Download 2 - College of Arts and Sciences

3/18/2014
Review Chap 11 Phase diagrams
Phase diagrams display the state of a substance at
various pressures and temperatures, and the places
where equilibria exist between phases.
The solid - liquid line marks the
melting point at each pressure
The liquid – vapor line is the
boiling point at that pressure
The high critical temperature
and critical pressure are due to
strong van der Waals forces
between water molecules
Phase Diagram of Water
Interpreting a Phase Diagram
Using the phase diagram for
methane CH4
What is the temperature and
pressure of the critical point?
The critical point is where
the liquid, gaseous, and
supercritical fluid phases
coexist.
It is marked point 3 and is
approximately 80C and 50
atm.
If solid methane at 1 atm is heated while the pressure
is held constant, will it melt or sublime?
Starting at P = 1 atm and
moving horizontally, liquid
region is reached at T 180C
Then into the gaseous region,
at T  160 C.
So, solid methane melts
In order for methane to
sublime, the pressure must be
below the triple point pressure
What is the temperature and pressure of the triple point
The triple point is the point
where the solid, liquid, and
gaseous phases coexist. It is
marked point 1 and is at
180 C and 0.1 atm
Is methane a solid, liquid, or
gas at 1 atm and 0 C?
The intersection of 0 C and
1 atm is marked point 2
It is in the gaseous region
If methane at 1 atm and 0 C is compressed until a
phase change occurs, in which state is the methane
when the compression is complete?
Moving vertically up from point
2,which is 1 atm and 0 C, the
first phase change we come to is
from gas to supercritical fluid.
This phase change happens when
the critical pressure (~50 atm) is
exceeded
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Chapter 13 Solutions and Their Properties
Solutions, Mixtures, & Colloids
• SOLUBILITY RULES – Chapter 4
• SOLUBILITY – factors that affect
– Pressure (HENRY’S LAW)
– Temperature
• CONCENTRATION
– Molecular weight
• COLLIGATIVE PROPERTIES
(a) A MIXTURE is heterogeneous
(b) A SOLUTION is homogeneous
(c) A colloidal dispersion is between a
– Molecular solutions
– Ionic solutions
solution and a mixture
• COLLOIDS
Types of Aqueous Solutions
• ELECTROLYTE – A substance that
dissolves in water to produce IONS
Example: HCl(aq), NaOH(aq), NaCl(aq)
• NON ELECTROLYTE A substance that
DOES NOT produce IONS, remain as
molecules, when dissolved in water.
Example: sugar
What is in an aqueous solution of
NaCl
Na+(aq) + Cl – (aq)
MgI2
Mg2+(aq) + 2 I-(aq)
Al(NO3)3
HClO4
(NH4)2SO4
Which of the following will be water soluble
NaCl
Yes – all sodium salts are water soluble
AgCl
No – Silver, Mercury & Lead are not
water soluble
Yes – “Like water” {polar compound
CH3OH
CH3Cl
Yes – “Like water” {polar compound
CCl4
No – Not “Like water” { not polar
Classify each of the following as a heterogeneous
mixture or homogeneous mixture (a solution).
Air ……………………. Mixture (homogeneous)
Tomato juice ………… Mixture (heterogeneous)
Al3+(aq) + 3 NO3-(aq)
Iodine crystals …….…. Pure substance
H+(aq) + ClO4-(aq)
Mud …….………….. Mixture (heterogeneous)
2NH4+(aq) + SO42-(aq)
18 Carat White Gold … Mixture (homogeneous)
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Predicting Solubility Patterns
Predict whether each of the following substances is
more likely to dissolve in the nonpolar solvent carbon
tetrachloride (CCl4) or in water:
C7H16
Na2SO4
HCl
I2
a hydrocarbon, so it is molecular and nonpolar.
therefore more soluble in the nonpolar CCl4
a compound containing a metal and nonmetals, is
ionic therefore more soluble in polar water
a polar diatomic molecule therefore more soluble
in polar water
a nonpolar diatomic molecule therefore more
soluble in the nonpolar CCl4
Mole Fraction (x) , Molarity (M)
and Molality (m)
XA 
Expressing Concentration (based on weight)
1
2
weight % 
ppm 
3
ppb 
weight of solute
x10 2
weight of solution
weight of solute
x106
weight of solution
weight of solute
x109
weight of solution
A solution is made by dissolving 4.35 g glucose
(C6H12O6) in 25.0 mL of water at 25 C. Calculate the
molality of glucose in the solution.
Moles of A
Total number of moles
(25.0 mL of water)(1.00 g/mL) = 25.0 g = 0.0250 kg solvent
Moles of solute
Molarity 
Liters of SOLUTION
Molality =
Moles of solute
Kilograms of SOLVENT
A solution with a density of 0.876 g/mL contains 5.0 g
of toluene (C7H8) and 225 g of benzene. Calculate the
molarity of the solution
Calculate the concentration of CO2 in a soft drink that
