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The Equilibrium Constant of an Ester Hydrolysis Reaction lab for chemistry 152. need help with the calculations as soon as possible, everything is ready. thank you!

CHM152LL LAB MANUAL THE EQUILIBRIUM CONSTANT OF AN ESTER HYDROLYSIS REACTION The Equilibrium Constant of an Ester Hydrolysis Reaction INTRODUCTION More than twenty million unique compounds have CAS1 registration numbers, most based on the covalent linkage of carbon atoms. This enormous diversity forced organic chemists to develop nomenclature rules based on recognizable patterns in the structure of the molecules called functional groups. Functional groups important for this experiment appear in Table 1. In this table, the symbol R represents a specified hydrocarbon residue. Table 1. Some organic functional groups Structure ROH RCOOH R1COOR2 Name Example R group alcohol C2H5OH = ethanol C2H5- carboxylic acid CH3COOH = acetic acid CH3- ester CH3COOC2H5 = ethyl acetate You will determine the equilibrium constant, Kc, for the acid-catalyzed reaction between an unknown ester and water to produce an unknown alcohol and an unknown carboxylic acid: [R OH][R1COOH] H+ R1COOR2 + H2O R2OH + R1COOH K c = 2 [R1COOR 2 ][H 2O] You will determine the equilibrium concentrations of the reactants and products by a combination of measurement and calculation. The organic molecules known as esters undergo hydrolysis, a reaction with water in which both the ester molecule and the water split and recombine to produce an acid and an alcohol. In this lab, you will determine the equilibrium constant, Kc, for the hydrolysis of an unknown ester into unknown acid and alcohol products. For the generalized reaction of interest: [alcohol]eq [acid]eq ester + water alcohol + acid , Kc = [ester]eq [water]eq where [acid]eq represents the molarity of the organic acid at equilibrium, and similarly for the other reactants and products. PROCEDURE Caution: Most esters are volatile, and some are toxic by inhalation or absorption through the skin. Whenever you handle your ester, work inside the fume hood and wear gloves. 1. Record the unknown ester/alcohol number in your lab notebook. Also record the density and molar mass of the ester and alcohol. 2. Determine the density of deionized water, and of the ~3 M HCl solution. 1 Chemical Abstracts Service, a division of the American Chemical Society REV 2016-06-14 THE EQUILIBRIUM CONSTANT OF AN ESTER HYDROLYSIS REACTION PAGE 1 OF 4 CHM152LL LAB MANUAL THE EQUILIBRIUM CONSTANT OF AN ESTER HYDROLYSIS REACTION 3. In the first week, prepare the reaction mixtures as given in Table 2. Measure and record the volumes to 0.01-mL precision. Cap each bottle and shake vigorously. Store the bottles in your lab drawer until the mixtures come to chemical equilibrium. Table 2. Reaction mixtures Bottle # 3 M HCl (mL) H2O (mL) ester (mL) alcohol (mL) 1 5 5 0 0 1A 5 5 0 0 2 5 0 5 0 3 5 1 4 0 4 5 3 2 0 5 5 2 2 1 4. In the second week, calculate the mass of sodium hydroxide needed to prepare 0.5 L of a 0.7 M NaOH solution. On a trip balance, quickly weigh out roughly this mass of NaOH pellets, and dissolve them in deionized water. Pour the solution into a polyethylene bottle and add water to bring the total volume of solution to about 500 mL. Cap the bottle tightly, to prevent atmospheric CO2 from entering and reacting with the NaOH. 5. Calculate the mass of potassium hydrogen phthalate (KHP) needed to neutralize 35 mL of 0.7 M NaOH solution. On a digital single-pan balance, weight out approximately this mass of KHP into a 150-mL Erlenmeyer flask. Record the mass to the full available precision. 