1. Chem. 1B – 9/1 Lecture

2. Announcements I Diagnostic Quiz
Should be receiving back in lab
Learn what material to review
SacCT – will be set up once adding is done
Bring a periodic table to class (I will provide on on exam days)
Today’s Lecture – note: not covered in same order as in text
More Review (I will stress what people had difficulty with)

3. Announcements II Today’s Lecture – cont.
Basic Equilibrium (and Questions)
Manipulating Reactions
KP vs. KC
Equilibrium Problems: STARTING AT EQUILIBRIUM
Equilibrium Problems: STARTING AT INITIAL CONDITIONS (if time)

4. Chem 1A Review A Few More Example Problems
Nomenclature
Cerium (IV) dichromate (cerium = Ce)
Type = ____________ formula = _____________
perchloric acid type = ______ formula = ____
Cu2SO4 type = ______ name = ____
How many g. of Cu2SO4 are needed to prepare 25.0 mL of a solution that is M Cu+?

5. Chem 1B First Topic Equilibrium
In Chapter 13 (near end of Chem 1A), we covered kinetics (how fast reactions occur)
Past assumption was that “→“ meant reaction went to completion (all products plus limiting reagent)
Also, unfavorable reactions, such as precipitation of CaCl2 from Ca2+ + Cl- are not supposed to occur at all
In some important cases, expected reactions will not proceed totally to completion and unexpected reactions can occur to some extent

6. Chem 1B - Equilibrium Introduction
Reactions that can occur in both the forward and reverse reactions to an appreciable extent are represented with a two way arrow:
A + B C or A + B ↔ C (easier to type this)
What does this mean?
From a molecular scale, reactions are occurring in both the forwards and backwards direction
When the forward rate (loss of A to form C) is equal to the backwards rate (gain of A from C), the system is said to be in equilibrium and concentrations don’t change

7. Chem 1B - Equilibrium Introduction
Graphic example of reaction:
A + B ↔ C (assume [A] = [B] at start)
In the beginning, forward rate is fast (as A and B are high). At the points the lines cross, [A] = [B] = [C], but the forward rate is still faster than the backwards rate. We can see that [C] is favored over [A] and [B] because its final conc. is higher.
Conc. (C)
Conc. (A and B)
Time

8. Chem 1B - Equilibrium Introduction
Some examples:
1. Carbon dioxide (CO2) is not very soluble (you are supposed to know it forms from acid + carbonates)
However, the acidity of rain in “pristine” conditions is supposed to be based on dissolution of CO2:
CO2 (g) + H2O (l) ↔ H2CO3 (aq) ↔ H+ + HCO3-
Nitrogen gas (N2) is difficult to react because of the strength of its triple bonds, however under certain conditions, beneficial and problematic reactions involve N2 where both reactants and products exist.
N2 (g) + 3H2 (g) ↔ 2NH3 (g) (used in fertilizer)
N2 (g) + O2 (g) ↔ 2NO (g) (source of smog from combustion)
Conditions are critical to make a) go but to limit b)

9. Chem 1B - Equilibrium Equilibrium Constant and Equation
Whether an equation favors reactants or products is determined by an EQUILIBRIUM CONSTANT (K) – as well as reaction stoichiometry
Generic reaction: aA + bB ↔ cC + dD
K = equilibrium constant and for above reaction,
A larger K value means products are more favored (bigger [C] and [D] vs. [A] and [B])

10. Chem 1B - Equilibrium Equilibrium Equation – Further Details
Technically, instead of concentration we have a concentration ratio to a “standard state” (e.g. 1 mol/L) – we can ignore this, but this is why K is unitless
Only species in gas phase or in solution will have concentrations. Solids and pure liquids are not included.
When gases are involved, there are two Ks: KC for concentration units (including gases) and KP in which pressure (atm) replaces concentration (assume K = KC unless specified)
Even further corrections (activity and fugacity) are needed under certain conditions but are beyond this class

11. Chem 1B - Equilibrium Equilibrium Equation – Example
N2 (g) + O2 (g) ↔ 2NO (g) or
At T = 25°C KC = KP (in this case) = 4 x 10-31
Almost no NO will form in air: PN2 = 0.8 atm , PO2 = 0.2 atm; rearranging the above equation, PNO = [KPPN2PO2]0.5 = 2 x atm
Equilibrium at room temp. is not realistic because kinetics is too slow (insufficient collision energy to break triple bond), so even less NO expected
At car engine temp., reactions are faster and K is much larger (even if <1). This leads to significant NO formation. Fuel rich conditions (low PO2 ) can limit NO formation (reducing PO2 reduces PNO).

12. Chem 1B - Equilibrium Equilibrium Equation – Questions I
If the following reaction is at equilibrium, which of the following statements is true:
2NO2 (g) ↔ N2O4 (g)
a) the concentrations of reactants and products are equal
b) the rate of formation of N2O4 (M/s) is equal to its loss (in M/s)
c) the rate of loss of NO2 in forming N2O4 is equal to the rate of loss of N2O4 in forming NO2
d) neither molecule is reacting at all as their concentrations are constant

13. Chem 1B - Equilibrium Equilibrium Equation – Questions II
For the following reactions, give an equilibrium equation:
CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)
H2(g) + I2(s) ↔ 2HI(g)
S(s) + 3F2(g) ↔ SF6(g)
AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Cl-(aq)
What are the units for the K value for the first reaction?
Given the reactions and K values, for which reaction is formation or products most likely?
4Cu(s) + O2(g) ↔ 2Cu2O(s) K = 3.9 x 1025
2Cu(s) + O2(g) ↔ 2CuO(s) K = 1.8 x 1022
2Cu(s) + H2O(l) ↔ Cu2O(s) + H2(g) K = 1.0 x 10-16

