CHEM 254 EXP 10 Chemical Equilibrium - Homogeneous and Heterogeneous Equilibrium

Transcription

1 Gibbs energy, G CHEM 254 EXP 10 Chemical Equilibrium Homogeneous and Heterogeneous Equilibrium A reaction at constant temperature and pressure can be expressed in terms of the reaction Gibbs energy. The reaction Gibbs energy is defined as the slope of the graph of the Gibbs energy plotted against extent of reaction (Figure 1). If r G<0 the forward reaction is spontaneous If r G>0 the reverse reaction is spontaneous If r G=0 the reaction is at equilibrium r G<0 r G=0 r G>0 extent of reaction, Figure 1. Variation of reaction Gibbs energy as a function of extent of reaction For a reaction with stoichiometric coefficients J, the amount of reactants and products change by and the change in Gibbs energy is where is the chemical potential of species J.

2 The chemical potential of species is related to its activity used where At equilibrium thus For a general equation; aa + bb = cc + dd where s are activities, s are activity coefficients and C 0 is the standard concentration, (1M) At infinite dilution activity coefficients are equal to one. Thus, for dilute solutions equilibrium constant may be calculated using concentrations instead of activities. When all the reactants and products are in the same phase then the equilibrium established is named as homogenous equilibrium. If more than one phase is present in the equilibrium mixture then a heterogeneous equilibrium is achieved. Purpose: The aim of this experiment is to study the homogeneous equilibrium process of an esterification reaction in the presence of HCl as catalyst. CH 3 COOH +C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O

3 and heterogeneous equilibrium of distribution of I 2 in two phases Apparatus and Chemicals Part I. Homogenous equilibrium I 2 (CHCl 3 ) I 2 (H 2 O) Apparatus: Nine 100 ml, glass stoppers, narrowmouth bottles, pipettes, burettes, flasks. Chemicals: Glacial acetic acid, ethanol, ethyl acetate, 3M HCI, 1.0 M NaOH, phenolphthalein. Part II. Heterogeneous equilibrium Apparatus: Six 250 ml glassstoppers, pipettes, burette, flasks, temperature (25 C). Chemicals: 0.01 M and 0.1 M KI, saturated Iodine Solution, 0.1 M and 0.02 M standard sodium thiosulfate Part I. Homogeneous Equilibrium In the esterification reaction, as stoichiometric coefficients are all equal to one, equilibrium constant can be also expressed in terms of number of moles of reactants and products; CH 3 COOH +C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O K = [Ester]. [water] [acid]. [ethanol] = n(ester). n(water) n(acid). n(ethanol) The enthalpy of the reaction is very close to zero. Thus, it can be assumed that the equilibrium constant is almost independent of temperature. In general, HCl is used as a catalyst to increase the rate of the reaction. Procedure Caution: The mixtures used in Part AHomogenous Equilibrium should be prepared in bottles as described in Table 1, one week before the assigned experiment day and left at room temperature for one week. The bottles should be shaken from time to time during this period. On the assigned experiment day, first prepare the solutions for Part IIHeterogeneous Equilibrium. Then, while waiting for the establishment of the equilibrium for Part II, continue the experiment on Homogenous Equilibrium starting from Part I Prepare the following mixtures in 9 different bottles and close the bottles and leave solutions for one week for establishment of equilibrium.

4 Table 1. Volume of each chemical that should be mixed to prepare the solutions Bottle number 3 M HCl, ml Ethanol, ml Acetic acid, ml Ester, ml Water, ml Record density of each reactant and product on your report sheet. 3. Transfer 5 ml of each solution into nine different erlenmeyers. 4. Add 1 drop of phenolphthalein to each solution. 5. Titrate with 1.0 M NaOH. Record the volume of NaOH consumed up to the end point (Report Sheet Table 1) Treatment of Data 1. Calculate the initial number of moles of acetic acid, HAc, ethanol, Et, and ester, Es. (Report Sheet, Table 1) n i,0 = m i / MWt i = d i x V i / Mwt i 2. Calculate the initial number of moles of water taking into account the amount of H 2 O introduced by adding HCl (Report Sheet, Table 1). Assume d solution = d water =1.0 g/ml m HCl = n HCl x MWt HCl =M HCl x V HCl x MWt HCl m sol = mass of solution = 1.0 g/ml x V m water = mass H 2 O added + mass of H 2 O due to HCl addition = m add + (m sol m HCl ) n water,0 = m water / Mwt