6. Dissolve the KHP in 50 mL of DI water. Add 2 drops of phenolphthalein indicator and titrate your NaOH solution to a pale pink endpoint. Repeat the weighing and titrating two more times, then calculate the average molarity of your NaOH solution. You have now standardized the solution. 7. In the third week, titrate the reaction mixtures. Transfer the contents of each numbered bottle to an Erlenmeyer flask, rinsing any remaining solution into the flask with deionized water. Add 2 drops of phenolphthalein indicator to the flask and titrate to a pale pink endpoint with the standardized NaOH solution prepared last week. NOTE: Some of these titrations may take more than 50 mL of base solution to reach the endpoint, and so will require you to refill your buret. Take care during titrations not to drain the buret below the lowest calibration mark. DATA ANALYSIS 1. Use the results of titrating bottles 1 and 1A to determine a precise concentration of the hydrochloric acid stock solution, which is only approximately 3 M. 2. Calculate the amount (in moles) of each compound initially present. In calculating the amount of H2O, remember to include the water contributed by the ~3 M HCl. Prepare a table of your calculated results. 3. Use your tabulated results and data to calculate the amount (in moles) of each compound present at equilibrium. Remember to account for the acid solution added at the start. REV 2016-06-14 THE EQUILIBRIUM CONSTANT OF AN ESTER HYDROLYSIS REACTION PAGE 2 OF 4 CHM152LL LAB MANUAL THE EQUILIBRIUM CONSTANT OF AN ESTER HYDROLYSIS REACTION 4. For each reaction mixture, calculate the molar concentration of each compound present at equilibrium, then calculate the equilibrium constant, Kc. 5. Calculate the average Kc for your unknown. Calculation Guidelines Consider the following ICE reaction table, representing the amounts of reactant and product initially present, the changes in reactant and product amounts, and the equilibrium amounts. Table 3. Ester hydrolysis reaction table Amounts (mol) ester + water Initial A B 0 C Change -x -x +x +x B-x x C+x Equilibrium A - x acid + alcohol You can determine the value of the equilibrium constant if you know the values in the last line of the ICE table. You will need to find the amounts of reactants and products as described below. 1. Calculate the initial amounts of reactant a. Ester (A): Given the mass (measured or calculated from volume and density) and molar mass of your unknown ester, find the initial amount (in mol) of ester. b. Water (B): Include both the pure water and the water in the HCl solution. i. Given the mass and molar mass of H2O, find the amount of H2O (in mol) added as water. ii. Use the volume and molarity of the NaOH solution, and the stoichiometric ratio of HCl to NaOH to find the amount of HCl in your 5-mL sample of HCl solution. Given the molar mass of HCl, find the mass of HCl in the solution. Subtract this mass from the mass of HCl solution to find the mass of solvent H2O. Use the molar mass of water to find the H2O in HCl (aq). iii. Add the results of calculations 1.b.i and 1.b.ii above. 2. Calculate the equilibrium amounts of product and reactant a. Acid (x): Given the volume and molar concentration of NaOH used in the reaction mixture titration, find the total amount of base used to titrate the total acid present (organic acid product + HCl). Subtract the results of the HCl-only titration (from 1.b.ii above) to find the amount of NaOH used to titrate organic acid alone. b. Alcohol (C + x): Use the x value calculated in 2.a above, except in bottle #5. c. Water (B - x): Use the value of B from 1.b.iii, and the value of x from 2.a. d. Ester (A - x): Use the value of A from 1.a and the value of x from 2.a above. 