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77 Cards in this Set

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[reactants] and [products] doesn't change; rate of forward reaction is equal to the rate of the reverse reaction
given aA +bB -><- cC + dD
Keq = [C]^c [D]^d/[A]^a [B}^b

brackets = molar concentraion at equilibrium
solids and liquids are not included (because concentraion doesnt change)
If gaseous, then can use partial pressure
mass action ration
K eq is constant
at given temp regardless of concentrations of product or reactant. Reaction will go in direction to equilibrium
If Keq < 1
reaction favors the reactant
If Keq = 1
reaction ba;ances the reacant and product
Keq > 1
products favored
Equilirium reached if
ration of products and reactants remains constant & soulution is saturated with solute (system no longer changes with time)
Q = reaction quotient
aA =bB -><- cC + dD
where Q = [C]^c [D]^d/[A]^a [B}^b
Q = Keq
no change
Q < Keq
Reaction proceed forward to produce more product
Q > Keq
Reaction proceeds in the reverse direction to attain more reactant
Le Chatliers
a system at equilibrium will try to neutralize any imposed change or stress (e.g., adding reactant) to reestablish equilibrium

Eg., add more reactant, system will react by favoring forward reaction to consume reactant
Changing volume of reaction container which is at equilibrium
If gas, and decrease volume (increase pressure), then reaction favors the soulte with least number of moles and vice cersa
+ change H
-chnage H
changing heat in a system at equilibrium
only effects endo thermic/ exothermic reaction

If add heat, and the process is exothermic, reaction proceeds left

If cool it down, reaction proceeds to right
Keq and changes in temp
changing temp will effect Keq; increases with endothermic, decreases with exothermic
addition of inert or unreactive gas to system in equilibrium
has no effect
adding a catalyst
increases reaction rate but does not affect the quilibrium; adding catalyst to a system that is already in quilibrium will have n oeffect
change in H is negative
exothermic and heat is on the product side
Change in H is positive
heat on reactant side
solubility product constant
the extent to which a salt will disolve in water can be determined from its solunility product constant Ksp
molar solubility
number of moles of a salt that will saturate a liter of water (asked to find out just how much of a solid can disosle in H20
given Ksp = 2.7 x 10 ^-9
for Li3PO4, how many moles of this salt would be required to form a saturated 1 liter aq solution
equil = li3PO4 -><- 3Li + PO4
let x = [PO4] and [Li] = 3x
Ksp = (3x)^2x = 27x^4= 2.7 x 10^-9 -> x = 3.2^-3 ol needed
Qsp < Ksp
more salt can be dissolved
Qsp = Ksp
solution is saturated
Qsp> Ksp
excess salt will precipitate
Common ion effect
reduce the solubility of ion is solution already has common ion; adding more ions that are in product will drive reaction to the left
Arrenhius acid
acids ionize in water to produce hydrogen H+ ions
HCL -> H+ + Cl-
Arrenhius base
ionize in H20 to produce OH- ions
NaOH -> Na+ +OH-
Bronset Lowry Acid
acids are proton donor
Bronsted Lowry Base
Bases are proton acceptors
Lewis Acid
electron pair acceptor
Lewis Base
electron pair donor
strong acid
on e that dissociates completly ( or nearly so ) in H20
E.g., HCl (aq) + H2) (l) -> H30 (aq) + Cl- (aq)

(HF is a example of weak acid)
Ka (equilirbium expression for acid dissocation)
"acid ionization or dissocation constant"
strength of acid is directly related to how much the products are favored over the reactants

Ka = [H30] [A-]/[HA]
Ka > 1
products favored
acid is strong
Ka < 1
reactants favored and acis is weak
Comparing the Ka strengths
The larger Ka, the stronger the acid
Common strong acids
HI, HBr, HCl, HClO3, HClO4, H2SO4, HNO3
Weak acid
any acid that is not a strong acid from the list
Base ionization constant
Kb = [HB][OH]/[B]
Larger Kb, stronger base
Strong base
Group 1 hydroxides (LiOH, NaOH, KOH, RbOH, CsOH)
Metal amids (NaNH2
Weak base
NH3 (ammonia) and amines
Conjugate base of a strong acid
has no basic properties (Cl- will not tend to accept proton, and doesnt act like a base
The conjugate base of a weak acis
weak base (F- has some tendency to accept proton)
has more than 1 proton to donatel; generally, everytime a polyprotic acid donates a proton, the resulting species will be a weaker acid than its predecessor
can act like a base or acid
autoionization (self-ionization)
of water, reacts with self in a Bronster Lowry acid-base reaction, 1 molecules acting as the acid and the other as athe base
=1.0*10^-14 = [H3O][OH]
varies with temp (like other equillibrium constants), this is at 25 degrees C
Kw is constatnt at a given temp regardless of the H3O concentration
If the introduction of an acid increases the concentration of H3O ions, then equilibrium is disturbed
the reverse reaction is favored, decreasing the concentration of OH- ions
If introduce base to increase its concentration, equilibium is disturbed
reverse reaction is favored, decreasing the concentration of H3O ions, however the produce to [H3O] and [OH ]will remain equal to Kw
Kw = [OH][H3O]
Increase one concentration
decreases the other
pH = -log[H+]

means that [H+] =

E.g., pH = 7 , [H] = 10^-7
trend for acidity
down and to the left
pOH = -log[OH]
[OH] = 10^-pOH
pH + pOH
[H] = 6.2x10^-5
pH is between 4 and 5
lowest concentration of H3O
has highes pH
solution has pH of 3.5, what is [H]
[H] = 10^-3.5 = 10^.5 x 10^-4 = sqrt(10) x 10^-4 = 3.2 *10^-4
pH calculation for strong acid
pH comes directly from molarity of solution; so .01 M solution of HCl will have [H} = .01 = pH 2
weak acid
concentration of a H ions will be much less than the undissociated ions
when K < 10 ^-4
x added to or subtracted from a number is neglible
with weak acids , use ICE
HCN -><- H+ + CN-
Initial: .2 0 0
at equil: (.2-x)M x x
know Ka = 4.9 x 10^-10

x^2/(.2-x) equivl to: x^2/.2 = 4.9 x 10^-10
take negative log
Ka1 < Ka2
So pKa1 > pKa2
Ka2 is more acidic
pKa + pKb
weak acid
may have a weak or strong conj base
10^x = 3.16
x = log(3.16)
x between 0 and 1 (10^0 or 10^1) so x equal .5
log (AB) = log (A) + log (B)
log (10^2) = 2
log (10^3) = 3
log (10^2 x 10^3) = 5
in a series of oxyacids
more acids means stronger acid
rate constant (kinetics) K and equilibrium (thermodynamics) are not related formally
are not related formally
If H2) is the solvent
do NOT include it in the equillibrium expression
is exothermic
there's no net consumption or production of H+ in an acetal reaction
H is a catalysit??
If the reacion is does not involve gases
ignore the effect of pressure (if it does include gases, do include pressure such that reaction will favor products or reactants that have least gas moles