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

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?

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.

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 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 Assuming 100.00 g of solution, calculate the volume: 100.00 g solution 1.025 g solution 1 mL solution 1000 mL solution 1 L solution x x = 0.09756 L solution Convert the mass of NaCl to moles: 3.50 g NaCl 58.4 g NaCl 1 mole NaCl x = 0.0599 moles NaCl Then, calculate the molarity: 0.09756 L solution 0.0599 moles NaCl = 0.614 M NaCl Copyright © 2010 Pearson Prentice Hall, Inc.

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: 100.00 g solution – 3.50 grams NaCl (solute) = 96.50 grams water (solvent) = 0.09650 kg Convert the mass of NaCl (solute) to moles: 3.50 g NaCl 58.4 g 1 mole NaCl x = 0.0599 moles NaCl Then, calculate the molality: 0.09650 kg Solvent 0.0599 moles NaCl = 0.621 m NaCl Copyright © 2010 Pearson Prentice Hall, Inc.

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)

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. 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

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)

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 = 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!

Freezing Point Depression Example When 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 m Kf 1.86 ⁰C m-1 0.100 kg water * 0.403 moles solute = 0.0403 moles solute 1.000 kg water Molar mass= 2.50 grams solute = 62.0 grams per mole 0.0403 moles solute

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)

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 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

Ways to Define the Rate of the Reaction Br2 (aq) + HCOOH (aq) 2Br- (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

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

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 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.

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

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 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.

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

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

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

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.

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.edu For follow-up questions, concerns, etc..