...Enzyme Kinetics: Inversion of Sucrose Abstract Enzymes are a class of proteins that catalyze practically all biochemical reactions. The enzymatic reactions were looked at by the use of two comparable reactions of enzyme invertase and a acidic form of the reaction. The order of the nonacidic enzymatic reaction was zero order due to substrate concentration and due to the fact that the reaction was time dependent. It is also due to having large R2 value at 0.9932 for D run and a smaller R2 value at 0.9902 for C run. The acidic runs had a first order reaction, which had a lower R2 values at 0.9028 for run D and 0.9028 for run C. Also, the percent error in run C at 33.31% was found to be much lower than the percent error in run D at 55.77%, which mean the concentration of run C is more effective than run D. I. Introduction In this experiment the chemical kinetics of the enzyme catalyzed inversion of sucrose was studied. The reaction that we will study in this experiment is the inversion of sucrose, catalyzed by the enzyme invertase that is derived from yeast: The rate of reaction of this reaction was compared to the same reaction that is to be catalyzed by hydrogen ions. Enzymes make up an important class of proteins that are used to catalyze a wide array of biochemical reactions. The enzyme that was used to catalyze the reactions in this lab experiment was the enzyme...
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...A hybird / digital software package for the solution of chemical kinetic parameter identification problems by ALAN M. CARLSON Electronic Associates, Inc. Princeton, New Jersey INTRODUCTION The modern hybrid computer offers many significant improvements over first generation hybrid systems These improvements include: 1. The increased speed of digital computers en- abling programs to be written in hybrid FORTRAN without drastically limiting hybrid solution rates. 2. The development of analog/hybrid software (e.g., hybrid simulation languages and analog set-up programs). The net result of these improvements has been an increase in the SCope and complexity of hybrid applications and a reduction in the effort required to program and debug hybrid problems. Unfortunately, the dev'elopment of hybrid applications software has not kept pace with recent hybrid improvements. Applications software for purposes of this discussion is defined as an integrated set of digital/hybrid programs capable of solving the majority of frequently occurring problems in a specific applications area. Based on this definition, little or no tangible information is currently available on the practicality of developing hybrid software packages although its benefits are obvious. In mid-1968, EAT's Princeton Computation Center initiated a development project to· determine the feasibility of hybrid applications software. The objectives of the project were to select a frequently occurring 733 application...
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...David O. Johnston David Lipscomb College I II The Methclnolysir of Acetal A chemical kinetics a n d g a r chromatography experiment Nashville, Tennessee The methanolysis of acetal and gas chromatography complement perfectly to provide a rare and interesting experiment for the physical chemistry student. The reaction of acetal with methanol is a consecutive reaction in which diethyl acetal (aretal) replaces its two ethoxy groups with methoxy groups forming methy1et)hylacetal first and dimethyl acetal in the second step. Investigating the reaction kinetics for this type of reaction mas almost impossible before the advent of gas chromatography (I). There are many advantages to using the methanolysis of acetal in this study: Sufficient data for a complete analysis can be collected within three hours. The reaction requires reagents which are readily available in high purity. The reactants, intermediate, and products are liquids at temperatures easily accessible for kinetic runs. The primary reactant, intermediate, and product are sepe rated rleanly on a typical gas chromatographic column. The peak heights are linear with concentration in the range studied (2). Analysis of the data affords an application of firshorder kin~ticsand the steady state hypothesis. The experiment presented herc is an adaptation of the detailed study of the reaction by Juvet and Chiu (2), who have shown it to be acid catalyzed and first order in diethyl acetal and methylethyl acetal. By decreasing...
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...Write your name here Surname Other names Centre Number Candidate Number Edexcel GCE Chemistry Advanced Unit 4: General Principles of Chemistry I – Rates, Equilibria and Further Organic Chemistry (including synoptic assessment) Monday 14 January 2013 – Afternoon Time: 1 hour 40 minutes You must have: Data Booklet Candidates may use a calculator. Paper Reference 6CH04/01 Total Marks Instructions Use black ink or ball-point pen. Fill in the boxes at the top of this page with your name, centre number and candidate number. Answer all questions. Answer the questions in the spaces provided – there may be more space than you need. Information The total mark for this paper is 90. The marks for each question are shown in brackets – use this as a guide as to how much time to spend on each question. Questions labelled with an asterisk (*) are ones where the quality of your written communication will be assessed – you should take particular care with your spelling, punctuation and grammar, as well as the clarity of expression, on these questions. A Periodic Table is printed on the back cover of this paper. Advice Read each question carefully before you start to answer it. Keep an eye on the time. Try to answer every question. Check your answers if you have time at the end. Turn over P41214A ©2013 Pearson Education Ltd. *P41214A0128* 7/6/5/5/4/ SECTION A Answer ALL the questions in this section. You should aim to spend no more than 20 minutes...
