Energy Science and Technology Introductory Laboratory (Physical Chemistry) Experiment No. 8 Reaction Kinetics Hydrolysis of ethyl acetate ester Abstract: The hydrolysis of an ester (ethyl acetate) in acidic solution will be studied for different temperatures. For each temperature, the progress of the reaction will be tracked by pH measurements after certain time steps, which will in turn yield the reaction rates. Using the Eyring model, the temperature dependent reaction rates will be used to calculate activation energy and entropy. Version 2.0, 31.01.2010, A. Engstfeld, H.E. Hoster EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 1. Theory 1.1 General Remarks 2 The reaction of ester hydrolysis is chosen for this experiment for practical reasons. First, it takes place with rather low rates, so that changes can be observed on time scales of minutes. Second, it requires an acid as catalyst. The amount of products can thus be followed as a function of time by extracting small amounts (= aliquots) from the reacting solution at certain time intervals. Diluting these aliquots by water will virtually stop the reaction and allow for an analysis without any hurry. Third, the product itself is an acid, so that its concentration can be easily and precisely determined by titration. In the following, we will first summarize the most important facts about acid-base reactions and the procedure of titration, i.e., the tool to follow the progress of our model reaction. We will then have a closer look at the reaction itself. The final two sections of theory concern the determination of the rate constant and the interpretation of its behaviour as a function of temperature. 1.2 Acid-Base Reactions Arrhenius defined an acid as a substance that increases the hydrogen ion (H+) concentration of an aqueous solution, and a base as a substance that increases the hydroxide ion (OH-) concentration. According to the Brønsted-Lowry theory of acids and bases, any substance that gives up a proton is an acid (proton donor), and any substance that can combine with a proton and remove it from solution is a base (proton acceptor). In aqueous solution, acids (base) are classified as either strong or weak. Strong acids (bases) are completely dissociated or ionized. Weak acids (bases) dissociate only partially. When an acid (base) loses its proton (hydroxide), it becomes the conjugate base (acid). Ionized means that a (H+) is transferred from the acid molecule to a water molecule, to form a hydronium ion (H3O+). Completely ionized means that each acid molecule transfers its acidic hydrogen atom as ion to a water molecule. For the case of incomplete ionization, only a fraction of the acid molecules loses their acidic hydrogen atoms. If an acidic solution is mixed with a basic solution, an acid-base reaction occurs. If a solution of any strong acid (e.g. HCl) is mixed with a solution of any strong base (e.g. NaOH) it is often called a neutralization reaction. This means if an equal amount of substance of a strong acid and a strong base is mixed with each other, they will neutralize to form water and a corresponding salt (e.g. NaCl), as shown in general in EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 3 equation (1.1) and for the example of the reaction between NaOH and HCl in equation (1.2). Acid (aq) + Base (aq) ↔ Water (l) + Salt (aq) reaktion (0.1) HCl (aq) + NaOH (aq) ↔ H2O (l) + NaCl (aq) reaktion (0.2) Other types of acid-base reactions besides reactions of strong acids and bases are reactions of strong acids/bases and weak bases/acids (see any basic textbook of chemistry for further information). The acidity or basicity of a solution is given by the pH value. It can be calculated via + + pH = - log10 ([H3O ]), where [H3O ] denotes the concentration of the hydronium ions and is given in mol / litre. A pH value of 7 describes a neutral solution. Is the value lower than 7, the solution becomes acidic, and if it is bigger than 7, the solution becomes alkaline. The amount of acid (base) present in a solution can for example be determined by titration. This involves adding a standardized solution called titrant (from a burette) to a flask containing the sample, called analyte, until the solution is neutralized. Neutralization means determining the stoichiometric point (also known as equivalence point), at which the amount of substance of the protons (or proton acceptors) n1 of the titrant equals the amount of substance of the proton acceptors (or protons) n2 of the analyte. The stoichiometric point can for instance be optically detected, by adding a so called indicator to the analyte. Depending on the pH value of the analyte, the chemical structure of the indicator changes and hence it changes the colour of the solution. A common indicator, also used in this experiment, is phenolphthalein, shown in figure 1. It turns colourless in acidic solutions and pink in basic solutions. If the concentration of indicator is particularly strong, it can appear purple. colourless (Acid form) Figure 1. Acidic and basic structure of phenolphtalein pink (Base form) EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 1.3 4 Hydrolysis of ethyl acetate in acidic solution An ester is a chemical condensed compound of an oxoacid and