5 3. For Solutions 13 CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O initial [Ester] 0 [water] 0 at equilibrium x x [Ester] 0 x [water] 0 x x = (n Hac ) eq = n ethanol (n HAc ) eq = M NaOH x V titration (n Es ) eq = (n Es ) 0 x (n water ) eq = (n water ) 0 x 4. For Solutions 47 CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O initial [HAc] 0 [ethanol] 0 [water] 0 at equilibrium [HAc] 0 x [ethanol] 0 x x [water] 0 + x (n Hac ) eq = M NaOH x V titration x = (n HAc ) 0 (n HAc ) eq (n Et ) eq = (n Et ) 0 x (n Es ) eq = x (n water ) eq = (n water ) 0 + x 5. Calculate the equilibrium moles of reactants and products in solutions Calculate K for each solution, and the average value. Determine the standard deviation. Discuss possible sources of errors. Part II. Heterogeneous Equilibrium The reaction between I 2 and I is an homogenous equilibrium in aqueous solution. I 2 + I I 3 As the amount of I 2 that dissolves in water is very small, the equilibrium concentrations are very small and thus can be used instead of activities. When CHCl 3 is added to the above equilibrium solution, molecular iodine I 2 is distributed in water and CHCl 3 phases according to the following heterogeneous equilibrium process. The two immiscible phases, water (polar) and CHCl 3 (nonpolar) are separated as shown in Figure 2.

6 I 2 (water) I 2 (CHCl 3 ) Figure 2. Separation of phases Procedure 1. Prepare the following solutions Solution 1: 35 ml I 2 saturated CHCl 3 (CHCl 3 phase) ml H 2 O (aqueous phase) Solution 2: 35 ml I 2 saturated CHCl 3 (CHCl 3 phase) + 75 ml 0.01 M KI (aqueous phase) Solution 3: 35 ml I 2 saturated CHCl 3 (CHCl 3 phase) + 35 ml 0.1 M KI (aqueous phase) 2. Place the solutions into water bath at 25 C and wait for 40 min. Shake the solutions at every 5 min. Then wait for 10 min for the separation of two phases. While waiting for establishment of equilibrium, continue on the experiment Homogenous Equilibrium, Part I From each solution withdraw volumes of samples from each aqueous and CHCl 3 phases as given in the following Table into six different erlenmeyer flasks. Caution: While withdrawing sample from the CHCl 3 layer, be careful not removing sample from the aqueous layer Blow very slowly air through the pipette to minimize contamination. Table Solution 1 Solution 2 Solution 3 V aq/ki withdrawn, ml V CHCl3 withdrawn, ml

7 4. To each aqueous phase: i. Add 10 ml 0.10 M KI (KI is added in excess amount to convert almost all I 2 to I 3 in the aqueous phase to avoid loss of I 2 during the experiment by evaporation.) ii. Titrate with 0.02 M S 2 O 3 2 until the color turns to very light Brown (color of light tea). I S 2 O 3 2 3I + S 4 O 6 iii. Add 2 ml starch and continue titration with S 2 O 3 2 until the solution becomes colorless. Record the volume of S 2 O 3 2 (Report Sheet Table 3). 5. To each CHCl 3 phase: i. Add 10 ml 0.10 M KI. ii. Titrate with 0.10 M S 2 O 3 2 until the color turns to color of light tea. iii. Add 2 ml starch as indicator and continue titration with S 2 O 3 2 until the solution becomes colorless. Record the volume of S 2 O 3 2 (Report Sheet Table 3). Treatment of Data 1. Calculate K D from the amount of I 2 distributed in each phase in solution 1 For both aqueous and CHCl 3 phases n(i 2 ) = n( I 3 ) = 1/2 x n(s 2 O 2 3 ) = M S2O3 x VS2O3 [ I 2 ] = n(i 2 ) / V where V is the volume of the aqueous or CHCl 3 phase used in the titration Calculate K D = 2. Calculate the equilibrium concentrations of I, I 2 and I 3 in solutions 2 and 3 For both solutions, when the equilibrium is established in the aqueous phase I 2 + I I 3 [I 2 ] aq,eq [I ] 0 x x [ I 2 ] aq,eq in the aqueous phase can be calculated using the heterogenous equilibrium process [ I 2 ] aq,eq = K D x [ I 2 ] CHCl3 [ I 2 ] CHCl3 = n(i 2 ) / V =1/2 x n(s 2 O 3 2 ) / V CHCl3 = M S2O3 x V S2O3 / V CHCl3