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...Kinetics by the Initial Rates Method Introduction Rate laws are mathematical expressions that describe the relationship between reactant concentrations and the rates of reaction, taking the form, Rate = k[reactant 1]m[reactant 2] n[reactant 3]p… [M/s] (1) The proportionality constant k is the rate constant; any given reaction has a specific value of k for a given set of conditions, such as temperature, pressure, and solvent; the rate constant, in contrast to the rate of reaction, does not depend on the concentration of the reactants. Exponents m, n, p, …, are the reaction orders, and indicate the degree to which the reaction rate depends on the concentration of the associated reactant. “There is no simple correlation between the stoichiometry of the reaction and the rate law”; the reaction orders, and thus the rate constant, must be determined experimentally, as in this experiment between acetone (CH3COCH3), the hydronium ion (H3O+), and iodine (I2), called the iodination of acetone. The reaction in this experiment is shown in Figure 1 and Eq.2.2 Figure 1. Reaction of acetone, iodine, and hydronium ion. CH3COCH3 (aq) + I2 (aq) CH3COCH2I (aq) + HI (aq) (2) Thus, the rate law in this experiment is shown in Eq. 3. Rate = k[CH3COCH3]m[H3O+] n[I2]p (3) The reaction rate expressions are shown in Eq. 4. Rate = -Δ[CH3COCH3]/Δt = -Δ[H3O+]/Δt = -Δ[I2]/Δt (4) Spectrophotometry can be used here to measure the reaction rate expressed as the disappearance...
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...Experiment 5 Enzyme Kinetics: The Effect of Yeast Alcohol Dehydrogenase and Coenzyme on the Rate of Oxidation of Ethanol. Results Raw data: Effect of Enzyme Concentration | Reagent | 1 | 2 | 3 | 4 | Buffer B | 2.5 ml | 2.4 ml | 2.3 ml | 2.2 ml | NAD+ | 200 μl | 200 μl | 200 μl | 200 μl | Dilute ADH | 200 μl | 300 μl | 400 μl | 500 μl | 6 M Ethanol | 100 μl | 100 μl | 100 μl | 100 μl | Time 0 sec. | 0.022 | 0.026 | 0.069 | 0.073 | Time 5 sec. | 0.028 | 0.038 | 0.110 | 0.114 | Time 10 sec. | 0.038 | 0.051 | 0.141 | 0.150 | Time 15 sec. | 0.048 | 0.064 | 0.176 | 0.187 | Time 20 sec. | 0.058 | 0.078 | 0.206 | 0.222 | Time 25 sec. | 0.068 | 0.091 | 0.239 | 0.259 | Time 30 sec. | 0.078 | 0.104 | 0.270 | 0.291 | Time 35 sec. | 0.088 | 0.115 | 0.301 | 0.325 | Time 40 sec. | 0.099 | 0.131 | 0.331 | 0.358 | Time 45 sec. | 0.109 | 0.145 | 0.362 | 0.391 | Time 50 sec. | 0.119 | 0.158 | 0.392 | 0.424 | Time 55 sec. | 0.130 | 0.167 | 0.420 | 0.455 | Time 60 sec. | 0.140 | 0.179 | | | Effect of Coenzyme(NAD) Concentration | Reagent | 1 | 2 | 3 | 4 | Buffer B | 2.4 ml | 2.35 ml | 2.3 ml | 2.25 ml | NAD+ | 10 μl | 50 μl | 100 μl | 150 μl | Dilute ADH | 500 μl | 500 μl | 500 μl | 500 μl | 6 M Ethanol | 100 μl | 100 μl | 100 μl | 100 μl | Time 0 sec. | 0.009 | 0.020 | 0.032 | 0.102 | Time 5 sec. | 0.015 | 0.038 | 0.049 | 0.152 | Time 10 sec. | 0.023 | 0.056 | 0.068 | 0.190 | Time 15 sec. | 0.029 | 0.072 | 0.088 | 0.229 | Time...
