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					   CE 541




Solutions
“A solution is a homogeneous throughout and is
composed of two or more pure substances. They are
weakly bounded mixtures of a solute and a solvent”

Solute: is usually the component in less quantity
Solvent: is usually the component in greater quantity

The solute dissolves in the solvent and is considered
soluble in the solvent

Aqueous solutions: are solutions where water is used as
the solvent
           Types of Solutions


   Gas in a Liquid
   Liquid in a Liquid
   Solid in a Liquid
 Solute   Solvent                          Examples

Gas       Liquid    Carbonated beverages (CO2 in water)

Liquid    Liquid    Anti-freeze in car radiators (ethyl glycol in watr)

Liquid    Solid     Dental fillings (mercury in silver)

Solid     Liquid    Sugar in water

Solid     Solid     Solder (tin in lead)
                                        Solute
Examples
   of
Solutions           Gas                Liquid                 Solid

            Oxygen and other
                                Water vapor in air   Iodine sublimates into
    Gas     gases in nitrogen
                                (humidity)           air
            (air)
                                Ethanol (common      Sucrose (table sugar)
                                alcohol) in water;   in water; sodium
         Carbon dioxide in
S                               various              chloride (table salt) in
  Liquid water (carbonated
o                               hydrocarbons in      water; gold in
         water)
l                               each other           mercury, forming an
v                               (petroleum)          amalgam
e        Hydrogen dissolves
n        rather well in
t        metals; platinum has
                              Hexane in paraffin
         been studied as a                           Steel, duralumin,
   Solid                      wax, mercury in
         storage medium.                             other metal alloys
                              gold.
         This effect was used
         in the cold fusion
         experiments.
Factors Affecting Solubility and Rate
             of Solution
   1. Factors that affect the actual solubility of a
    given solute in a solvent:
       Properties of solute
       Properties of solvent
       Temperature
       Pressure
   2. Factors that affect the rate (how fast) at which a
    given solute dissolves in a given solvent:
       Particle size of solute
       Rate of stirring
       Temperature
Assignment 2: Discuss the two sets
of factors.
                   Example


Calculate the solubility in grams per liter of a
certain gas in water at a partial pressure of 3.5 atm
and 0 C. The solubility is 0.530 g/l at a total
pressure of 1.00 atm and 0 C.
                        Solution
Using Dalton's Law of Partial Pressure
Ptotal = Pgas Pwater
Pwater at 0 C is 0.006 atm
Pgas = 1.00 – 0.006 = 0.994 atm
Solubility1 = 0.53 g/l                    P1 = 0.994
Solubility2 = ?                           P2 = 3.50 atm
Solubility2 = Solbility1  Pressure Factor
               = 0.53 g/l  (3.5 atm / 0.994 atm) = 1.87 g/l
       Saturated, Unsaturated and
        Supersaturated Solutions
1. Saturated Solutions
Are solutions which are in dynamic equilibrium
() with undissolved solutes
They can be prepared by adding an excess of
solute to a given amount of solvent and allowing
sufficient time for a maximum amount of solute
to dissolve with excess solute present
In this case:
Rate of dissolution (dissolved solute) = Rate of
crystallization (undissolved solute)
2. Unsaturated Solutions
Are solutions in which the concentration of solute
is less than that of the Saturated (equilibrium)
Solutions, under the same conditions.

3. Supersaturated Solutions
Are solutions in which the concentration of solute
is greater than that possible in Saturated
(equilibrium) Solutions, under the same
conditions.
      Concentrations of Solutions
1. percent by mass
       % by mass = (mass of solute / mass of solution)  100
2. parts per million, ppm
       ppm = (mass of solute / mass of solution)  1,000,000
3. molarity
       M = molarity = (moles of solute / liter of solution)
4. Molality
       m = molality = (moles of solute / kilogram of solvent)
5. Normality
       N = normality = (equivalents of solute / liter of solution)
             Reaction Rates
The law of mass action states that:
"the rate of a chemical reaction is proportional
to the active mass of the reactants"

The active mass is related to relative molar
concentration of the reactants in moles per liter
for solutions
                  aA + bB  cC + dD
For the General Reaction
The overall rate of reaction is proportional to the
concentration of the reactants in moles per liter raised to
certain power
                    Rate  [A]x [B]y

