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Stephanie Yeung Per. 3 Semester 2 (Gas Laws (Kinetic Molecular Theory (Gas…
Stephanie Yeung Per. 3
Semester 2
Properities of Solutions
solutions are
homogenous
(uniform composition)
mixtures
unsaturated solution:
less than max amount of solute dissolved
supersaturated solution:
more than max amount of solute dissolved
saturated solution:
max amount of solute dissolved
solvent:
dissolving substance
solute:
dissolved substance in solution
calculations of solution concentration
grams per liter:
g of solute/ L of solution
in
dilute water-based solution
, 1 ml = 1 g
parts per million:
(g of solute/ g of solution) × (1 ⋅ 10^6)
percent composition:
% mass = (g of solute/ g of solution) × 100
molarity:
M = mol of solute/ L of solution
Thermochemistry
q = mC
Δ
T
Δ
T
= temperature change (°C) = T
final
- T
initial
m
= mass (g)
C
= specific heat (J/g°C)
q
= heat lost/gained (J)
mC
Δ
T
= -[
mC
Δ
T
]
heat gained = heat lost
Heat (enthalpy) change,
∆H
:
amount of heat energy released or absorbed during process
exothermic processes, -∆H:
processes where energy is released as it proceeds, and surroundings become warmer
endothermic processes, +∆H:
processes where energy is absorbed as it proceeds, and surroundings become cooler
∆H from ∆Hf:
standard enthalpy change of reaction =
sum
of standard molar enthalpies of formation of
products
multiplied by its coefficient, n, in balanced equation, minus corresponding
sum
of standard molar enthalpies of formation of
reactants
.
ΔH°
reaction
=∑ΔH°
f(products)
−∑ΔH°
f(reactants)
water phase changes
C
water
= 4.184 J/g°C
∆H
fus
= 333 J/g
heat of fusion:
energy that must be absorbed in order to convert 1 mol of
solid
to
liquid
at its melting point
C
ice
= 2.09 J/g°C
C
steam
= 1.87 J/g°C
∆H
vap
= 2240 J/g
heat of vaporization:
energy that must be absorbed in order to convert 1 mol of
liquid
to
gas
at its boiling point
specific heat:
amount of heat required to raise temperature of 1 g of substance by 1 °C
Joule:
measures heat
1
calorie
= 4.184
Joules
calorie:
heat required to raise temperature of 1 g of water by 1 °C
Gas Laws
pressure
= force/unit area; caused by collisions of molecules with walls of container
1 atm
= 101,325 Pa = 101.3 kPa = 760 mmHg = 760 torr
K
inetic
M
olecular
T
heory
Gas particles are in constant, rapid motion. They, therefore, possess kinetic energy, the energy of motion.
Collisions between gas particles and between particles and the walls of the container are elastic collisions.
No energy is lost in elastic collisions.
Gases consist of tiny particles that are far apart relative to their size.
There are no forces of attraction between gas particles.
The average kinetic energy of gas particles depends on temperature.
The temperature of a gas is proportional to the average kinetic energy of the gas particles.
deviations from ideal behavior
likely to behave nearly ideally:
gases at high temp and low pres, small non-polar gas molecules
likely not to to behave nearly ideally:
gases at low temp and high pres, large polar gas molecules
Boyle's Law:
P
1V
2 = P
2
V
2
V
= volume (same units for both)
P
= pressure (same units for both)
Charles's Law:
V
1/
T
1 =
V
2/
T
2
T
= temperature (K) K = °C + 273
V
= volume (same units for both)
The Combined Gas Law:
P
1
V
1/
T
1 =
P
2
V
2/
T
2
V
= volume (same units for both)
T
= temperature (K)
P
= pressure (same units for both)
Gay Lussac's Law:
P
1/
T
1 =
P
2/
T
2
P
= pressure (same units for both)
T
= temperature (K)
Avogadro's Law:
V
1/
n
1 =
V
2/
n
2
V
= volume (same units for both)
n
= moles
Dalton's Law of Partial Pressures:
P
total =
P
1 +
P
2 +
P
3 + ...
where
P
1,
P
2,
P
3 are partial pressures
S
tandard
T
emperature and
P
ressure
1 atm
0°C, 273 K
ideal gas:
molar volume = 22.4 L
Percent Composition, Empirical and Molecular Formulas
finding empirical formula
empirical formula:
lowest whole # ratio of atoms in compound
mass to mole
divide by small
Divide each value of moles by the smallest of the values
percent to mass
Assume there is 100 g of each compound.
multiply till whole
If necessary, multiply each # by same integer to get all whole #s
finding molecular formula
molecular formula:
true # of atoms of each element in formula of compound
Divide the molecular mass by the empirical mass
Multiply empirical formula by this # to get molecular formula
Find the empirical formula mass
finding empirical formula by combustion analysis
H2O g × 2.02g/18.02g = H g
N can be NH3, N2, NO2,...