3. Calculate the equilibrium constant a. Because our reaction has 1:1:1:1 stoichiometry and all of the reactants and products are dissolved in the same total volume, the volume of the solution cancels out of the REV 2016-06-14 THE EQUILIBRIUM CONSTANT OF AN ESTER HYDROLYSIS REACTION PAGE 3 OF 4 CHM152LL LAB MANUAL THE EQUILIBRIUM CONSTANT OF AN ESTER HYDROLYSIS REACTION calculations, and we do not need to know the total volume of the reaction mixture. Calculate Kc using the amounts (in mol) instead of the concentrations (in mol/L). Kc = x2 (A - x)(B - x) Sample Calculations Suppose that you prepared a reaction mixture in bottle #2 with 5.02 mL of 2.98 M HCl (aq) and 4.97 mL of ester (density given as 0.9003 g/mL, molar mass given as 88.11 g/mol). It takes 45.67 mL of 0.6789 M NaOH solution to reach a titration endpoint. 0.6789 mol NaOH 1 mol mixed acids 45.67 mL NaOH soln = 31.01 mmol mixed acids 1 L NaOH soln 1 mol NaOH 2.98 mol HCl 5.02 mL HCl soln = 14.96 mmol HCl initially present 1 L HCl soln 31.01 mmol mixed acids 14.96 mmol HCl = 16.05 mmol R1COOH product 1 mol alcohol 16.05 mmol acid produced = 16.05 mmol R 2OH product 1 mol acid 1.05 g HCl soln 5.02 mL HCl soln = 5.27 g HCl soln 1 mL HCl soln 36.46 g HCl 14.96 mmol HCl = 0.545 g HCl 1 mol HCl 5.27 g HCl soln - 0.545 g HCl = 4.72 g H2O 1 mol H 2O 4.72 g H 2O = 262 mmol H 2O from HCl solution 18.02 g H 2O 0.9003 g ester 1 mol ester 4.97 mL ester = 50.8 mmol ester initially present 1 mL ester 88.11 g ester 50.8 mmol ester initially present - 16.05 mmol reacted = 34.7 mmol ester at equilibrium Table 4. Calculation results Bottle # H2O (mmol) ester (mmol) organic acid (mmol) alcohol (mmol) 2 246 34.7 16.05 16.05 From these equilibrium amounts, you can calculate Kc, the equilibrium constant for the hydrolysis reaction. Remember that because we have 1:1:1:1 stoichiometry, the total solution volume cancels out of the Kc calculation. Check with your instructor and the lab schedule for the formal report due date. REV 2016-06-14 THE EQUILIBRIUM CONSTANT OF AN ESTER HYDROLYSIS REACTION PAGE 4 OF 4 The Equilibrium Constant of an Ester Hydrolysis Reaction Alcohol #2 Density :7.993 g /mol Mass: 46.07 g/mol Intro: 1. To determine the value of the equilibrium constant for an ester hydrolysis reaction. 2. Use acid-base titrations and solution stoichiometry in determining the equilibrium constant. Chemical principles: The equilibrium of a reaction occurs when a rate of a forward reaction is equivalent to reaction going backwards. Compounds in liquid and gas are constantly undergoing change. The reactions and the products are both being formed in an equal amount of time and they are both expressed by equilibrium constant K c . The state at which they stop reacting and remain at constant molar ratios is called equilibrium. The reaction being studied in the experiment is the hydrolysis, reaction with water, and unknown ester which was ( Alcohol 2. With density 0.7993 g /molmass of 46.07 g/mol . For the equilibrium equation we can specify an equilibrium constant, K c , that relates the concentrations of all product and reactant: ester + H 2 O alcohol+acid K c= [ alcohol ] eq [ acid ]eq [ ester ]eq [ water ] eq In the experiment, the acid catalyzed hydrolysis of an ester will be studied to form an alcohol and an acid. To express the ratio of product vs. reactant at equilibrium we find the rate constant. We calculate the equilibrium constant by determining the concentration of a single species at equilibrium when we know the initial concentration of all species. For the reaction, the equilibrium constant will equal: K c= products reactants K c= [ R2 OH ] [R 1 COOH ] [ R1 COO R 2 ] [H 2 O] R1 COO R2 + H 2 O R 2 OH + R1 COOH When we know the initial concentration of each component, we can determine the equilibrium concentration of one component to calculate the equilibrium constant. Methods: 1. I recorded the unknown ester/alcohol number with the density and mass of ester alcohol ( 2: density=0.7993 g/mol , mass=46.07 g /mol ) 2. Determine the density of DI water~3 M HCl solution. 