[A] = concentration of A in moles / liter
[B] = concentration of B in moles / liter
x and y = whole number, fractional numbers, negative
numbers, or zero as determined by experimentation
Then:
              Rate = k [A]x [B]y
k = a proportionality constant, called the specific
rate constant

Sometimes x and y are equal to the coefficients of
the balanced equation; that is a and b. The values
of x and y have to be determined experimentally.
The value of x and y is the reaction order of each
reactant. The sum of x and y is the overall
reaction order.
                        Example
Given the following chemical equation and rate equation, determine
the reaction order of each reactant and the overall reaction order.
                   Cl2 + CHCl3  HCl + CCl4
                       Rate = [Cl2]0.5[CHCl3]
Chlorine + Chloroform  Hydrochloric Acid + Carbon tetra-
chloride
The reaction is half order for chlorine and first order for
chloroform, with the overall reaction order being 1.5.

                             A+BC
                         Rate = k [A]2[B]3
The reaction is second order for A and third order for B, with
overall reaction order being 5.
    Factors Affecting the Rate of a
         Chemical Reaction


   Nature of Reactants
   Concentration of Reactants
   Temperature
   Catalysts
Assignment 3: Discuss the Four
           Factors
                Chemical Equilibria
 (Reversible and Irreversible Reactions)
Some reactions are irreversible in practice, meaning that
the chemical equilibrium is not established and that the
reaction is complete. When can this happen?
      Products are removed
      Rate of reverse reaction is very slow (negligible)


What products act as a driving force for a reaction to go
irreversibly?
      Gas
      Precipitate
      Non-ionized or partially ionized substance, such as water
                      GAS
Gas removed as soon as it forms

   MgCO3 + 2HCl  MgCl + H2O + CO2



If gas remains in contact with the reactants, as in a
closed container, then a reversible reaction occurs
and an equilibrium is established
             PRECIPITATE
The precipitation of a substance acts to remove it
from the reaction

      AgNO3 + HCl  AgCl  + HNO3

The reaction is reversible as long as the precipitate
is in contact with the solution but equilibrium
favors the products
                 WATER

        NaOH + HBr  NaBr + H2O



The equilibrium is established but the reaction
strongly favors the products
Reversible Reactions
  A+BC+D
The system is in chemical equilibrium when:

Rate at which C and D molecules react to form A and
B molecules = Rate at which A and B molecules
react to form C and D molecules

For any equilibrium reaction, a constant known as the
equilibrium constant (K) can be obtained experimentally if
all quantities in the expression can be determined.
For the Law of Mass Action
     Rate Forward  [A][B] = kf [A][B]
      Similarly, Rate Reverse = kr [C][D]

kf and kr are the specific rate constants for the
forward and reverse reactions, respectively.

At equilibrium:
Rate of forward reaction = Rate of reverse
reaction
                    k f [ A][ B ]  k r [C ][ D ]


        kf   [C ][D]
then       
        k r [ A][B]

since kf and kr are constants, then (kf / kr) is also constant.
                       [C ][ D]
                   K
                       [ A][ B ]
K is the equilibrium constant which has a certain value at
a given temperature for a given reaction.
Generally,
If aA  bB  cC  dD

then,       [C ]c [ D] d
         K
            [ A] a [ B ]b
                 Example
Write the expression of K for the following
reactions:


       CO  Cl2  COCl2 .....( )
                             1
       H 2  I 2  2 HI .....( )
                             2
       CaCO3  CaO  CO2 .....( )
                              3
                   Solution
                    [COCl ]
                K             .....( 1)
                   [CO ][Cl2 ]
                     [ HI ]2
                K                .....( 2)
                   [ H 2 ][ I 2 ]
                K  [CO2 ].....( 3)

in equation (3), since CaCO3 and CaO are solids,
they are not considered in the equilibrium
expression because their concentrations are
constant at a given temperature and hence they are
included in the value for the constant K.
        Le Chatelier’s Principle


“If an equilibrium system is subjected to a change
in conditions of Concentration, Temperature, or
Pressure, the system will change to a new
equilibrium position, where possible, in a direction
that will tend to restore the original conditions.”
                  Concentration
   When the concentration of one of the substance in a
    system at equilibrium is increased, the principle predicts
    that the equilibrium will shift so as to use up partially
    the added substance.
   Decreasing the concentration of one substance in a
    system at equilibrium will cause the equilibrium to shift
    so as to replenish partially the substance removed.
   In all cases, the equilibrium constant, K, will remain
    constant with the concentration of the reactants or
    products varying.
    If we have:


