Colorimetric Analysis & Determination of the Equilibrium for a Chemical reaction Help Notes AP Chemistry.

Slides:

Advertisements

Similar presentations

Unit 5 - Chpt 13 & 17 - Equilibrium and Thermochemistry Part II

Advertisements

O “K” : Manipulation of K Kp verse Kc.  Write an equilibrium constant expression for any chemical reaction. The concentrations of solids and solvents.

Chapter 9 Combining Reactions and Mole Calculations.

Chapter 13 Chemical Equilibrium.

Equilibrium Pre-lab Procedure and Instructions. Experimental Design Fe 3+ (aq) + SCN 1- (aq) FeSCN 2+ (aq) 1.In the first reaction the concentration of.

Equilibrium Constant Turn in CV Final Report and Pre-lab in folders

Chemical Equilibrium Chapter 18. Chemical Equilibrium Happens to any reversible reaction in a closed system Happens to any reversible reaction in a closed.

Equilibrium The condition of a system in which competing influences are balanced. The condition of a system in which competing influences are balanced.

1 Chemical Equilibrium Chapter 13 AP CHEMISTRY. 2 Chemical Equilibrium  The state where the concentrations of all reactants and products remain constant.

CHEMISTRY Please take out:  Intro to Keq worksheet (gray)  Intro K eq vs. Q #2 worksheet (yellow)  Your notebook ready for the warm-up Ask your partner,

The Equilibrium Constant 7.3. Opposing Rates and the Law of Chemical Equilibrium The Law of Chemical Equilibrium: At equilibrium, there is a constant.

Chapter 18: Chemical Equilibria Irreversible reactions: The reactants combine to form products and the reaction stops when the limiting reagent has been.

Chemical Equilibrium The reversibility of reactions.

5.3 – ICE Box Problems Unit 5: Equilibrium.

Chemical Equilibrium Introduction to Chemical Equilibrium Equilibrium Constants and Expressions Calculations Involving Equilibrium Constants Using.

Chapter 13 Chemical Equilibrium. Section 13.1 The Equilibrium Condition Copyright © Cengage Learning. All rights reserved 2 Chemical Equilibrium  The.

Making Dilutions from Solutions

Topic: Dilution Do Now:

Chemistry Chapter 12 – Quantitative Equilibrium Teacher: H. Michael Hayes.

Precipitation Titration Calculations Video Example Here we’ll be given some data from a titration and asked to use this data to calculate the concentration.

Equilibrium. How do we write the equilibrium constant expression for the following reaction? 2SO 2 (g) + O 2 (g)  2SO 3 (g)

Dilutions. Solve problems involving the dilution of solutions. Include: dilution of stock solutions, mixing common solutions with different volumes and.

When Compounds do not React in a 1:1 Molar Ratio The molar ratios affect how the concentrations change during a reaction. We must account for these ratios.

Exp. 20: Spectrophotometric Analysis: Determination of the Equilibrium Constant for a Reaction Exp videoExp video(time: 41:13 minutes)

Section 8.3—Reaction Quotient

Solution Stoichiometry. We have already explored gravimetric stoichiometry (mass-to-mass) However, most industrial reactions take place in solution –Easier.

Buffer or Common Ion problems. Identify the weak acid or base in the problem. There has to be one or the other to create a buffer. Write the ionization.

Buffers 1986 A.

Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9.

Determination of the Equilibrium Constant. Theory Beer’s Law: Concentration is proportional to Absorbance The reaction: Fe +3 + SCN - [Fe(SCN)] +2 Kc.

Fe3+ (aq) + SCN- (aq) Fe(SCN)2+ (aq) colorless colorless.

Measuring K for.

Chapter Thirteen: CHEMICAL EQUILIBRIUM.

Chemical Systems & Equilibrium

EQUILIBRIUM CONSTANT DETERMINATION

Here, we’ll show you how to calculate the initial concentration of a weak acid, given the pH and the Ka of the acid. In this particular example, we’ll.

1. - ICE Tables - 2  Previous chemistry courses, stoichiometric calculations were straightforward when reactions proceed to completion (called quantitative.

Determination of the Equilibrium Constant

CHAPTER 13 Acids and Bases 13.3 Acid-Base Equilibria.

DILUTION CALCULATIONS Molarity of Mixture = total moles of chemical in which we are interested total volume of mixture Dilute Solution – a solution with.

