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notes 6.1-6.4, 7.1-7.5, and 8.1
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6.1 bronsted lowery
acids ____ protons
bases ____ protons
donate
accept
6.1 the more stable the conjugate base the more ___ the acid
strong
6.1 counterions/spectater ions
mask charges (+ on a -)
first two tables of periodic tables
*sometimes halogens (Cl-, Br-, I-)
6.2 using pka values to compare strenghts
strong acid = ____ pka
weak acid = ___ pka
weaked pka =
strongest pka =
LOW
HIGH
50
-10
6.2 pka to pkb
weaker base = ___ stable
stronger base = ____ stable
strong acids are willing to lose protons because..
MOST
LEAST
they have a place to put the electrons that is stable
6.2
strong acids pka is..
moderately strong acids pka is..
weak acids pka is..
very weak acids pka is..
EXTREMLY weak acids pka is..
<1 (HCl, HBr, HI)
1-3
3-5
5-15
>15 (H2O)
6.3 CARIOS
C=
A=
R=
I=
O=
S=
charge
atom
resonance
induction
orbital
solvation
6.3 CHARGE
positively changed acids are
protonated, H+
6.3 CHARGE
carbon tail
longer carbon tail less acidic
6.3 CHARGE
more acidic the molecule, ___ charge
more basic the molecule, ___ charge
+
-
6.3 ATOM
____ atoms are better for stabilizing negative charge
electronegative
6.3 ATOM
larger atoms ____ acidic
smaller atoms ____ acidic
more acidic, more room for e-
less acidic, less room for e-
(dif. in size more important that dif. in charge!!)
6.3 RESONANCE
resonance stabilizes the negative charge by…
spreading it out into partial charges
6.3 RESONANCE
more stable the conjugate base..
more likely the protons taken away
6.3 INDUCTION
induction stabilizes negative charge by..
spreading it out (like resonance)
6.3 INDUCTION
if there is a highly negative atom on the molecule undergoing protonation..
the negatively charged atom will share the negative charge through resonance and pull the charge towards itself, stabilizing the molecule
6.3 INDUCTION
induction causes the atom to be..
more acidic, induction = highly e- species in proximity to e-
(F, Br, Cl)
6.3 ORBITAL
negative charges need to be ___ to the nucleus because ___
closer, because it holds protons (positive charge)
6.3 ORBITAL
sp3
sp2
sp
least acidic (furthest away bc. least s character)
middle
most acidic (closest together bc. more s character)
6.3 SOLVATION
in simular atoms, branching causes …
the solvent to be blocked from the proton
6.3 SOLVATION
solution ___ the negative charge
stabilizes the …
stabilizes
transitions states, intermediate, and/ or products to allow reaction
6.4 what difference causes a non reversible protenation
difference in pka greater than 10
6.4 a stronger acid wants to what?
give away a proton
6.4 how to find the difference in amount of products vs. reactants
large pka-small pka= n 10^n is the difference
6.4 how to choose a strong reagent (base)
protonate the base (negative) so that the equilibrium will favor the products
6.4 how to protonate the base
choose an acid more acidic than the conjugate acid of the base, arrow right favors products = good reagent
6.4 how to choose a reagent (acid)
deprotenate the acid (positive) so that the equilibrium will favor the products
6.4 how to deprotenate the acid
choose a base with a conjugate acid more acidic than the original acid
6.4 a solvent should ___ react with itself
NOT, it is there to facilitate collisions of the acid and base
6.4 acids greater than ___ cannot exist in water
-1.74, H3O+ (conjugate base H2O), it will favor reactants not products
6.4 bases strong than ___ cannot be used in water
OH-, its conjugate acid is H2O
7.1
exergonic reaction
endergonic reaction
thermodynamically favored, reactants higher in energy than products, - delta H/G
unfavored, products higher in energy than reactants, + delta H/G
7.1
energy is delta ___
potential energy is delta ___
H
G
7.1
transition state is the
intermediate is the
humps, energy maxima, not observable
valleys, energy minima, oberservable
7.1
in an exergonic reaction, the transition state is closer to the
in an endergonic reaction, the transitions state is closer to the
reactants
products
7.1
small EA
larger EA
kinetically favored, fast
unfavored, slow
7.1 heterolytic
bond cleavage breaks and goes to one molecule, creates ions X—Y —> X+ + Y-
7.2 homolytic
bond cleavage breaks and separates to both evenly, creates radicals
X—Y —> X* + Y*
7.2 BDE
bond disassociation energy for homolytic bond cleavage
7.2 - delta H is
exothermic, bond energy gained > energy needed to break, temp of surroundings increases, stable products
7.2 + delta H is
endothermic, energy needed to break > bond energy gained, temp of surroundings decreases, unstable products
7.2 stronger bonds need __ energy to be broken
more
7.2