AP Chem Ch 13-14 Flashcards
Dynamic vs static with regards to equilibrium
Eq is dynamic, not static, meaning that there is no net change, but there are still products and reactants being produced.
Equilibrium
The state where the concentration of the reactants and the concentration of the products remain constant with time
Rates at equilibrium
Rate of forward reaction = rate of reverse reaction
Position of equilibrium depends on:
- Initial concentration
- Relative energy of reaction and products
- Relative degree of organization of our reactants and products
GOAL is more entropy and less energy
Calculating keq
Keq = [product A]^coefficients * [prod B]^coef / [reactants]^coefficients
Note: ONLY INCLUDE AQ AND GASSES!!!
Dependent on temperature, not the initial amount
Kp
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
Q
Calculated the same as k, but done not at equilibrium
Used to determine which way the reaction will shift to get to equilibrium
If Q>K?
Then too many products, shifts to the left to make more reactants
If Q<K?
Then not enough products, shifts to the right to favor the forward reaction
Le chateliers principle
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.
How does adding more of a solid influence equilibrium?
NO INFLUENCE!
Changes in pressure:
- Add or remove a gaseous starting material/product
- Add inert gas (if at constant volume, no affect. If at constant pressure, then same as increasing volume)
- Changing volume of container
Decrease in volume, increase pressure, shifts to the side of the reaction with fewer moles of gas
How does changing temperature influence equilibrium
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
ICE table
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
Approximation when using ICE
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%
Rate law for elementary steps
K * concentration of products ^ coefficients
Arrhenius concept
Acids produce H+ and bases produce OH-
Bronsted Lowry definition
Acids are proton donors (donates H+)
Bases are proton acceptors
Ka
Equilibrium constant for acids
HA (acid) + H2O –> <– H3O+ + A - (conjugate base)
Ka = [H3O+] [A-]/[HA]
What does value of Ka tell us
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
Strong acids
HCl, HNO3, H2SO4, HClO4
Completely dissociate
Ka is so large, can’t be determined
Amphoteric
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
Kw
[H+][OH-]
At 25 degrees, = 1.0*10^-14
pH and pOH
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
Calculating pH for weak acid
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%
Percent dissociation
= 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
Base
Generate OH-, proton acceptor
Stronger the base, better proton acceptor
Strong bases
Group 1 and group 2 hydroxides
Kb
Describes the reaction between a base and water to generate the conjugate acid of the base and a hydroxide ion
[OH-][HB+] / [Base]
Polyprotic acid
More than one hydrogen
Such as H2co3
Two dissociations
For conjugate acid base pair Ka and Kb relation
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
How to tell if a proton is acidic
How polar is the bond?
High polarity is more acidic
How strong is the bond
Want a weak bond to be more acidic
HClO4 vs HClO3 acid strength
As you gain an oxygen, stronger polarization, more electron density, so more acidic
H-O-X bond
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+
Lewis acid and base model
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
