1. Chemistry Lab 2010 Presenter: John R Kiser, MS
Hickory Regional Director
State Supervisor, Chemistry Lab

2. Introductions Topics for 2010: Kinetics and Solutions
Regional vs State Topics
Safety Requirements
Must bring calculator!
Need to know topics
Formula Writing/Nomenclature
Mole & Stoichiometry Calculations

3. Solution Terminology Solution: Homogeneous mixture
Solvent: Component in greater/greatest amount
Solute: Component(s) in lesser/least amount

4. Factors that influence solubility
Polarity of Solute and Solvent –
“Like Dissolves Like”
Polar solutes dissolve in polar solvents
Nonpolar solutes dissolve in nonpolar solvents
Nonpolar solutes do not dissolve well in polar solvents
Temperature
Solubility of most solids in water increases with temperature
Solubility of gases in water decreases with temperature
Gas Pressure
As the pressure of a gas above a solution increases, the solubility of the gas in the solution increases. (Henry’s Law)
Sweet Tea and Soft Drinks

5. Amounts of Solute in Solution
Saturated: The maximum amount of solute is dissolved in the solvent
Unsaturated: Less than maximum amount of solute is dissolved in the solvent
Supersaturated: More than the maximum amount of solute is dissolved in solvent
To obtain a supersaturated solution, you heat solution until all solute dissolves. Carefully and slowly cooling the solution keeps all the solute dissolved in solvent.
Solubility curves show maximum amount of solute that can be dissolved in 100 mL of water at a particular temperature. Above curve = supersaturated, Below curve = unsaturated

6. Saturation What is the solubility of Sodium Acetate at 60 oC?
What mass of sodium acetate will dissolve in 250 mL of water at 60 oC?
Is 40 grams of sodium acetate in 50 mL of water at 60 oC saturated, unsaturated, or supersaturated?

7. Units of Concentration
Chapter 11: Solutions and Their Properties
Units of Concentration
9/18/2018
Liters of solution
Moles of solute
Molarity (M) =
Total mass of solution
Mass of component
x 100%
Mass Percent =
Mass of solvent (kg)
Moles of solute
Molality (m) =
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

8. Units of Concentration
Chapter 11: Solutions and Their Properties
Units of Concentration
9/18/2018
Assuming that seawater is an aqueous solution of NaCl, what is its molarity? The density of seawater is g/mL at 20 °C, and the NaCl concentration is 3.50 mass % mass % = 3.50 grams of salt in grams of solution
Assuming g of solution, calculate the volume:
g solution
1.025 g solution
1 mL solution
1000 mL solution
1 L solution x x
= L solution
Convert the mass of NaCl to moles:
3.50 g NaCl
58.4 g NaCl
1 mole NaCl x
= moles NaCl
Then, calculate the molarity:
L solution
moles NaCl
= M NaCl
Copyright © 2010 Pearson Prentice Hall, Inc.

9. Units of Concentration
Chapter 11: Solutions and Their Properties
9/18/2018
Units of Concentration
In the previous example, what was the MOLALITY (m) of sodium chloride in seawater? Assume seawater contains only sodium chloride and water.
Calculate the mass of water (solvent) in kg:
g solution – 3.50 grams NaCl (solute) = grams water (solvent) = kg
Convert the mass of NaCl (solute) to moles:
3.50 g NaCl
58.4 g
1 mole NaCl x
= moles NaCl
Then, calculate the molality:
kg Solvent
moles NaCl
= m NaCl
Copyright © 2010 Pearson Prentice Hall, Inc.

10. Concentrations - ppm
50 ppm means a solution contains 50 grams of solute in 106 grams of solution (50 mg in 1 kg solution) In dilute aqueous solutions at 25 oC, ppm is also equivalent to mg solute in 1 L solution
Total mass of solution
Mass of component
x 106
Parts per million (ppm)=
(Mass based)

11. Solving for Unknown Concentration: Density
Density of solution increases as solute concentration increases.
The plot of density of solution versus concentration of solution should be linear.
Can be used to solve for an unknown concentration.
Example: Sugar concentration in Juice

12. Solving For Unknown Concentration: Titrations (Volumetric Analysis)
In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Indicator – substance that changes color at (or near) the
equivalence point
Equivalence point – the point at which the reaction is complete
Sometimes called stoichiometric point
Endpoint – The point at which the indicator changes color
Slowly add
reactants
UNTIL
the indicator
changes color
4.7

13. Steps for Solving Titration Problems
STEP 1: Write the correct chemical equation
STEP 2 Determine moles of starting compound
STEP 3: Determine moles of desired compound
STEP 4: Solve the problem
Example Problem: Titration of Citric Acid in Fruit Juice
3 NaOH (aq) + H3C6H5O7 (aq) → 3 H2O (l) + Na3C6H5O7 (aq)

14. Solving for Unknown Concentration: Lambert-Beer
Lambert-Beer Law: A = εbc
A = Absorbance (unitless) b = path length (cm)
c = concentration (M)
ε = Molar absorbtivity (constant, units M-1 cm-1)
Lambert-Beer or Beer’s Law Plot:
Provided all absorbance measurements are made on the same spectrophotometer with the same cell, a graph of absorbance vs. concentration will be linear.
Example: Copper (II) ion concentration

15. Using Concentrations to Find Molar Mass: FP Depression
Adding a solute to a solvent decreases the freezing point
∆Tf = kf*m ∆Tf = decrease in freezing point
kf = Freezing point constant (1.86 oC m-1 for water)
m = molality
Assumes ideal behavior and that solute is NOT ionic.
How to use to find molar mass of solute:
Use ∆Tf and kf to find molality of solution
Use mass of solvent to find moles solute present
Mass solute dissolved divided by moles solute gives the molar mass of solute!

