Part I: Colligative Properties and Colloids

1
Part I Colligative Properties and Colloids 

• CHM 102
• Monday, July 9th

2
Colligative Properties

• Colligative properties depend upon the AMOUNT of
  solute particles dissolved in solution, not on
  the IDENTITY of the solute.
• So its not the what, its the how much.
• Colligative properties include vapor pressure,
  boiling point, and freezing point.

3
Vapor Pressure

• Recall a volatile substance is one that shows a
  measurable vapor pressure, and a nonvolatile
  substance doesnt.
• Vapor pressure ALWAYS LOWERS when a nonvolatile
  solute is dissolved into a volatile solvent.
• Raoults Law states that
• Psolvent in solutionXsolventPpure
  solvent

4
Practice Raoults Law

• The partial pressure of water vapor above a
  solution of sugar and water with a mole fraction,
  Xsucrose0.200, is 14.0 mmHg. What is the
  partial pressure of water vapor in pure water
  under these conditions?
• Psolvent in solutionXsolventPpure
  solvent

5
Boiling-point elevation

• When nonvolatile solute is added to a volatile
  solvent (aka sugar in water), the boiling point
  of the resulting solution is higher.
• The increase in boiling point is described the
  equation DTb iKbm
• i is the Vant Hoff factor, and is how many
  pieces the solute breaks into in solution.
  NaCl ionizes into 2 ions, K2CO3 into 3, C6H12O6
  into 1, etc. Pay attention to whether solutes
  are electrolyes or not!

6
Practice Boiling-point elevation

• What is i for the following solutes
• a)NH4ClO4 b)Na3PO4
• c)C6H6 d)Ca(C2H3O2)2
• What is the increase in boiling point for a
  solution in which 2.5 g of K2SO4 is dissolved in
  250 g of water?
• ( Kb0.51oC/m, DTb iKbm)

7
Freezing-point Depression

• When nonvolatile solute is added to a volatile
  solvent (aka sugar in water), the freezing point
  of the resulting solution is lower.
• The decrease in freezing point is described the
  equation DTf -iKfm

8
Practice Molar Mass with Freezing-point
Depression

• What is the molar mass of nonelectrolyte (i1)
  substance X if 3.06 g is dissolved into 650 g of
  water to result in a DTf of -1.78oC?
• (Kf 1.86oC/m, DTf -iKfm)

9
Colloids

• Colloids are essentially solutions with very
  large solute particles.
• For example, a solution of hemoglobin (MW 64,500
  amu) is a colloid. So is whipped cream, butter,
  fog, etc.
• In each case, youve got solute-like particles
  that are much larger than the molecular-level
  solutes weve seen so far (i.e. sucrose, NaCl,
  etc).

10
Colloids the Tyndall Effect

• When light passed through a colloid, it is
  scattered by the colloidal particles, creating
  the Tyndall effect.

11
Part II Reaction Rates, Dependence of Rate on
Concentration  

• CHM 102
• Monday, July 9th

12
Kinetics

• Now were moving on to Chapter 14, and well be
  working with reaction rates.
• This chapter deals with the speed at which
  reactions take place, and the factors that can
  affect the kinetics of a chemical reaction.
• Were also going to work on predicting the
  kinetics of chemical reactions.

13
Reaction Rates

• The speed of a car is measured in miles/hr. The
  speed of a chemical reaction is measured in M/s.
  
• The reaction rate is really a measure of how
  concentration of reactants or products change
  over time.
• In the reaction 2A ? B, we could measure the
  disappearance of A, or the the appearance of B.

14
Reaction Rateswhy do we care?

• DEMO sparkler
• Why does the reaction rate of this sparkler
  matter?
• Would it sell as well if the reaction rate was
  tripled?

15
Other important reaction rates

• Epoxies setting to form fast fix-its.
• Steel rusting to render a useful tool useless.
• The spoiling of foods such as milk, etc.
• The consumption of gasoline by an automobile
  engine.
• Quick Dry super glue vs. Oops super glue.

16
What effects reaction rates?

• The physical state of the reactants- example
  reacting a gas with metal
• Concentration of the reactants-
• example burning something in oxygen
• Temperature at which the reaction occurs-
• example milk spoiling
• Presence of a Catalyst-
• example catalytic converters, lactase

17
Average Reaction Rates

• Avg. Rates are calculated just as a slope of
  the line between the two time points would be
• For t4 to t24
• Avg. Rate -DA/Dt
• Avg Rate of disappearance of A
• -(-.15M)/20s 0.0075 M/s

18
Average Reaction Rates and Stoichiometry

• If we took the average rate of disappearance of A
  in the reaction A ? 2B to be 0.05 M/s in the
  first 20 seconds, what would be the average rate
  of appearance of B in the first 20 seconds?

19
Instantaneous Rates

• Lets look at the previous plot a different way.
• Instantaneous as opposed to average rates are
  rates at a specific point in a reaction, not over
  a range of times.
• Well calculate these later!

20
Concentration and Rate

• Chemists use the rate law to relate rate to
  concentration through a proportionality constant
  k, called the rate constant.
• Rate laws are of the general form
• Rate kAmBn
• where A and B are concentrations of reactants, m
  and n are typically small whole numbers (i.e.
  0,1,2). The sum of m and n is the overall order
  of the reaction.

21
Determining exponents m and n

• A good typical rule for finding the exponents for
  a rate law using a data table similar to the one
  below is
• If concentration doubles, and the rate stays the
  same ? zero order.
• If the concentration doubles, and the rate
  doubles ? first order.
• If the concentration doubles and the rate
  quadruples ? second order

22
Determining exponents m and n

• A typical problem will give you something similar
  to the data below
• Look at how the A changes when B is constant, as
  well as how B changes when A is constant.

23
Now that we know m and n

• Since we know m is 1 (the order for A) and n is 0
  (the order for B), we can now take it one step
  further for our rate law and find k.
• We know that the initial rate was 0.005 M/s when
  the initial A and B were 0.1 M.
• Fill it in! Rate kA1B0
• Rate kA1
• 0.005 M/s k0.1 M
• so k .1/.0051000 s-1

24
Now that we know m and n

• Since we know m is 1 (the order for A) and n is 0
  (the order for B), we can now take it one step
  further for our rate law and find k.
• We know that the initial rate was 0.005 M/s when
  the initial A and B were 0.1 M.
• Fill it in! Rate kA1B0
• Rate kA1
• 0.005 M/s k0.1 M
• so k .1/.00520 s-1 at 25oC!

25
k and temperature

• So k 20 s-1 at 25oC, and we can now complete
  our rate law as
• Rate 20 s-1A1
• How is k related to temperature? Notice that the
  experiment we just found k for was at 25oC. We
  had to specify that because k is temperature
  dependent! As temperature goes up, k goes up!
• What effect will a higher k have on rate?

26
Units of k

• If Rate laws are of the general form
• Rate kAmBn
• we need to be able to figure out units of k so
  the unit of rate works out to M/s.
• Units of k (units of rate)/(units of conc)(mn)
• First order? Units of k (M/s)/(M)1 s-1
• Second Order ? Units of k (M/s)/(M)2 M-1s-1
• Third Order ? Units of k (M/s)/(M)3 M-2s-1