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

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Presentation on theme: "Chemistry Lab 2010 Presenter: John R Kiser, MS Hickory Regional Director State Supervisor, Chemistry Lab."— Presentation transcript:

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 o C? What mass of sodium acetate will dissolve in 250 mL of water at 60 o C? Is 40 grams of sodium acetate in 50 mL of water at 60 o C saturated, unsaturated, or supersaturated?

7 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 component x 100% Molarity (M) = Liters of solution Moles of solute

8 Units of Concentration 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 3.50 g NaCl = 0.0599 moles NaCl 58.4 g NaCl 1 mole NaCl x 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 solution x 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:

9 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. 3.50 g NaCl = 0.0599 moles NaCl 58.4 g 1 mole NaCl x 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:

10 Concentrations - ppm 50 ppm means a solution contains 50 grams of solute in 10 6 grams of solution (50 mg in 1 kg solution) In dilute aqueous solutions at 25 o C, ppm is also equivalent to mg solute in 1 L solution Parts per million (ppm)= (Mass based) Total mass of solution Mass of component x 10 6

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. Equivalence point – the point at which the reaction is complete Sometimes 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 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) + H 3 C 6 H 5 O 7 (aq) → 3 H 2 O (l) + Na 3 C 6 H 5 O 7 (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 ∆T f = k f *m∆T f = decrease in freezing point k f = Freezing point constant (1.86 o C 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 ∆T f and k f 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 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 K f 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

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 (N 2 and O 2 ) are broken, and new bonds (NO) form. Energy needed to start the reaction (break reactant bonds) is called the Activation Energy (E a )

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 = ∆[B] ∆ t rate = - ∆[A] ∆ t 1 2 aA + bB cC + dD rate = - ∆[A] ∆ t 1 a = - ∆[B] ∆ t 1 b = ∆[C] ∆ t 1 c = ∆[D] ∆ t 1 d

20 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = - ∆[Br 2 ] ∆ t = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent 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

21 Chapter 12/21 Reaction Rates 2N 2 O 5 (g)4NO 2 (g) + O 2 (g)

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 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 Determining Order of Reaction: Initial Rates Method

29 Determine the rate law and calculate the rate constant for the following reaction from the following data: NO 2 (g) + CO (g)NO (g) + CO 2 (g) As NO 2 doubles, rate quadruples, so second order with respect to NO 2 As CO doubles, rate is not changed, so zeroth order with respect to CO Rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 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 [NO 2 ] M[CO] M Initial Rate (M/s) 10.10 2.2 x 10 -3 x 10 -3 30.20 8.8 x 10 -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, 0 th order with respect to A, slope = -k If ln [A] vs t is linear, 1 st order with respect to A, slope = -k If [A] -1 vs t is linear, 2 nd 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! Please email me at For follow-up questions, concerns, etc..

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