8 The amount I 3 in aqueous phase is the sum of the amount of I 3 produced by the equilibrium process n([i 3 ])water, eq and the amount of I 3 produced by the conversion of I 2 upon addition of KI before the titration by n(i 3 )due to reaction of iodine with KI added. n( I 3 ) aq, total = 1/2 x n(s 2 O 3 2 ) = M S2O3 x V S2O3 [ I 3 ] aq, total = n( I 3 ) aq, total / V aq = [ I 3 ] aq, eq + [ I 3 ] [ I 3 ] = [ I 2 ] aq,eq [ I 3 ] aq, eq = [ I 3 ] aq, total [ I 2 ] aq =x= [ I ] eq = [ I ] 0 x Calculate K for the reaction I 2 + I I 3 Questions 1. State Le Chateliers s Principle. 2. State the factors that alter the composition of an equilibrium mixture based on Le Chateliers s Principle. 3. Explain the effect of the changes on reactant or product when a stress is applied to a system. 4. Determine the reaction direction when a system at equilibrium reacts to a stress applied to the system including changes in concentrations, pressure and volume, or temperature. 5. Describe the effect of adding a catalyst to a system at equilibrium. In our system, how does the strong acid function as the catalyst? 6. Explain the factors that affect the equilibrium constant.

9 DATA SHEET Group Number: Experiment 11 Homogeneous and Heterogeneous Equilibria Date: Assistant name and signature: Part I. Homogeneous Equilibria 1. Initial concentrations n HAc = n Et = n Es = 2. m H2O = m add + (m sol m HCl ) = n H2O = Table 1. Initial number of moles of reactants and products and V NaOH consumed Solution n Hac n Et n Es Water(mL) V NaOH (ml) For Solutions 13 CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O [Ester] 0 [water] 0 initial x x [Ester] 0 x [water] 0 x at equilibrium

10 x = (n HAc ) eq = (n Etl ) eq (n HAc ) eq = (n Es ) eq = (n water ) eq = 4. For Solutions 47 CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O [HAc] 0 [ethanol] 0 [water] 0 initial [HAc] 0 x [ethanol] 0 x x [water] 0 + x at equilibrium (n HAc ) eq = x = (n Et ) eq = (n Es ) eq = (n water ) eq = 5. For Solutions K for each solution Table 2. The number of moles of reactants and products

11 Solıtion n(hac) eq n(et) eq n(es) eq n(h 2 O) eq K K (average) = Standard deviation = Part II. Heterogeneous Equilibria Table 3. VS 2 O 3 (ml) Solution 1 Solution 2 Solution 3 aqueous phase CHCl 3 phase aqueous phase CHCl 3 phase aqueous phase CHCl 3 phase 1. n( I 2 ) aq = [ I 2 ] CHCl3 = n( I 2 ) CHCl3 = [ I 2 ] aq = K D = 2. For solution 2 [ I 2 ] CHCl3 = n(i 2 ) / V =1/2 x n(s 2 O 3 2 ) / V(CHCl 3 ) = M(S 2 O 3 2 ) x V(S 2 O 3 2 )/ V(CHCl 3 ) =

12 [ I 2 ] aq,eq = K D x [ I 2 ] CHCl3 = n( I 3 ) aq, total = 1/2 x n(s 2 O 3 2 ) = M(S 2 O 3 2 ) x V(S 2 O 3 2 ) = [ I 3 ] aq, total = n( I 3 ) aq, total / V aq = [ I 3 ] aq, eq + [ I 3 ] due to reaction of iodine with KI added [ I 3 ] due to reaction of iodine with KI added = [ I 2 ] aq,eq [ I 3 ] aq, eq = [ I 3 ] aq, total [ I 2 ] aq = x= [ I ] eq = [ I ] 0 x 3. For solution 3 [ I 2 ] CHCl3 = [ I 2 ] aq,eq = n( I 3 ) aq, total = [ I 3 ] aq, total = [ I 3 ] due to reaction of iodine with KI added = [ I 3 ] aq, eq = [ I ] eq = 4. Calculate the equilibrium constant K Solution 1 Solution 2 [ I 2 ] aq,eq = [ I 2 ] CHCl3 [ I 3 ] aq, total [ I ] eq K 5. Compare K values obtained for solutions 2 and 3. What are the possible reasons for the difference in these two values.