AP Chem Ch 13-14

Question 0

Dynamic vs static with regards to equilibrium

Answer 0

Eq is dynamic, not static, meaning that there is no net change, but there are still products and reactants being produced.

Question 1

Equilibrium

Answer 1

The state where the concentration of the reactants and the concentration of the products remain constant with time

Question 2

Rates at equilibrium

Answer 2

Rate of forward reaction = rate of reverse reaction

Question 3

Position of equilibrium depends on:

Answer 3

1. Initial concentration
2. Relative energy of reaction and products
3. Relative degree of organization of our reactants and products
   GOAL is more entropy and less energy

Question 4

Calculating keq

Answer 4

Keq = [product A]^coefficients * [prod B]^coef / [reactants]^coefficients
Note: ONLY INCLUDE AQ AND GASSES!!!

Dependent on temperature, not the initial amount

Question 5

Kp

Answer 5

PV= nRT
P=n/V * RT
P=CRT
C=P/RT
So, Kp = [pressure of products]^coef / [pressure of reactants]^coef
Pressure is related to coencentration by equation given
Keq = [P prod / RT]^coef / [P reactant / RT]^coef

Question 6

Q

Answer 6

Calculated the same as k, but done not at equilibrium

Used to determine which way the reaction will shift to get to equilibrium

Question 7

If Q>K?

Answer 7

Then too many products, shifts to the left to make more reactants

Question 8

If Q<K?

Answer 8

Then not enough products, shifts to the right to favor the forward reaction

Question 9

Le chateliers principle

Answer 9

If a change is implied on a system at eq, the position of the eq will shift in a direction that tends to reduce that change.

For example, say we add more of a reactant, then we have too much of the reactant on the left side, shift to the right, favors the forward reaction. More of the reactant allows for more reactions, shifts right.

Question 10

How does adding more of a solid influence equilibrium?

Answer 10

NO INFLUENCE!

Question 11

Changes in pressure:

Answer 11

1. Add or remove a gaseous starting material/product
2. Add inert gas (if at constant volume, no affect. If at constant pressure, then same as increasing volume)
3. Changing volume of container
   Decrease in volume, increase pressure, shifts to the side of the reaction with fewer moles of gas

Question 12

How does changing temperature influence equilibrium

Answer 12

For an exothermic reaction, add energy to right side
Increase temp, then too much on the right, shifts to the left

Endothermic is reverse–shift to the right

Question 13

ICE table

Answer 13

Shows the concentration of the reactants and products initially, the change, and then at equilibrium.
Determine Q and that will tell you if to add X or subtract x

Question 14

Approximation when using ICE

Answer 14

If you’re subtracting x from the reactants and K is really small, then you can assume X is negligible. Check by dividing (initial-x)/initial needs to be less than 5%

Question 15

Rate law for elementary steps

Answer 15

K * concentration of products ^ coefficients

Question 16

Arrhenius concept

Answer 16

Acids produce H+ and bases produce OH-

Question 17

Bronsted Lowry definition

Answer 17

Acids are proton donors (donates H+)

Bases are proton acceptors

Question 18

Ka

Answer 18

Equilibrium constant for acids
HA (acid) + H2O –> <– H3O+ + A - (conjugate base)

Ka = [H3O+] [A-]/[HA]

Question 19

What does value of Ka tell us

Answer 19

If it’s large, then products are favored. This means it’s a strong acid and weak conjugate base
Stronger acids have larger Ka values

Question 20

Strong acids

Answer 20

HCl, HNO3, H2SO4, HClO4
Completely dissociate
Ka is so large, can’t be determined

Question 21

Amphoteric

Answer 21

Act as acid or base
Water is amphoteric in that it produces H+ and OH- auto ionizations of water
Stronger base than A- (conjugate base) when Ka is large

Question 22

Kw

Answer 22

[H+][OH-]

At 25 degrees, = 1.0*10^-14

Question 23

pH and pOH

Answer 23

pH = - log [H+] 
pOH = - log [OH-]
pH+pOH = 14
1-14 scale
1 is most acidic, 14 most basic 
Number sig figs in the concentration is how many decimals should be in pH or pOH

Question 24

Calculating pH for weak acid

Answer 24

Use an ice table
1. Major components at equilibrium
2. Which species generate H+ and which is the dominate producer (larger Ka value)
Use ice table and solve for concentration of H+ and then take -log to get pH
Can use 5% rule– if Ka is small then assume X is 0 on the denominator of expression. Then divide value - X / value and make sure less than 5%

Question 25

Percent dissociation

Answer 25

= amount dissociated material (M) / Initial (M)
As a weak acid becomes more dilute, the percent dissociation increases.
Diluting the system has the equilibrium system shift to the right because too many reactants, so this increases H+ and dissociation

Question 26

Base

Answer 26

Generate OH-, proton acceptor

Stronger the base, better proton acceptor

Question 27

Strong bases

Answer 27

Group 1 and group 2 hydroxides

Question 28

Kb

Answer 28

Describes the reaction between a base and water to generate the conjugate acid of the base and a hydroxide ion

[OH-][HB+] / [Base]

Question 29

Polyprotic acid

Answer 29

More than one hydrogen
Such as H2co3
Two dissociations

Question 30

For conjugate acid base pair Ka and Kb relation

Answer 30

Ka * Kb = Kw
Finding one allows us to find the other (Kw is known– 1.0*10^-14)

If Ka is bigger, acid
Smaller, base
Equal, neutral

Question 31

How to tell if a proton is acidic

Answer 31

How polar is the bond?
High polarity is more acidic

How strong is the bond
Want a weak bond to be more acidic

Question 32

HClO4 vs HClO3 acid strength

Answer 32

As you gain an oxygen, stronger polarization, more electron density, so more acidic

Question 33

H-O-X bond

Answer 33

If x has a high enough EN (non-metal), then the O-X bond will be covalent and strong, acidic species in solution

If x has low EN, then more ionic characteristic, OH- + X+

Question 34

Lewis acid and base model

Answer 34

Lewis acid: electron pair acceptor
Lewis base: electron paid donor

Example:
H+ + NH3 –> NH4 +
The NH3 gives a lone pair to the H+, making NH3 a Lewis base