7.3 The Equilibrium Constant: Measuring Equilibrium Concentrations Equilibrium Calculations (ICE Charts) Qualitatively Interpreting the Equilibrium Constant.

Section 8.2—Equilibrium Constant How can we describe a reaction at equilibrium?

Equilibrium. Equilibrium is a state in which there are no observable changes as time goes by. Although there are still changes occurring, they are not.

Equilibrium Jeopardy ConceptsEquilibrium expression Finding KeqFinding QLe Chatelier’s

Chapter 15 & 16: Applications of Aqueous Equilibrium.

The Determination of an Equilibrium Constant Chemistry 117.

K eq calculations Here the value of K eq, which has no units, is a constant for any particular reaction, and its value does not change unless the temperature.

11-5 RICE Problems And you. A common problem in chemistry is to mix several chemicals, allow them to reach equilibrium, and then calculate the final concentrations.

13.1 EQUILIBRIUM CONDITION CHEMICAL EQUILIBRIUM The state where the concentrations of all reactants and products remain constant with time. On the molecular.

EQUILIBRIUM. Equilibrium Constant (K Values)  The equilibrium constant (Keq) is a number showing the relationship between the concentration of the products.

Experiment #5 DETERMINATION OF AN EQUILIBRIUM CONSTANT.

Molarity Molarity is defined as the amount of moles of a compound dissolved in an amount of solvent (usually water). It can be solved with the equation:

4.4 Oxidation-Reduction Reactions

Chapter 7.6 Solubility Equilibria and the Solubility Product Constant

Acids and Bases.

Making Dilutions from Solutions

KSP = Solubility product constant

DO NOW Pick up notes. Turn in Concentration and Reaction Rate lab.

Pick up notes. Get out Study Guide for Content Mastery. DO NOW.

The Determination of Keq for [FeSCN2+]

Solving Equilibrium Problems

Colorimetric Determination of Keq

Chapter 17: Chemical Equilibrium

Solution Limiting Reactants

The Extent of a Reaction

EXPERIMENT 3 – DETERMINATION OF THE Keq

Chemical Equilibrium AP Chemistry, LFHS.

Concentration = # of moles volume (L) V = 1000 mL V = 1000 mL

Presentation transcript:

Colorimetric Analysis & Determination of the Equilibrium for a Chemical reaction Help Notes AP Chemistry

PART II How to determine Concentration of FeSCN 2+ at equilibrium. In this step you are determining the concentration of the product at equilibrium using the absorbance reading you measured in the data table. Use the calibration curve you generated in Part I to determine the concentration of your samples. Plug in the y value for absorbance to find the x value for concentration.

PART II How to determine Concentration of FeSCN 2+ at equilibrium. Here is one set of sample data generated in class. Your absorbance values should be

DETERMINING y WHEN GIVEN x Steps for TI Graphing Calculators The directions below are form your Arrhenius made easy notes. You will use the same steps to find x except instead of looking for Kelvin temperature your x will represent the concentration of your sample at equilibrium.

How to determine the initial concentrations of diluted stock standard. In this step you are determining the INITIAL concentrations of reagents added to the reaction. You will use the calibration curve you generated in Part I to determine the equilibrium concentrations AFTER the reaction has completed. Tube # M Fe(NO 3 ) 3 (mL) M SCN- (mL) H 2 O (mL) Initial [Fe 3+ ] Initial [SCN - ] ?? ?? ?? ?? ??

How to determine the initial concentrations concentration of diluted stock standard. FIRST: Determine the total volume of the new, diluted concentration. To do this, add up all the amounts of standard added. Tube # M Fe(NO 3 ) 3 (mL) M SCN- (mL) H 2 O (mL) Initial [Fe 3+ ] Initial [SCN - ] ?? ?? ?? ?? ?? Note that for every sample the total volume is 10 ml = 0.01 L.

How to determine the initial concentrations concentration of diluted stock standard. Second: Calculate the molar concentration of Fe(NO 3 ) 3 in each tube. Tube # M Fe(NO 3 ) 3 (mL) M SCN- (mL) H 2 O (mL) Initial [Fe 3+ ] Initial [SCN - ] ?? ?? ?? ?? ?? Note that this concentration will be the same for ALL samples because you add the same amount with the same final volume each time.