               A B C  D
    1.   increase in concentration of either   A or B will shift the
         equilibrium to the products side
    2.   increase in concentration of either   C or D will shift the
         equilibrium to the reactants side
    3.   decrease in concentration of either   A or B will shift the
         equilibrium to the reactants side
    4.   decrease in concentration of either   C or D will shift the
         equilibrium to the products side
                     Temperature
    “If the temperature of a system at equilibrium is
     changed, the equilibrium will shift so as to change the
     temperature towards its original value.”
    A. Exothermic Reactions

    A  B  C  D  heat.energy
    1.   increase in temperature will shift the equilibrium to
         reactants side
    2.   decrease in temperature will shift equilibrium to products
         side
    B. Endothermic Reactions


    A  B  C  D  heat.energy
    1.   increase in temperature will shift the equilibrium to
         the products side
    2.   decrease in temperature will shift equilibrium to
         the reactants side
    The equilibrium constant, K, will change when
     temperature is changed.
                    Pressure
   Increasing the pressure on a system at
    equilibrium will shift the equilibrium in the
    direction which will decrease the volume.
    Decreasing the pressure will have the opposite
    effect.
   If no change in volume in going from reactants
    to products, pressure will have no effect on the
    equilibrium. The equilibrium constant, K, does
    not change with change in pressure.
Examples
                Concentration
   Changing the concentration of an ingredient will
    shift the equilibrium to the side that would
    reduce that change in concentration.
   This can be illustrated by the equilibrium of
    carbon monoxide and hydrogen gas, reacting to
    form methanol.
               CO + 2 H2 ⇌ CH3OH
Suppose we were to increase the concentration of
CO in the system. Using Le Chatelier's principle
we can predict that the amount of methanol will
increase, decreasing the total change in CO. If we
are to add a species to the overall reaction, the
reaction will favor the side opposing the addition
of the species. Likewise, the subtraction of a
species would cause the reaction to fill the “gap”
and favor the side where the species was reduced.
                   Temperature
   Let us take for example the reaction of nitrogen gas
    with hydrogen gas. This is a reversible reaction, in
    which the two gases react to form ammonia:

             N2 + 3 H2 ⇌ 2 NH3 ΔH = -92kJ

   This is an exothermic reaction when producing
    ammonia. If we were to lower the temperature, the
    equilibrium would shift in such a way as to produce
    heat. Since this reaction is exothermic to the right, it
    would favor the production of more ammonia.
                  Total Pressure
   Once again, let us refer to the reaction of nitrogen gas
    with hydrogen gas to form ammonia:

             N2 + 3 H2 ⇌ 2 NH3 ΔH = -92kJ

   Note the number of moles of gas on the left hand side,
    and the number of moles of gas on the right hand side.
    We know that gases at the same temperature and
    pressure will occupy the same volume. We can use this
    fact to predict the change in equilibrium that will occur
    if we were to change the total pressure.
Suppose we increase total pressure on
the system: now, by Le Chatelier's
principle the equilibrium would move to
decrease the pressure. Noting that 4
moles of gas occupy more volume than
2 moles of gas, we can deduce that the
reaction will move towards the products
if we were to increase the pressure.
a. Effect of Adding an Inert Gas
   An inert gas (or noble gas) such as helium is one which
    does not react with other elements or compounds. To
    add an inert gas into a closed system at equilibrium may
    or may not result in a shift. For example, consider
    adding helium to a container with the following
    reaction:

                    N2 + 3H2 ⇌ 2NH3

   The main effect of adding an inert gas to a closed
    system is that it will increase the total pressure or
    volume. An inert gas would not be directly involved in
    the reaction, but could result in a shift.
     b. Volume Held Constant
If volume is held constant, the individual
concentrations of the above gases do not
change. The partial pressures also do not
change, even though we have increased the
total pressure by adding helium. This means
the reaction quotient does not change, so the
system is still at equilibrium and no shift
occurs.
  c. Volume Allowed to Increase
If the volume is allowed to increase, the
concentrations, as well as the partial pressures, all
decrease. Because there are more stoichiometric
moles on the lefthand side of the equation, the
decrease in concentration affects the lefthand side
more than the righthand side. Therefore, the
reaction would shift to the left until the system is
at equilibrium again.

				
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