14. Chem 1B - Equilibrium Equilibrium Equation – Manipulating Equations
This is similar to Hess’s Law (used to calculate DH for a reaction which can be made from combinations of other reactions)
Rules:
Flipping Directions
If for A ↔ B, K = K1, then for B ↔ A, K = 1/K1
Multiplication
½N2 (g) + ½O2 (g) ↔ NO (g) K = K1
N2 (g) + O2 (g) ↔ 2NO (g) = 2·RXN1, then K = K12

15. Chem 1B - Equilibrium Equilibrium Equation – Manipulating Equations
Rules: - cont.
Adding Reactions:
N2 (g) + O2 (g) ↔ 2NO (g) K = K1
2NO (g) + O2 (g) ↔ 2NO2 (g) K = K2
N2 (g) + 2O2 (g) ↔ 2NO2 (g) K = K3
Rxn 3 = rxn 1 + rxn 2 K3 = K1·K2
Why is math different than Hess’s Law?
Covered in detail in Ch. 17, but short reason:
DG (somewhat like DH) = -RTlnK
And from math we know ln(K1·K2) = lnK1 + lnK2
Thus addition in reaction becomes multiplication in K

16. Chem 1B - Equilibrium Equilibrium Equation – Manipulating Equations
Example Problem:
If the following reactions have the given equilibrium constants:
Ag+ + 2NH3(aq) ↔ Ag(NH3) K = 1.70 x 107
NH3(aq) + H2O(l) ↔ NH4+ + OH- K = 1.76 x 10-5
H2O(l) ↔ H+ + OH K = 1.0 x 10-14
Determine the equilibrium constant for the following reaction:
Ag(NH3)2+ + 2H+ ↔ Ag+ + 2NH4+

17. Chem 1B - Equilibrium Equilibrium Constants – KP vs. KC
Are KP and KP the same? If not how are they related?
Q1 – No (except for some reactions)
Q2 – How is P related to C?
Ideal gas law: PV = nRT
And we know [] is moles gas/volume = n/V
We can rearrange the ideal gas law to: P = (n/V)RT = []RT

18. Chem 1B - Equilibrium Equilibrium Constants – KP vs. KC
Example: 2NO2 (g) ↔ N2O4 (g) KP = PN2O4/PNO22
Or KP = ([N2O4]RT)/([NO2]RT)2 = KC(RT)-1
General Rule: KP = KC(RT)Dn where Dn = change in number of moles (moles gas product – moles gas reactants) = 1 – 2 = -1 in above example

19. Chem 1B - Equilibrium Most Common Types of Problems
Conditions given
Question Asking for K
Question Asking for Equilibrium Concentrations
Only at equilibrium
Requires knowledge of concentrations (or pressures) of all species
Requires knowledge of K and concentrations of all but one species
Initial conditions (requires use of ICE table)
Usually requires equilibrium concentration of at least 1 species
Requires knowledge of K

20. Chem 1B - Equilibrium Equilibrium Problems – AT EQUILIBRIUM
In this case the equilibrium equation is used with concentrations (or pressures) given AT EQUILIBRIUM
These types of problems are very important for environmental chemistry, but underemphasized in text
For example, an atmospheric chemist measured “high” NO in air near fresh lava. He wondered if it came from the N2(g) + O2(g) ↔ 2NO(g) reaction. If KP(T = 1000 K) = 7 x 10-9, calculate PNO in equilibrium with N2 and O2 in air.

21. Chem 1B - Equilibrium Equilibrium Problems – AT EQUILIBRIUM
2nd Example Problem:
A rich chemist wants to measure KC for the reaction: N2O4 (g) ↔ 2NO2 (g)
He puts N2O4 in a container at the temperature he wants to measure KC. He measures [NO2] and [N2O4] (using an expensive mass spectrometer) until the concentrations stop changing. He finds [NO2] = M and [N2O4] = M. What is KC?

22. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions
In this case an initial concentration or pressure is given (typically of reactants)
The reaction then proceeds to equilibrium
The student calculates K or the concentration of a reactant or product
An important part of working out this problem is to make an ICE table
ICE stands for initial change equilibrium

23. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions
To understand how an ICE table works, let’s start with a reaction that goes 100% to completion (covered in Chem 1A)
Example: 1.00 mol/L H mol/L O2 going to H2O in a container at 200°C.
reaction H2(g) O2(g) → 2H2O(g)
initial conc mol/L mol/L
change mol/L mol/L mol/L
completion 0 mol/L 0.50 mol/L mol/L
mol/L O2 lost = (1.00 H2 mol /L)(1 mol O2/2 mol H2)
limiting reagent
remember: for every 2 mol H2 we use 1 mol O2

24. Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions
Example Problem: The rich chemist lost his research grant and had his mass spectrometer repossessed. He still has a UV-Visible spectrometer to measure [NO2] (it’s a brown gas – while N2O4 is invisible).
Can he still calculate K?
Yes, but we need to define the experiment more carefully
Initially, the chemist puts mol N2O4 into a 5.0 L container and sets T. He measures [NO2]. When the concentration stops increasing, he finds [NO2] = M. What is K?