a hydroxyl compound, such as an alcohol. In our experiment, we will take a closer look at the acid catalysed hydrolysis of the ethyl acetate, forming acetic acid and ethanol. Figure 2 depicts the molecular structure of the ester. O CH3 C O CH 2 CH3 Figure 2. Structure of the ethyl acetate + The formal reaction equation of the H -ion catalysed hydrolysis of the ethyl acetate is shown in Figure 3, where k1 and k2 describe the reaction rate constants of the forward and reverse reaction, respectively. O CH3 C + H 2O + H+ O k1 CH3 + H3CCH 2OH + H + C k2 O CH 2CH3 OH Figure 3. Reaction scheme for the H+-ion catalysed hydrolysis of ethyl acetate. In our experiment, H2O will be available in excess, which shifts the equilibrium virtually fully to the right hand side. For completeness, Figure 4 shows the reaction mechanisms with all its intermediate steps. EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis O CH3 O OH OH H+ C CH3 H2O C O CH 2CH3 5 CH3 CH 2CH3 C H O OCH 2CH3 H OH OH + H+ H CH 3 C CH 3 OCH 2 CH 3 C OCH 2 CH 3 H OH OH OH - H3CCH2OH CH3 C CH3 OH OH OH CH3 C OH O H+ CH3 C OH Figure 4. Intermediate steps of the acid hydrolysis of ethyl acetate. C OH EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 1.3 6 Determination of the rate constant The rate equation of the ester hydrolysis in Figure 3 may be written as: d[EA] = −k1[EA][H2O][H+ ] + k 2 [S][HAc][H+ ] dt (0.3) [EA] = concentration of ethyl acetate, [HAc] = concentration of acetic acid, [A] = concentration of alcohol. Since the experiment is carried out in a solution of a strong acid (HCl) that serves as + catalyst, the concentration of [H ] will not change during the reaction. Though the generation of acetic acid (HAc; weak acid) from the hydrolysis goes along with a + slight increase in the [H ] concentration, this effect is too little to have an effect on the reaction rate. The equation (0.3) can thus be simplified to: d[EA] = −k1[EA][H2O] + k 2 [HAc][A] dt (0.4) Also water is present in such excess that its concentration remains unchanged during the reaction and can thus be included into the rate constant. On the other hand, this excess reduces the possibility of the reverse reaction so that the term k2[HAc][A] can be neglected. Thus, the rate constant equation (0.4) becomes a "pseudo first order" reaction under these circumstances. d[E] = −k '[E] dt k ' = k1[H2O] (0.5) EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 1.4 7 Determination of activation energy and entropy The temperature dependence of the reaction rate constant can be expressed with the Arrhenius equation: −Ea k = A ⋅ e RT (0.6) Likewise, it can also be expressed using the Eyring-Polanyi equation: k= kBT # K h (0.7) kB = Boltzmann's constant h = Planck's constant # K = equilibrium constant of the transition state of the reaction For the transition state the Gibbs energy can be written as follows: ∆G0# = ∆H0# − T∆S0# = −RT lnK # (0.8) By insertion of equation (0.8) in equation(0.7), the reaction rate constant k can be expressed as follows: ∆S0# ∆H0# kBT k= exp exp − h R RT (0.9) ∆H #0 and ∆S0# are the enthalpy and entropy of reaction, respectively. By setting up two equations for the rate constants k(T1) and k(T2), at the different temperatures, T1 and T2, and by dividing those two equations, the following expression is obtained: ∆S # exp k 2 (T2 ) T2 R = k1(T1 ) T2 ∆S # exp R ∆H# exp − RT2 ∆H# xp e − RT1 (0.10) ∆H# 1 1 k 2 (T2 ) T2 = exp − k1(T1 ) T2 R T1 T2 From equation (0.10), ∆G 0# , ∆H #0 und ∆S0# for the transition state can be calculated. EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 2. 8 Experimental Procedure In this experiment, the hydrolysis of ethyl acetate (ester) is achieved by mixing a defined amount of the ester with a defined amount of HCl of known concentration. At defined time steps, aliquots (= samples) of the mixed solution will be taken. These aliquots will be diluted in water in order to slow down the reaction. The amount of acetic acid (analyte) formed from the hydrolysis of the ester can be determined by titration with a standard sodium hydroxide (NaOH) solution (titrant). Experimental steps: 1. Check the exact concentration of the NaOH (titrant) at room temperature. Therefore, at least two titrations of 5 ml (1 molar = 1 M) of HCl with the NaOH should be carried out. The concentration of NaOH will be calculated from the average volumes of titrant (Vx = VNaOH) needed for the titration of the HCl. 2. For the actual measurement, a pipetted volume of 130 ml of a 1 M HCl and 8 ml of ethyl acetate are placed in clean, dry conical flasks, which are then immersed in water-baths at 25 °C. 3. To determine the exact concentration of the heated HCl, a pipetted volume of 5 ml will be titrated with the standard NaOH. 4. From the remaining HCl, a pipetted volume of 100 ml will be transferred into a new conical flask, and also immersed in a water-bath at 25 °C. After about 5 minutes, when thermal equilibrium should be attained, 5 ml of ethyl acetate from the 25 °C bath is taken by pipette and added to the 100 ml HCl. A timer is started when the pipette is half drained into the HCl. The conical flask should quickly be closed with a glass stopper and well stirred. 