Chemistry Lab 2010

Presenter:John R Kiser, MS

Hickory Regional DirectorState Supervisor, Chemistry Lab

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

Solution Terminology

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

Factors that influence solubility• Polarity of Solute and Solvent –

“Like Dissolves Like”Polar solutes dissolve in polar solventsNonpolar solutes dissolve in nonpolar solventsNonpolar solutes do not dissolve well in polar solvents

• TemperatureSolubility of most solids in water increases with temperatureSolubility of gases in water decreases with temperature

• Gas PressureAs 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

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

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?

Copyright © 2010 Pearson Prentice Hall, Inc.

Units of Concentration

Mass Percent =

Molality (m) =Mass of solvent (kg)

Moles of solute

Total mass of solution

Mass of componentx 100%

Molarity (M) =Liters of solution

Moles of solute

Units of ConcentrationAssuming that seawater is an aqueous solution of NaCl, what is its molarity? The density of seawater is 1.025 g/mL at 20 °C, and the NaCl concentration is 3.50 mass %. 3.50 mass % = 3.50 grams of salt in 100.00 grams of solution

3.50 g NaCl= 0.0599 moles NaCl

58.4 g NaCl

1 mole NaClx

Convert the mass of NaCl to moles:

100.00 g solution

1000 mL solution

1 L solution= 0.09756 L solution

1.025 g solution

1 mL solutionx x

Assuming 100.00 g of solution, calculate the volume:

0.09756 L solution

0.0599 moles NaCl= 0.614 M NaCl

Then, calculate the molarity:

Units of ConcentrationIn the previous example, what was the MOLALITY (m) of sodium chloride in seawater? Assume seawater contains only sodium chloride and water.

3.50 g NaCl= 0.0599 moles NaCl

58.4 g

1 mole NaClx

Convert the mass of NaCl (solute) to moles:

100.00 g solution – 3.50 grams NaCl (solute) = 96.50 grams water (solvent) = 0.09650 kg

Calculate the mass of water (solvent) in kg:

0.09650 kg Solvent

0.0599 moles NaCl= 0.621 m NaCl

Then, calculate the molality:

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 tomg solute in 1 L solution

Parts per million (ppm)=(Mass based) Total mass of solution

Mass of componentx 106

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

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.

Equivalence point – the point at which the reaction is completeSometimes called stoichiometric point

Endpoint – The point at which the indicator changes color

Indicator – substance that changes color at (or near) the equivalence point

Slowly add reactants

UNTIL

the indicatorchanges color

4.7

Steps for Solving Titration ProblemsSTEP 1: Write the correct chemical equationSTEP 2 Determine moles of starting compound STEP 3: Determine moles of desired compoundSTEP 4: Solve the problem

Example Problem: Titration of Citric Acid in Fruit Juice3 NaOH (aq) + H3C6H5O7 (aq) → 3 H2O (l) + Na3C6H5O7 (aq)

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

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 = molalityAssumes 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!

Freezing Point Depression ExampleWhen 2.50 grams of a covalent compound is dissolved in 0.100

kg of water, the freezing point is determined to be -0.750 ⁰C. What is the molar mass of the compound? (Assume Ideal Behavior)

Molality =∆T = 0.750 ⁰C = 0.403 mKf 1.86 ⁰C m-1

0.100 kg water * 0.403 moles solute = 0.0403 moles solute1.000 kg water

Molar mass= 2.50 grams solute = 62.0 grams per mole0.0403 moles solute

Chemical KineticsCollision 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)

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.

Kinetics – Measuring Rates• The Rate of a reaction is measured by:

Change in concentration divided by change in time

Reaction Rates and Stoichiometry2A B

Two moles of A disappear for each mole of B that is formed.

rate = ∆[B]∆ t

rate = -∆[A]∆ t

12

aA + bB cC + dD

rate = -∆[A]∆ t

1a

= -∆[B]∆ t

1b

=∆[C]∆ t

1c

=∆[D]∆ t

1d

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)

average rate = -∆[Br2]

∆ t= -

[Br2]final – [Br2]initial

tfinal - tinitial

slope oftangent

slope oftangent slope of

tangent

instantaneous rate = rate for specific instance in time (slope of tanget line)Initial rate = rate at very start of experiment

13.1

Ways to Define the Rate of the Reaction

Chapter 12/21

Reaction Rates2N2O5(g) 4NO2(g) + O2(g)

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.

Collision Theory and RateThe 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.

CatalystsA 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

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.

Expressing the Rate Law

First order: Exponent is 1Second order: Exponent is 2Zeroth 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.

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

• 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 quadruplesRemember 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

Determining Order of Reaction: Initial Rates Method

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 2k = 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 0.10 2.2 x 10-3

2 0.20 0.10 8.8 x 10-3

3 0.20 0.20 8.8 x 10-3

13.2

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 = kConsiderations• [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.

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”

Linear Plot Method Activity

Zero and First order reactions

Thank you!

Please email me at jkiser@wpcc.eduFor follow-up questions, concerns, etc..