𝐶𝐻 3 𝑂𝐻(𝑠𝑙𝑛)+ 𝐶 4 𝐻 9 𝑂𝐻(𝑠𝑙𝑛) 𝑐𝑎𝑡𝑎𝑙𝑦𝑡𝑖𝑐 𝐻 + 𝐶 4 𝐻 9 𝑂𝐶𝐻 3 (𝑠𝑙𝑛)+ 𝐻 2 𝑂(𝑠𝑙𝑛)
⟹ 𝐶 4 𝐻 9 𝑂𝐶𝐻 3 𝐻 2 𝑂 𝐶𝐻 3 𝑂𝐻 𝐶 4 𝐻 9 𝑂𝐻
𝐶𝑂 2 (𝑔)+𝐶(𝑠)⇌2𝐶𝑂(𝑔)
⟹ 𝐶𝑂 𝐶𝑂 2

6. I know a thing or two about K’s
Equilibrium Constant
Value representing the reaction quotient when the system is at equilibrium
K > 1 -> favors products
K < 1 -> favors reactants
Can be huge or tiny, but is never negative
Different values for concentrations:
𝐾 𝑐 = 𝐶 𝑐 𝐷 𝑑 𝑋 𝑥 𝐵 𝑏
𝐾 𝑝 = 𝑃 𝐶 𝑐 𝑃 𝐷 𝑑 𝑃 𝑋 𝑥 𝑃 𝐵 𝑏
𝐾 𝑝 = 𝐾 𝑐 (𝑅𝑇 ) ∆𝑛
Arrived here from the ideal gas law (see gchem)
I know a thing or two about K’s

7. Manipulating K
There are 3 main ways we manipulate reactions and wish to know the new K
Reverse the reaction
𝐴+𝐵⇌𝐶 𝐾 1 ⟹𝐶⇌𝐴+𝐵
𝐾 2 = 1 𝐾 1
Multiply stoichiometric coefficients
𝐴+𝐵⇌𝐶 𝐾 1 ⟹ x𝐴+𝑥𝐵⇌𝑥𝐶
𝐾 2 = 𝐾 1 𝑥
Add subsequent reactions
𝐴+𝐵⇌𝐶 𝐾 1 ;𝐶⇌𝐷 𝐾 2 ⟹𝐴+𝐵⇌𝐷
𝐾 3 = 𝐾 1 𝐾 2

8. Q and K
Remember: Q is at any point in the reaction and moves, but K is only at equilibrium and is fixed (it depends on ΔG)
Comparing Q and K let’s us figure out in what direction a reaction will proceed, if at all
Q = K -> at equilibrium (forward rate = reverse rate)
Q < K -> Too many reactants (reaction produces more products)
Q > K -> Too many products (reaction produces more products) K Q

9. ΔG and Equilibria At equilibrium, ΔG = 0 ∆𝐺= ∆𝐺 𝑟 ° +𝑅𝑇𝑙𝑛 𝑄
The sign on ΔG tells you which direction the reaction will move
ΔG < 0: Reaction moves to products
ΔG > 0: Reaction moves to reactants
∆𝐺= ∆𝐺 𝑟 ° +𝑅𝑇𝑙𝑛 𝑄
At any instant in the reaction
∆𝐺 𝑟 ° =−𝑅𝑇𝑙𝑛 𝐾
At Eq
𝐾= 𝑒 −∆ 𝐺 𝑟 ° 𝑅𝑇

10. Let’s look at this equation more
𝐾= 𝑒 −∆ 𝐺 𝑟 ° 𝑅𝑇
If ∆ 𝐺 𝑟 ° =0, K=1
Temperature can change the equilibrium point!
Direction depends on ∆ 𝐻 𝑟𝑥𝑛
Endothermic (ΔH > 0), increased T increases K
Exothermic (ΔH < 0), increased T decreases K

11. RICE tables How we solve for changes in a system towards equilibrium
R: Reaction
I: Initial Conditions
C: Change
E: Equilibrium
Example: 𝐴+𝐵⇌2𝐶 initially at .1 M in A and B (no C present)

12. This reaction is important in steel production!
Practice
𝐹𝑒𝑂 𝑠 +𝐶𝑂 𝑔 ⇌𝐹𝑒 𝑠 + 𝐶𝑂 2 𝑔 ; 𝐾 𝑝 1000 𝐾
𝑃 𝐶𝑂,𝑖 =1.000 𝑎𝑡𝑚; 𝑃 𝐶 𝑂 2 ,𝑖 =.500 𝑎𝑡𝑚
Find the equilibrium partial pressures
𝐾 𝑝 =.259= 𝑃 𝐶𝑂 2 𝑃 𝐶𝑂 = .500+𝑥 1.000−𝑥
.0259−.0259𝑥=.500+𝑥
⟹𝑥= − =−𝟎.𝟏𝟗𝟏
→ 𝑃 𝐶𝑂 =1.000− −.191 =𝟏.𝟏𝟗𝟏 𝒂𝒕𝒎; 𝑃 𝐶𝑂 2 =.500+ −.191 =.𝟑𝟎𝟗 𝒂𝒕𝒎
Rxn
FeO (s) +
CO (g) ⇌
Fe (s) +
CO2 (g)
Intl P (atm) -
1.000
.500
Change (atm) -x +x
Eq P (atm) x x
Screenshot taken by me in Fallout 4
This reaction is important in steel production!

13. More practice!
NO is a pollutant produced in car engines from the reaction:
𝑁 2 (𝑔)+ 𝑂 2 (𝑔)⇌2𝑁𝑂(𝑔)
𝐾 𝑐 =1.7∗ 10 −3 @ 2300 𝐾
If the initial concentrations of N2 and O2 are both 1.40 M, what are the equilibrium concentrations of all 3 compounds?
𝐾 𝑐 =1.7∗ 10 −3 = 𝑁𝑂 𝑁 2 𝑂 2 = (2 𝑥) −𝑥 1.40−𝑥 = (2 𝑥) −𝑥 2
→ 1.7∗ 10 −3 = 2𝑥 1.40−𝑥
Rxn
N2 (g) +
O2 (g) ⇌
2NO (g)
Intl conc (M)
1.40
Change (M) -x
+2x
Eq conc (M) x
0 + 2x

14. More practice!
NO is a pollutant produced in car engines from the reaction:
𝑁 2 (𝑔)+ 𝑂 2 (𝑔)⇌2𝑁𝑂(𝑔)
→ 1.7∗ 10 −3 = 2𝑥 1.40−𝑥
−𝑥 =2𝑥
.058−.0412𝑥=2𝑥→2.041𝑥=.058
𝑥=.𝟎𝟐𝟖 𝑴
𝑁 2 = 𝑂 2 =1.40−.028=𝟏.𝟑𝟕𝑴; 𝑁𝑂 = =.𝟎𝟓𝟔𝑴
Rxn
N2 (g) +
O2 (g) ⇌
2NO (g)
Intl conc (M)
1.40
Change (M) -x
+2x
Eq conc (M) x
0 + 2x

15. Takeaways Products/Reactants Keep track of your units
Q is at any time, K is at equilibrium
Q seeks K
The ideal gas law is used to convert between Kp and Kc
ΔGrxn = 0 at equilibrium