3. I followed the reaction mixture table to prepare the reaction mixture, once done measuring and recording volumes, I capped the bottle and shake it so it can dissolve. I stored the bottle so it can reach equilibrium. Reaction mixture Bottle # 3 M HCl(mL) Ester (mL) H 2 O(mL) Alcohol (mL) 1 5 5 0 0 1A 5 5 0 0 2 5 0 5 0 3 5 1 4 0 4 5 3 2 0 5 5 2 2 1 4. In week 2, we calculate the mass of sodium hydroxide that we need to prepare 0.5 L of a 0.7 M NaOH solution. We weigh out a mass of NaOH ballets, and dissolve them in DI water. Pure the solution into a polyethylene bottle and add water to bring the mL . Cab the bottle tightly for preventing total volume of solution to about 500 atmospheric C O2 from interring the reaction. 5. We calculate the mass of potassium hydrogen phthalate needed to neutralize 35 mL of 0 .7 M NaOH solution. We weigh out the mass of KHP into a 150 mL Erlenmeyer flask and we calculate the mass. 6. We dissolve the KHP in 50 mL of DI water and we add 2 drops of phenolphthalein indicator and titrate my NaOH solution to a pale pink endpoint. We repeat the titration process two more times, then we calculate the average molarity of NaOH . We now have a standardized solution. 7. For week 3, we titrate the reaction mixtures and transfer contents of each numbered bottle to an Erlenmeyer flask raising any remaining solution into the flask with DI water. Than we add 2 drops of phenolphthalein indicator to the flask and titrate until it gets to a pale pink point with standardized NaOH solution was prepared a week before. Observations and calculations: Week 1 data Bottle # Mass3 M HCl g HCl mL mass H 2 O g H 2 O Mass of mL Ester g Este r mL Mass of Alcohol Alcoho mL l - 4.39 5 1 3.54 4 3.05 3 1.78 2 1.95 2 1.76 2 0.77 1 17.26 1 5.17 5 4.97 5 - 19.86 1A 5.21 5 4.99 5 - 17.40 2 5.19 5 - - 17.70 3 5.19 5 1 17.67 4 5.26 5 17.64 5 5.23 5 Calculation: Mass of NaOH : 0.7 mol NaOH Mass of KHP : 0 .035 L 0.5 39.98 g NaOH =13.993 g NaOH 1 mol NaOH 1 mol KHP 0.7 mol 1 mol KHP 204.2 gKHP =5.00339 gKHP L 1 mol NaOH 1mol KHP Week2 data: g KHP 5.007 g Trail # 1 Initial 11.0 mL Volume Delivered Final 34.1 mL 45.1 mL 5.009 g 2 Initial 10.6 mL Volume Delivered Final 34.3 mL 44.9 mL 5.015 g 3 Initial 11.4 mL Volume Delivered Final 34.1 mL 45.5 mL Week 3 Bottle # 1 Initial 12.8 Final 30.3 Change 17.5 1A 16.3 33.8 17.5 2 1.0 50.0 49.0 3 1.9 50.0 48.1 4 1.5 39.6 38.1 5 9.1 43.0 33.9 Calculations. Mass of NaOH : 0.7 mol NaOH Mass of KHP : 0 .035 L 0.5 39.98 g NaOH =13.993 g NaOH 1 mol NaOH 1 mol KHP 0.7 mol 1 mol KHP 204.2 gKHP =5.00339 gKHP L 1 mol NaOH 1mol KHP 1mol KHP 1 mol NaOH 1 Trail 1: 5.007 gKHP 204.2 g KHP 1 mol KHP 0.0341 L NaOH =.7191 mol /L NaOH 1 mol KHP 1 mol NaOH 1 Trail 2: 5.009 gKHP 204.2 g KHP 1 mol KHP 0.0343 L NaOH =.7152 mol /LNaOH 1 mol KHP 1 mol NaOH 1 Trail 3: 5.015 gKHP 204.2 gKHP 1 molKHP 0.0341 L NaOH =.7223 mol /LNaOH Average molarity of NaOH : 0.7191mol / L+ 0.7152mol / L+ 0.7223 mol/ L =0.7188 mol /L 3 Molarity of HCl : Trail 1 Trail 1A Average Molarity of HCl: 0.7188 17.5 =2.419 5.2 0.7188 17.5 =2.419 5.2 2.419+ 2.419 2.419 mol HCl 2