How to determine the initial concentrations concentration of diluted stock standard. Finally: Calculate the molar concentration of SCN- in each tube at each concentration. Tube # M Fe(NO 3 ) 3 (mL) M SCN- (mL) H 2 O (mL) Initial [Fe 3+ ] Initial [SCN - ] x ? x ? x ? x ? x ? For sample #1 it is 0.00 because you didn’t add any SCN-.

How to determine the initial concentrations concentration of diluted stock standard. Finally: Calculate the molar concentration of SCN- in each tube at each concentration. Tube # M Fe(NO 3 ) 3 (mL) M SCN- (mL) H 2 O (mL) Initial [Fe 3+ ] Initial [SCN - ] x x ? x ? x ? x ? For sample #2-5 find the amount of stock solution and calculate for a new final volume of 10 mL (0.01 L)

How to determine the initial concentrations concentration of diluted stock standard. Finally: Calculate the molar concentration of SCN- in each tube at each concentration. Tube # M Fe(NO 3 ) 3 (mL) M SCN- (mL) H 2 O (mL) Initial [Fe 3+ ] Initial [SCN - ] x x x x ? x ? x ? You will have to calculate the concentration for each sample since the amounts vary. However, final volume will remain 10 mL for each sample.

PART II: Pre-Lab Questions (a)Write the balanced net ionic equation for the reaction taking place: (see pg 6) Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq)

PART II: Pre-Lab Questions (b) Write the equilibrium expression for the reaction taking place. Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Remember, [products] coeff. / [reactants] coeff.

PART II: Post Lab Questions 3. Use a RICE table along with the initial concentrations of the reactants you calculated in Part II and equilibrium concentrations to determine the value of K eq for each experimental trial of this experiment.

PART II: Post Lab Questions RICE tables are a way of organizing your data to quickly solve equilibrium problems. R: Write the balanced net ionic equation. I: List the initial concentrations for each species (product and reactants) C: On this row you will figure out what the change in concentration for the reaction is. E: Here you will list the equilibrium concentrations for each species.

PART II: Post Lab Questions For our Lab:. R: Write the balanced net ionic equation. I: C: E:

PART II: Post Lab Questions For our Lab:. R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I: C: E:

PART II: Post Lab Questions For our Lab:. R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I: List the initial concentrations. These are the values you calculated in table 2 on page 7. C: E:

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: E:

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: E: the equilibrium concentrations are the concentrations you measured in table 1, page 7. You calculated these concentrations from absorbance and your calibration curve.

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: E: 4.8x M

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: E: 4.8x M Do you remember why we are doing the RICE table? Our goal is to find K eq, the equilibrium constant. To do this we need to know the concentrations of ALL species at equilibrium.

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: E: 4.8x M We are missing the equilibrium concentrations of the reactants. BUT, we know the change in concentration and the ratios of those changes so we can calculate the final concentrations of our reactants.

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: That’s where “C” comes in. This represents the change in concentration as equilibrium is reached. E: 4.8x M

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: Use the coefficients to determine the mole ratio changes during the reactions. Our reaction is 1:1:1 so this problem is relatively simple. E: 4.8x M

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: Notice that if we produce one mole of Product then one mole of each reactant is consumed. E: 4.8x M

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: Let “x” represent this amount. E: 4.8x M

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: -x -x +x One mole of each reactant was consumed (-x) to produce one mole of product (+x) E: 4.8x M

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: -x -x +x E: 4.8x M Now determine the equilibrium concentrations after the “C” values are subtracted or added. Evaluating the product first we see the x = 4.8x10-3

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: -x -x +x 6x10 -4 M 4x x x10 -4 E: 2.2x x x10 -4 M Now we can subtract “X” from our starting concentrations of reactants

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: -x -x +x 6x10 -4 M 4x x x10 -4 E: 2.2x x x10 -4 M Do you remember our goal? Find K eq. Now all we need to do is plug our equilb. Concs. Into the equlibrium expression!

PART II: Post Lab Questions For our Lab Test tube 2: R: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I : 6x10 -4 M 4x10 -4 M 0 (initially there is no product!) C: -x -x +x 6x10 -4 M 4x x x10 -4 E: 2.2x x x10 -4 M [FeSCN 2+ ] Keq = [Fe 3+ ][SCN - ] = plug and chug!