CO2 g × 12.01g/44.01g = C g
subtract total mass by mass of other elements to get O
Ideal Gas
PV = nRT
n
= mole
m/M
, where m = mass M = molar mass
R
= proportionality constant, 0.08206 L⋅atm/mol⋅K
V
= volume (L)
T
= temperature (K)
P
= pressure (atm)
at
STP
,
gas density = molar mass/22.4 L
D = MP/RT
P
= pressure
R
= gas constant
D
= mass/volume
M
= molar mass
T
= temperature
Chemical Kinetics
Factors affecting rate
increased surface area:
increases rate.
particles collide more
increased concentration:
usually increases rate.
particles collide more
increased temp:
always increases rate.
particles collide more
presence of catalysts:
lower reaction rate by providing alternate pathways
Collision Model
collisions must have
sufficient energy
to produce reaction
colliding particles must be
correctly oriented
to one another in order to produce reaction
rate expression
rate = -∆[reactant]/∆
t
rate = ∆[product]/∆
t
coefficient in balanced equation is multiplied to ∆
t
reaction rate:
rate at which chemical reaction takes place; measured by rate of formation of product or rate of disappearance of reactants
rate law:
describes way in which reactant concentration affects reaction rate
rate =
k
[reactant]^
n
[] = M of reactant
n
= order of reactant
k
= rate constant
Acids and Bases
Bronsted-Lowry
conjugate acid:
forms when base gains proton (H+)
conjugate base:
forms when acid loses proton (H+)
base:
accepts proton (H+)
acid:
donates proton (H+)
Arrhenius
base:
produce hydroxide ions (OH-) in aqueous solutions
acid:
produce hydronium ions (H3O+) in aqueous solutions
neutralization:
acid and base react to produce salt and water
pH and pOH calculations
[OH-] = 10^-pOH
[H3O+] = 10^-pH
pH + pOH = 14
pOH = -log[OH-]
pH = -log[H3O+]
weak acids and bases
ICE table, solve for x, which is also [H3O+] or [OH-]
5% Rule:
If K
a
/K
b
is 1/1000 smaller than [salt], remove the change in x for [salt]
strong acids and bases
pH directly from concentration of acid
pOH directly from concentration of base
at 25 °C,
K
w
= [H3O+][OH-] = (1 × 10^-7)(1 × 10^-7) =
1 × 10^-14
Classifying salts as acidic, basic, or neutral
strong acid + weak base:
acidic
weak acid + strong base:
basic
strong acid + strong base:
neutral
weak acid + weak base:
which K
a
/K
b
is greater
Strong Acids:
HCl, HBr, HI, HCIO4, HClO4, HNO3, H2SO4
Strong Bases:
KiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
Chemical Equilibrium
Le Chatelier's Principle:
when system at equilibrium is placed under stress, system will undergo change in such way to relieve that stress
remove concentration from one side of the reaction:
system shifts to
that
side
increase temp of exothermic forward reaction:
system shifts to reactants
add more concentration to one side of the reaction:
system shifts to the
other
side
increase temp of endothermic forward reaction:
system shifts to products
increase pressure of reaction:
system shifts to side with fewer moles in balanced equation
equilibrium expression
Keq
= [product]/[reactant]
raise each substance'a concentration to power equal to substance's coefficient in balanced equation
Keq
= equilibrium constant
[] = M of substance
concentration of any solid or pure liquid that takes part in reaction is left out because these concentrations never change
equilibrium expression involving pressure
Kp
=
Pproduct/Preactant
where
Pproduct,Preactant
are equilibrium partial pressures
reaction quotient
Draw a # line from least to greatest. Plot approximate values for
Q
and
K
the # line.
Q
ueen to
K
ing.
Q
→
K
. The direction of the arrow determines the direction the system shifts until equilibrium is achieved.
Q
= [product]/[reactant]