5. Immediately after stirring, an aliquot of 5 ml of the mixed solution (reaction solution) is taken by pipette and added to a new conical flask with about 100 ml of water (diluted sample). The time of sampling is noted when the pipette containing the mixed solution is half drained into the water. The diluted sample is immediately titrated with the NaOH using 3 drops of phenolphthalein as indicator. 6. The procedure described in point 5 is repeated after about 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110 and 120 minutes. The experiment from point 2 to 6 is repeated at a temperature of 35 °C. In this case, samples are extracted at 0, 5, 10, 15, 20, 25, 30, 40, 50 and 60 minutes. Further reading on how to titrate can be found on the following homepage: http://www.dartmouth.edu/~chemlab/techniques/titration.html EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 3. 9 Experimental data collection and calculation At any time (t) the volume of the titrant (Vt, NaOH) is measured (with a burette) and noted. For a complete reaction, a volume V∞ for the titrant can be determined. Since a complete hydrolysis can not be achieved experimentally, it has to be determined theoretically. The final volume V∞ (equation (1.11)) contains on the one hand the volume necessary to neutralise the HCl (VHCl) and on the other hand the volume to neutralise the acetic acid formed from the ethyl acetate (VHAc). V∞ = VHCl + VHAc (0.11) Let’s first concentrate on the volume VHCl, which is needed to neutralise the HCl included in the 5 ml samples (from the mixed solution: HCl + Ester). The volume can be expressed by equation (1.12), where Vx stands for the prior determined titrant volume to neutralise 5 ml of the heated HCl (as described under point 3 of the experimental procedure). VS stands for the volume of the reaction solution. Due to the non-ideal reaction solution its total volume is not 105 ml, but should be assumed to be only 104.6 ml. VHCl = VX ⋅100 ml VS (0.12) To determine the titrand volume VE needed to neutralise the acetic acid formed after complete hydrolysis in a aliquot of 5 ml of the mixed solution, it is only necessary to know the initial concentration of our ethyl acetate. All the needed data for further calculations at different temperatures are given in table 1. Ethyl acetate V [ml] 5 M [g/mol] 88.11 ρ [g/ml] (25°C) 0.8945 ρ [g/ml] (35°C) 0.8818 EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 10 The amount of substance (n) of the initial ethyl acetate is given by equation (1.13). n= m V ⋅ ρ 5ml ⋅ρ = = M M M (0.13) The amount of substance of ethyl acetate in a 5 ml sample (nE) pipetted from the reaction solution is given by equation (1.14). nE = 5ml 5ml 5ml ⋅ρ ⋅n = ⋅ VS VS M 25ml ⋅ρ nE = VS ⋅ M (0.14) To calculate the volume of the needed titrant, nE has to be equal to the amount of substance of the NaOH (nOH), which leads to equation (1.15). Here cOH is the concentration of the NaOH, obtained in step 1 of the experimental procedure and VHAc is the volume of NaOH needed to neutralise the formed acetic acid. n E = n OH = cOH * VHAc (0.15) Combining equation (0.14) and (0.15), VHAc can be expressed with equation (1.16). VHAc = 1 cOH ⋅ nE = 1 cOH ⋅ 25ml ⋅ρ VS ⋅ M (0.16) Now the final titrand volume V∞ can be calculated and results in: V∞ = VX ⋅100 ml 1 25ml ⋅ ρ + ⋅ VS cOH VS ⋅ M VHCl (0.17) VHAc The remaining ester at a certain time step (t) is proportional to ( V∞ − Vt ) . For the evaluation of your data, please develop a method to derive the reaction rate constant from a set of Vt values at increasing t by a linear regression. EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 11 Results Sheet Date: _______________________ Names: Team: _______________________ _______________________ _______________________ _______________________ 1. Concentration of NaOH (at least two titrations to calculate average value) Used NaOH: Vtitr_1 = ______mL; Vtitr_2 = ______mL; (Vtitr_3= ______mL) Vtitr_average = ________mL; ⇒ [NaOH] = ________ mol L-1 2. Hydrolysis at T1 = 298.15 K Titration of HCl at working temperature: Vtitr = ______mL Determination of V∞: V∞ = ______mL t / min 10 20 30 40 50 60 80 100 120 80 100 120 used NaOH Vtitr / mL 3. Hydrolysis at T1 = 308.15 K Titration of HCl at working temperature: Vtitr = ______mL Determination of V∞: V∞ = ______mL t / min 10 20 30 40 50 60 used NaOH Vtitr / mL supervisor’s signature: _____________________ EST Introductory Laboratory (Physical Chemistry), Exp. No. 8: Kinetics | Ester Hydrolysis 12 Colloquium topics Basics of chemical kinetics • • • • • • Definition of the reaction rate Extend of reaction, order and molecularity of a reaction Rate law’s (zero, first, second order) Determination of the reaction order Relation between kinetics and thermodynamics Temperature dependence of the reaction rate Theory of chemical kinetics • • • Kinetic theory of gases Transition state theory (Eyring) Comparison of the Arrhenius-Equation, Equation derived from the kinetic gas theory and the Eyring-Equation Basic theory on acid-base reactions Experiment data collection and calculation • • • • Reaction scheme of ester hydrolysis Mechanism of the acid catalysed ester hydrolysis Experimental Procedure, especially titration Experimental evaluation Further reading Titration: http://www.dartmouth.edu/~chemlab/techniques/titration.html