ap chem equilibrium

chemical equilibrium

state where concentrations of products and reactants remain constant

• top curve reactants

• bottom curve products

+ forward reaction rate equals reverse reaction rate

equilibrium reaction rate vs time

top curve: forward rate

bottom curve: reverse rate

big K

K_c = K = K_eq

K changes with temperature (not [ ] or pressure)

size of K and time needed to reach equilibrium unrelated

Shortcuts

reversed products and reactants: K_c’ = 1/K_c

original reaction multiplied by some factor (coefficient n in front of it): K_c’ = K_cⁿ

2 step reaction: K_{3 ⁼} K₁ * K₂

K_p TTR

DONT USE BRACKETS

measured in atm

K_c and K_p relationship

if diff amount of gas moles in prod/reactant:

K_p = K_c(RT) ^Δn        R = 0.08206 Latm/molK
                                T = Kelvin temperature
Δn = #moles gaseous product - # moles gaseous reactant

if same amount of gas moles in prod/reactant:

K_c = K_p

dont include what in equilibrium expressions

pure solids (s) and pure liquids (l), these have no effect on position of equilibrium

(known as Heterogeneous Equilibria)

K>1 and K<1

K>1 means that the equilibrium is far to the right (mostly products).

K<1 means that the equilibrium is far to the left (mostly reactants).

**Values for K in the range of 0.001 to 1000 describe reactions where reactants and products are both present in significant quantities at equilibrium.

big Q

calculated like Kc, but uses initial concentrations instead of equilibrium concentrations

Q=K

then system is at equilibrium (no shift occurs)

Q > K

too much products, [product]/[reactant] is too large (system shifts to produce more reactants or the reverse reaction is favored)

Q < K

too much reactants, [product]/[reactant] is too small (system shifts to produce more products, or the forward reaction is favored)

what happens to K in LeChatelier’s

DOES not change when there are stressors, remains equal

K only changes with temp

• not concentration, pressure/volume, catalysts, inert gasses

What’s happening in the graph?

T₁: equilibrium is established, flattening

T₂: Nitrogen injected into system

between T₂ - T₃: reaction is favoring the formation of products to reestablish equilibrium

T₃: equilibrium is re-estalished

exothermic

reactants → products + heat

• delta H is negative

• An increase in temperature of an exothermic reaction will cause equilibrium to favor the reverse reaction → K will decrease. ([reactants] will increase, [products] decrease)

• A decrease in temperature of an exothermic reaction will cause equilibrium to favor the forward reaction → K will increase. ([reactants] will decrease, [products] increase)

endothermic

reactants + heat → products

• delta H is positive

• An increase in temperature of an endothermic reaction will cause equilibrium to favor the forward reaction → K will increase.  ([reactants] will decrease, [products] increase) 

• A decrease in temperature of an endothermic reaction will cause equilibrium to favor the reverse reaction → K will decrease. ([reactants] will increase, [products] decrease) 

changes in pressure only affect

equilibrium system with GASEOUS products/reactants

• Increasing the pressure of a gaseous system will cause equilibrium to shift to the side with fewer gas particles.

• Decreasing the pressure of a gaseous system will cause equilibrium to shift to the side with more gas particles.

• If the system has the same number of moles of gas on each side, changes in pressure do not affect equilibrium.

• Adding an inert gas does not affect equilibrium since the partial pressures of the gases in the reaction are not affected.

inert gases

noble gases (row 18)

adding a catalyst

doesn’t affect equilibrium position, only affects the time it takes for a reaction to reach equilibrium

lechat secondary reactions

Common secondary reactions:

1. Reaction forms a precipitate

2. Acid/Base reactions:  Acids (for now compounds starting with H) = produce, H⁺ Bases(for now ionic hydroxide compounds) = produce OH^-

  Net ionic equation:  H⁺(aq) +OH^-(aq) → H₂O

review SNAP solubility rules

ionic compounds containing sodium Na⁺, nitrate NO₃⁻, ammonium NH₄⁺, and potassium K⁺ are always soluble in water

• no precipitate formed

• usually spectator ions in net ionic eq.

solubility

max quantity of a substance that dissolves to form a saturated solution at a given temperature

molar solubility

max number of moles of the solute that dissolves to form a liter of saturated solution; given in mol/L. (S = molar solubility)

saturated solution

solution that contains the max amount of solute for a given temperature.

• The rate at which the solid solute dissolves equals the rate at which the dissolved solutes precipitate

Ksp (solubility product) - “insoluble salts are actually slightly soluble”

the equilibrium constant for the equilibrium between an ionic solid and its saturated solution (no unit of measurement)

or a special type of equilibrium constant that measures how much of a poorly soluble ionic solid can dissolve in water at a specific temperature

• ksp often only contains aq products side

below, equal, and above to ksp meaning; or comparing Q to Ksp to see if ppt forms

below (<): unsaturated, no solid precipitates

equal (=): exactly saturated, no solid precipitates (yet)

above (>): supersaturated, solid precipitates

smaller ksp values indicate

lower solubility (reactants dominate final reaction mixture, more solid present)

… so larger ksp values indicate more solubility (products dominate final reaction mixture, more ions present)

Ksp values cannot be directly compared if

if the compounds dissociate into different numbers of ions

• ex. AgI and Ca₃(PO₄)₂

• need to solve for S (solubilities) in order to compare

even if the volume of water changes in a saturated solution, the concentration of ions

will stay the same

proof:

• If you add water (increasing the volume), the system is temporarily diluted, so some of the extra solid dissolves until the solution becomes saturated again.

• If you remove water (evaporating it down), some dissolved ions precipitate out as solid.

• In both cases, once the system re-reaches equilibrium at the same temperature, the concentration of the ions goes right back to its original solubility limit

common ion

when an ionic compound, which is added to an aq. solution, that has an ion in common with the equilibrium reaction —> decrease the solubility of a compound

when to ignore +x or +2x of common ion

• small Ksp (less than 10^-4) meaning very little solid dissolves at equilibrium

• The initial concentration of the common ion ex. 0.050 M I^- is significantly larger than the amount that will dissolve.

dont ignore it when: Ksp is greater than 10^-4

dilution shortcut

when volumes are equal and then combined, concentrations are halved

ex. if 400 mL of each solution in the previous problem were combined, the resulting concentrations would be

[Ba²⁺] = 0.0400/2 = 0.0200M

and
[SO₄^2-] = 0.0800/2 = 0.0400M

For solids, as temperature increases, solubility...

increases: the added heat supplies the energy needed to break the bonds holding the solid particles together

pH and solubility, Compounds with anions exhibiting basic properties…

(e.g., Mg(OH)₂ / OH^–, CaCO₃ / CO₃^2–, CaF₂ / F^–)

inc in solubility as solution becomes more acidic.
H+ removes the OH^-, leads to shift right, solubility increases

if you add like NaOH what happens

if H⁺ or like H₃O⁺ is on product side OH^- removes it and makes H₂O, so rxn will shift right to make up for lost products