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...A Kinetics Experiment The Rate of a Chemical Reaction: A Clock Reaction Andrea Deoudes February 2, 2010 Introduction: The rates of chemical reactions and the ability to control those rates are crucial aspects of life. Chemical kinetics is the study of the rates at which chemical reactions occur, the factors that affect the speed of reactions, and the mechanisms by which reactions proceed. The reaction rate depends on the reactants, the concentrations of the reactants, the temperature at which the reaction takes place, and any catalysts or inhibitors that affect the reaction. If a chemical reaction has a fast rate, a large portion of the molecules react to form products in a given time period. If a chemical reaction has a slow rate, a small portion of molecules react to form products in a given time period. This experiment studied the kinetics of a reaction between an iodide ion (I-1) and a peroxydisulfate ion (S2O8-2) in the first reaction: 2I-1 + S2O8-2 I2 + 2SO4-2. This is a relatively slow reaction. The reaction rate is dependent on the concentrations of the reactants, following the rate law: Rate = k[I-1]m[S2O8-2]n. In order to study the kinetics of this reaction, or any reaction, there must be an experimental way to measure the concentration of at least one of the reactants or products as a function of time. This was done in this experiment using a second reaction, 2S2O3-2 + I2 S4O6-2 + 2I-1, which occurred simultaneously with the reaction under investigation. Adding...
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...Ch. 15: Kinetics 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9 Week 15 Reaction Rates Rate Laws Forms of Rate Laws Integrated Rate Law Rate Laws: A Summary Reaction Mechanisms Steady-State Approximation A Model for Chemical Kinetics Catalysis CHEM 1310 - Sections L and M 1 Reaction Rates: Example Recap: Avg. Reaction rate = +∆[Product]/∆t Avg. Reaction rate = -∆[Reactant]/∆t Butane, C4 H 10, burns in oxygen to give CO2 and H2 O vapor. If the butane concentration is decreasing at a rate of 0.20 M/s, what is the rate at which the oxygen is decreasing? What is the rate at which the products are increasing? 2C4H10(g) + 13O2(g) Week 15 8CO2(g) + 10H2O(g) 2 CHEM 1310 - Sections L and M Reaction Rates: Example Butane, C4 H 10, burns in oxygen to give CO2 and H2 O vapor. If the butane concentration is decreasing at a rate of 0.20 M/s, what is the rate at which the oxygen is decreasing? What is the rate at which the products are increasing? 2C4H10(g) + 13O2(g) 0.20 M/s For oxygen: 0.20 mol C4 H 10 Ls Week 15 8CO2(g) + 10H2O(g) ? ? ? x 13 mol O2 2 mol C4 H 10 CHEM 1310 - Sections L and M = 1.3 M O2 /s 3 1 Reaction Rates: Example Butane, C4 H 10, burns in oxygen to give CO2 and H2 O vapor. If the butane concentration is decreasing at a rate of 0.20 M/s, what is the rate at which the oxygen is decreasing? What is the rate at which the products are increasing? 2C4H10(g) + 13O2(g) 0.20 M/s For CO2: 0.20 mol...
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...CHAPTER 14 CHEMICAL KINETICS PRACTICE EXAMPLES 1A (E) The rate of consumption for a reactant is expressed as the negative of the change in molarity divided by the time interval. The rate of reaction is expressed as the rate of consumption of a reactant or production of a product divided by its stoichiometric coefficient. A 0.3187 M 0.3629 M 1min rate of consumption of A = = = 8.93 105 M s 1 t 8.25 min 60 s rate of reaction = rate of consumption of A2 = 8.93 105 M s 1 4.46 105 M s 1 2 1B (E) We use the rate of reaction of A to determine the rate of formation of B, noting from the balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (–2 moles A). (Recall that “M” means “moles per liter.”) 0.5522 M A 0.5684 M A 3moles B rate of B formation= 1.62 104 M s 1 60s 2 moles A 2.50 min 1min 2A (M) (a) The 2400-s tangent line intersects the 1200-s vertical line at 0.75 M and reaches 0 M at 3500 s. The slope of that tangent line is thus 0 M 0.75 M slope = = 3.3 104 M s 1 = instantaneous rate of reaction 3500 s 1200 s The instantaneous rate of reaction = 3.3 104 M s 1 . (b) At 2400 s, H 2 O 2 = 0.39 M. At 2450 s, H 2 O 2 = 0.39 M + rate t At 2450 s, H 2 O 2 = 0.39 M + 3.3 10 4 mol H 2 O 2 L1s 1 50s = 0.39 M 0.017 M = 0.37 M 2B (M) With only the data of Table 14.2 we can use only the reaction rate during the...