is bottled with a partial pressure of CO2 of 4.0 atm over
the liquid at 25 C. The Henry’s law constant for CO2
in water at this temperature is 3.4  102 mol/L-atm
Henry’s law:
The volume of the solution is obtained from the mass
of the solution= 5.0 g + 225 g = 230 g and its density
Sg = k Pg
SCO2 = kPCO2 = (3.4  102 mol/L-atm)(4.0 atm)
SCO2 = 0.14 mol/L
SCO2 = 0.14 M
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Colligative Properties
van’t Hoff i Factor
i represents the number of ions in solution
Vapor Pressure LOWERING
∆P = i x2 P°1
i=1
sugar, ethylene glycol, etc
Boiling Point ELEVATION
∆Tb = i Kb m
i=2
NaCl ; KNO3 ; CsC2H3O2 ; etc
Freezint Point LOWERING
∆Tf = i Kf m
i=3
MgCl2 ; (NH4)2ClO4 ; etc
i=4
Can you think of examples ?
Osmotic Pressure
π = i MRT
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a
density of 1.26 g/mL. Calculate the vapor pressure of a
solution made by adding 50.0 mL of glycerin to 500.0
mL of water at 25 C. The vapor pressure of pure water
at 25 C is 23.8 torr
Raoult’s law:
Psolution = Xsolvent Po solvent
Automotive antifreeze consists of ethylene glycol,
CH2(OH)CH2(OH), a nonvolatile nonelectrolyte.
Calculate the boiling point and freezing point of a 250g
of ethylene glycol in 750 g water
∆Tb = kb m
and ∆Tf = kf m
∆Tb = kb m = (0.51)(5.37) = 2.7
therefore Tb = 102.7
∆Tf = kf m = (1.86)(5.37) = 10.0
so Tf = -10.0 0C
The osmotic pressure of an aqueous solution containing
3.50 mg of a protein dissolved in sufficient water to
form 5.00 mL of solution. The osmotic pressure of the
solution at 25 C was found to be 1.54 torr. Treating the
protein as a nonelectrolyte, calculate its molar mass.
Osmotic pressure = MRT
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Chapter 14
• Factors that affect reaction rates
• Reaction Rates
• Rate Law Expression
Rate Law :
How rate depends on concentration of reactants
rate α (concentration of reactants)
Rate Constant:
A constant of proportionality between the reaction
rate and the concentration of reactants.
– Rate constant (units)
– Exponents in Rate Law (order of reaction)
rate =
k
x (concentration of reactants)
---------------------------------------------
Reaction Order
The EXPONENTS in the rate law
rate = k x (concentration of reactants)X
Determine the rate law
the order of reaction
and the rate constant
Reaction Order
the sum of the powers to which all reactant
concentrations in the rate law are raised
Reaction order is determined experimentally
Example: for the reaction
2–(aq)
S2O8
+ 3 I–(aq) 2SO42–(aq) + I3–(aq)
Determine rate law, order and rate constant
Rate = k [S2O82–] X [I–] Y
How many unknowns in rate equation?
“Run” the reaction
NH4+(aq) + NO2−(aq)  N2 (g) + 2 H2O (liq)
at least three (3) times
Comparing Experiments 1 and 2, when [NH4+] doubles
the initial rate doubles
Therefore X = 1 in
Rate = k [NH4+]x [NO2−] y
for
NH4+(aq) + NO2−(aq)  N2 (g) + 2 H2O (liq)
The rate law is
Rate = k [NH4+]x [NO2−] y
Find the three (3) unknowns
k, X, and Y
Using the method
of Initial Rates
NH4+(aq) + NO2−(aq)  N2 (g) + 2 H2O (liq)
Rate = k [NH4+]x [NO2−] y
Comparing Experiments 5 and 6 when [NO2-] doubles
the initial rate doubles
Therefore Y = 1 in
Rate = k [NH4+]1 [NO2−] y
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3/18/2014
NH4+(aq) + NO2−(aq)  N2 (g) + 2 H2O (liq)
Rate = k [NH4+]x [NO2−] y
Reaction of peroxydisulfate ion (S2O82–) with iodide ion (I–)
is S2O82–(aq) + 3 I–(aq) 2SO42–(aq) + I3– (aq)
Rate = k [NH+] 1[NO2−] 1
Rate = k [S2O82–] X [I–] Y
The equation is called the rate law,
the reaction is 2nd order
and k is the rate constant
Which is found from any of the experiments
Exp
1
2
3
4
[S2O8]
0.018
0.027
0.036
0.050
[I-]
0.036
0.036
0.054
0.072
2–] X
Rate = k [S2O8
Use the following data to Determine
– the rate law
– the order
– and rate constant
Rate(M/s)
2.6 x 10-6
3.9 x 10-6
7.8 x 10-6
1.4 x 10-5
Rate1 = 2.6 x 10-6 = k [0.018]x [0.036]y
Rate2 = 3.9 x 10-6 = k [0.027]x [0.036]y
Rate3 = 7.8 x 10-6 = k [0.036]x [0.054]y
Rate4 = 1.4 x 10-5 = k [0.050]x [0.072]y
[I–] Y
Divide Rate 2 by Rate 1 to find X
Rate1 = 2.6 x 10-6 = k [0.018]x [0.036]y
Rate2 = 3.9 x 10-6 = k [0.027]x [0.036]y
Rate3 = 7.8 x 10-6 = k [0.036]x [0.054]y
Rate4 = 1.4 x 10-5 = k [0.050]x [0.072]y
Rate1 = 2.6 x 10-6 = k [0.018]x [0.036]y
Rate2 = 3.9 x 10-6 = k [0.027]x [0.036]y
Rate3 = 7.8 x 10-6 = k [0.036]x [0.054]y
Rate4 = 1.4 x 10-5 = k [0.050]x [0.072]y
How do you find y ?
Remember that X is no longer unknown !
Rate = k [S2O82–]1 [I-]y
Divide Rate 4 by Rate 1 to find Y
R4 14 x 10- 6
k .050 .072


