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AP Chemistry
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Laboratory report rubric
Notes
1. Chemical foundations
2. Atoms, molecules, and ions
3. Stoichiometry
4. Types of chemical reactions and solution stoichiometry
5. Gases
6. Thermochemistry
7. Atomic structure and periodicity
8. Bonding: general concepts
9. Covalent bonding: orbitals
10. Liquids and solids
11. Properties of solutions
12. Chemical kinetics
13. Chemical equilibrium
14. Acids and bases
15. Acid-base equilibria
16. Solubility and complex ion equilibria
17. Spontaneity, entropy, free energy
18. Electrochemistry
Drug unit
Basics
Analgesics
Antacids
Anesthetics
Depressants
Stimulants
Antibiotics
Antiviral drugs
Mind-altering drugs
Textbook (incomplete)
1. Chemical foundations
2. Atoms, molecules, and ions
3. Stoichiometry
4. Types of chemical reactions and solution stoichiometry
5. Gases
6. Thermochemistry
7. Atomic structure and periodicity
8. Bonding: general concepts
CED
1. Atomic structure and properties
2. Compound structure and properties
3. Properties of substances and mixtures
4. Chemical reactions
5. Kinetics
6. Thermochemistry
7. Equilibrium
8. Acids and bases
9. Thermodynamics and electrochemistry
Notes
12. Chemical kinetics
Reaction rates
kinetics
: study of the rate of a reaction
rate
units
molarity over time
calculating
always negative
example: what is the rate of NO₂ from 0 seconds to 50 seconds?
Differential rate law
only with reactants (not products)
k:
rate constant (1/timeⁿ)
units change depending on the order
e.g. first order: 1/s
figure out by solving for k in equation (use M/s for rate)
[A]: concentration (M)
n:
order
of reaction
units: M/s or [M]
example: what are the rates of H₂ and H₂O?
use ratios to solve
H₂ = - 4.0 M/s
H₂O = + 4.0 M/s
O₂ = 2.0 M/s
Changing reaction rates
concentration
temperature
pressure/volume
surface area
catalyst
Rate laws from experiment
order can only be determined/calculated by experiments
example: find order for each reactant, overall order, k
three experiments, changing concentrations (only worry about reactants)
0.100 M and 0.0050 M
0.100 M and 0.010 M
0.200 M and 0.010 M
pick two experiments where one concentration does not change and one does (e.g. 1 and 2)
NO₂⁻ molarity doubles
rate doubles
pick two experiments where the other concentration changes
NH₄⁺ molarity doubles
rate doubles
overall order
: all individual orders added together
overall order: 1 + 1 = 2
k is a constant, so pick one experiment to plug in and solve for k
example: find individual and overall orders
0.10 M, 0.10 M, 0.10 M
0.20 M, 0.10 M, 0.10 M
0.20 M, 0.20 M, 0.10 M
0.10 M, 0.10 M, 0.20 M
solving for a
experiments 1 and 2: molarity doubles, rate doubles
a = 1
solving for b
experiments 2 and 3: molarity doubles, rate doubles
b = 1
solving for c
experiments 1 and 4: molarity doubles, rate quadruples
c = 2
overall order: 1 + 1 + 2 = 4
Integrated rate laws
[A]: concentration at time t
k: rate constant
t: time (s)
[A]₀: initial concentration
lines up with line equation (y = mx+ b)
pick the graph with a straight line, then check units
half-life
: the time it takes for half of the concentration of the reactants to decay
radioactive decay: first order
order
zero
first
second
rate law
rate = k
rate = k[A]
rate = k[A]²
integrated rate law
plot needed to give a straight line
[A] vs. t
ln[A] vs. t
1/[A] vs. t
relationship of rate constant to slope of straight line
slope = - k
slope = - k
slope = k
half-life
There are no rows in this table
Reaction mechanisms
factors that determine reaction rate
collide with enough energy to break bonds
collide with the correct orientation
catalyst
grabs reactants and puts them in the correct orientation
example:
multiple steps that cancel to form reaction
(slow)
(fast)
reaction is as fast as the slowest step
look at slowest rate
write the reactants
intermediate
: molecule in the product of the first step and the reactant of the second step
very
low concentration
(does not exist for very long)
example: find rate law and intermediate
(slow)
(fast)
intermediate: F
example (second step is slow)
(fast)
reversible
think of it like an equals sign (A + B = C)
(slow)
substitute first reaction into second reaction
A + B + D ⟶ E
rate = k[A][B][D]
the higher the “hill” the slower the reaction
exothermic: high → low
endothermic: low → high
catalyst will shrink the size of the hill
number of hills = number of steps
catalysts never get used up in the reaction
present in both reactants and products
Reaction rates
Differential rate law
Changing reaction rates
Rate laws from experiment
Integrated rate laws
Reaction mechanisms
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