16. Freezing Point Depression Example
When 2.50 grams of a covalent compound is dissolved in kg of water, the freezing point is determined to be ⁰C. What is the molar mass of the compound? (Assume Ideal Behavior) Molality = ∆T = ⁰C = m Kf 1.86 ⁰C m kg water * moles solute = moles solute kg water Molar mass= 2.50 grams solute = 62.0 grams per mole moles solute

17. Chemical Kinetics Collision Theory A chemical reaction occurs when
Collisions between molecules have sufficient energy to break the bonds in the reactants.
Molecules collide with the proper orientation.
Bonds between atoms of the reactants (N2 and O2) are broken, and new bonds (NO) form.
Energy needed to start the reaction (break reactant bonds) is called the Activation Energy (Ea)

18. What Prevents Collisions From Not Resulting in a Reaction?
A chemical reaction does not take place if the
Collisions between molecules do not have sufficient energy to break the bonds in the reactants.
Molecules do not collide with proper orientation.

19. Kinetics – Measuring Rates
The Rate of a reaction is measured by:
Change in concentration divided by change in time
Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
rate = -
∆[A]
∆ t 1 2
rate =
∆[B]
∆ t
aA + bB cC + dD
rate = -
∆[A]
∆ t 1 a
= -
∆[B]
∆ t 1 b =
∆[C]
∆ t 1 c =
∆[D]
∆ t 1 d

20. Ways to Define the Rate of the Reaction
Br2 (aq) + HCOOH (aq) Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
average rate = -
∆[Br2]
∆ t
= -
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time (slope of tanget line)
Initial rate = rate at very start of experiment
13.1

21. Chapter 12: Chemical Kinetics
Reaction Rates
9/18/2018
2N2O5(g)
4NO2(g) + O2(g)
Copyright © 2010 Pearson Prentice Hall, Inc.

22. How Can Collision Theory Be Used to Increase Rate
Increasing the concentration of reactants
Increases the number of collisions.
Increases the reaction rate.
Increasing the temperature of reaction
Increases average kinetic energy of molecules
Increases the force of collisions
Increases collisions with enough energy to break reactant bonds.

23. Collision Theory and Rate
The States of Reactants tend to also affect reaction rate:
All solid reactants, reactants held firmly in place, little collisions can take place.
Gas, Liquid, or Aqueous – Particles can move more freely to have collisions.
Related is Surface Area
Increasing surface area of solid generally increases rate.
More solid surface is ready to react, more collisions, increasing the rate.

24. Catalysts A catalyst : Increases rate of a reaction.
Lowers the energy of activation.
Lower activation energy means that more collisions occur that break reactant bonds
Is not used up during the reaction
Biological catalysts are called enzymes

25. Rate Laws Rate laws are always determined experimentally.
Reaction order is always defined in terms of reactant (not product) concentrations.
General Form of Rate Law:
Rate = k [Reactant 1 ]x [Reactant 2]y etc…
k = rate law constant (only at particular temperature)
With respect to individual reactants, the exponents in the rate law represent the order with respect to that reactant.

26. Expressing the Rate Law
First order: Exponent is 1 Second order: Exponent is 2 Zeroth order: Exponent is 0 (not in the RL!) Overall order: Sum of exponents
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

27. The Rate Constant
The rate law constant k is only valid at a particular temperature. k = Rate of Reaction [Reactant 1 ]x [Reactant 2]y … If the overall order of the reaction is z, units for k are M 1-z s-1. Results in rate of reaction having units of M s-1

28. Determining Order of Reaction: Initial Rates Method
One method to determine the exponent (order of reaction) is to change initial concentrations of each reactant.
Measure how initial rate changes as concentration of each reactant is changed.
Zeroth order = concentration doubles, rate unchanged
First order = concentration doubles, rate doubles
Second order = concentration doubles, rate quadruples
Remember that rate is change in concentration divided by time. Just looking at time changes can lead to wrong answer!
When two experiments are being compared, remember to make sure only one reactant has a change of initial concentration

29. As NO2 doubles, rate quadruples, so second order with respect to NO2
Determine the rate law and calculate the rate constant for the following reaction from the following data:
NO2 (g) + CO (g) NO (g) + CO2 (g)
As NO2 doubles, rate quadruples, so second order with respect to NO2
As CO doubles, rate is not changed, so zeroth order with respect to CO
Rate = k [NO2]2[CO]0 = k [NO2]2
Overall order is 2+ 0 or 2
k = 2.2 x 10-3 M s-1 = 0.22 M-1 s-1
(.10 M)2
Experiment
[NO2] M
[CO] M
Initial Rate (M/s) 1
0.10
2.2 x 10-3 2
0.20
8.8 x 10-3 3
13.2

30. Determining Order of Reaction: Linear Plot Method
A second method for determining order of reaction is linear plot method
Monitor concentration of one reactant (A) versus time
If [A] vs t is linear, 0th order with respect to A, slope = -k
If ln [A] vs t is linear, 1st order with respect to A, slope = -k
If [A]-1 vs t is linear, 2nd order with respect to A, slope = k
Considerations
[A] should be relatively small compared to the concentration of other reactants.
Unless A is the only reactant, k is called “pseudo-rate constant“
To find order with respect to other reactants, vary their concentration and notice change of A. Use initial rate method.

31. Integrated Rate Law Obtained from rate law via calculus
Can also be determined from linear plot
Use to predict concentration of a reactant at a certain point in time
Please see “Kinetics Reference Sheet”

32. Linear Plot Method Activity
Zero and First order reactions

33. Thank you!
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