 (2.78)(2) y
R1 2.6 x 10 -6 k .0181 .036Y
1
5.38 
R2 3.9 x 10 - 6 k .027 .036
.027  (1.5) X



R1 2.6 x 10 -6
k .018X .036Y .018X
X
1 .5 
Y
X
Therefore X = 1
For the Reaction
S2O82–(aq) + 3 I–(aq) 2SO42–(aq) + I3–(aq)
Rate law : Rate = k [S2O82–] [I-]
Order :
X+Y=1+1= 2
Rate constant :
2.2 x 10-6 = k (0.018)(0.036)
Y
k = 1.5 x 10-3 Units ???
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3/18/2014
For First Order Reactions:
final

Extra
1. Integrated form of the rate equation
2. Half life

initial
1st
ln
[ A]0
kt
[ A]

t
0
k dt
[A]0 = initial concentration at time t = 0
[A] = concentration at time t
ln
For
Order Reactions
d [ A]

[ A]
AB
[ A]0
kt
[ A]
When ln P is plotted as a function of time, for
reaction CH3NC  CH3CN get a straight line
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order,
a plot of ln [A] vs. t
will yield a straight line,
and the slope of the line will be -k.
II. Reaction Rates & Half-Life
The half-life (t1/2) of a reaction
is the TIME required for the
concentration of a reactant to
decrease to one half its initial value
Therefore CH3NC  CH3CN is a 1st order process
and k is the negative of the slope = 5.1  10-5 s−1.
Half-life {t1/2 }
is defined as the time required for one-half of a
reactant to react
For
AB
at t1/2 ,
the concentration of A is one-half the
initial concentration of A
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Half-life is defined as the time required
for one-half of a reactant to react
All radioactive decay is 1st order
[ A ]0
 kt
[ A]
At t1/2 concentration is ½ initial
ln
[A] at t1/2 is one-half
of the original [A],
at 13,000 sec
[CH3NC] = 0.5 [A]0
ln
[ A]0
 k t1/2
2  A0
1
ln 2 = k t1/2
Plants take up atmospheric CO2 by
photosynthesis, and are ingested by animals, so
every living thing is constantly exchanging 14C
with its environment as long as it lives
Once it dies this exchange stops, and the amount of 14C
gradually decreases through radioactive beta decay
What is the rate constant for the
radioactive C14 decay?
Carbon-14 has a half-life of 5,730 ± 40 year
All radioactive decay is 1st order
ln 2 = k t1/2
k = ln 2 / t1/2 = 0.693 / 5730
k = 1.21 x 10 – 4 yr -1
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