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...After it was first discovered in 1886 by Hans Heinrich Landolt, the iodine clock reaction has become one of the most used classroom experiments for demonstrating kinetics (SuperchargedScience, 2018). A clock reaction is a reaction where one of the chemical species (clock chemical) starts off with a low concentration and then has a rapid increase which can then lead to a dramatic colour change (UnitverityNottingham, 1999). The iodine clock works by using iodine as the clock chemical that then reacts with clear starch to produce a deep blue colour. For the Landolt chemical clock in two clear liquids are mixed with one containing KIO3 and the other NaHSO4 they then follow the following chemical reactions. Equation 1 is known as the rate determining...
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...Introduction Enzymes are a key aspect in our everyday life and are a key to sustaining life. They are biological catalysts that help speed up the rate of reactions. They do this by lowering the activation energy of chemical reactions (Biology Department, 2011). In chemical reactions bonds must be broken and new bonds must be formed. In order for this to occur the bonds must be made less stable. For bonds to become less stable a small input of energy is required and this is called the activation energy. In simpler terms, in order for a reaction to begin and proceed spontaneously a small input energy is required to give the reaction a push and get it started (Cooper, 2000). As said before catalysts are chemical agents used to speed up the rates of reactions. The biological catalyst is a group of proteins called enzymes. Enzymes work by lowering the activation energy and making it easier for the reactants to obtain the necessary energy to break the kinetic barrier. Even though enzymes speed up the rate of reaction, they do not change the free energy of the reactants and the products (Russel et al., 2010). Enzymes work by combining with reacting molecules at the active site. Each enzyme is specific to only one kind of molecule and can only bind to its specific molecule. The active site is a groove in the enzyme where the molecule will bind to; this is formed by the enzyme folding into a specific shape. When the enzyme is done and the molecules are then in the transitional...
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...UNIT 3 SELF-QUIZ (Page 412) 1. False: A physical change usually involves a larger enthalpy change than does a chemical change. 2. True 3. False: The potential energy of the products is smaller than the potential energy of the reactants in an exothermic change. 4. True 5. False: An endothermic reaction absorbs heat from the surroundings. 6. True 7. True 8. False: Three-quarters of a radioisotope will have changed after two half-lives. 9. False: In an endothermic reaction, only the potential energy of the chemical system increases. 10. True 11. (b) 12. (c) 13. (e) 14. (a) 15. (e) 16. (c) 17. (b) 18. (d) 19. (b) 20. (e) 21. (b) 22. (a) 23. (c) 24. (c) 25. (e) 26. (c) 27. (b) 28. (d) 29. (c) 30. (d) UNIT 3 REVIEW (Page 414) Understanding Concepts 1. qwater = mc∆T = 1500 g ϫ 4.18 J/(g•ºC) ϫ (75 – 20)ºC qwater = 340 kJ 2. MCl = 70.9 g/mol 2 1 mol nCl = 2250 g ϫ ᎏᎏ 2 70.9 g nCl = 31.7 mol 2 ∆H = nCl ∆Hvap 2 = 31.7 mol ϫ 20.7 kJ/mol ∆H = 657 kJ 218 Unit 3 Copyright © 2003 Nelson 3. (a) qwater = mc∆T = 100.0 g ϫ 4.18 J/(g•ºC) ϫ 5.6ºC qwater = 2341 J nNaOH = MV = 0.700 mol/L ϫ 0.600 L nNaOH = 0.420 mol n∆Hneut = qwater qwater ∆Hneut = ᎏᎏ n 2341 J = ᎏᎏ 0.420 mol ∆Hneut = 5573 J/mol, or 5.57 kJ/mol Because the reaction is exothermic, ∆Hneut = –5.57 kJ/mol. (b) The assumption is that the reaction went to completion and that all heat from the reaction was absorbed by the water in the calorimeter and not...
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...CAcT HomePage Rate and Order of Reactions Skills to develop Derive the integrated rate laws from differential rate laws. Describe the variation of concentration vs. time for 1st order reactions. Describe the variation of concentration vs. time for 2nd order reactions. Figure out order of reaction from concentration vs. time plots. Rate and Order of Reactions The rate of a chemical reaction is the amount of substance reacted or produced per unit time. The rate law is an expression indicating how the rate depends on the concentrations of the reactants and catalysts. The power of the concentration in the rate law expression is called the order with respect to the reactant or catalyst. This page deals specifically with first- and second-order reaction kinetics, but you should know that other orders such as zeroth-order and 3rd order may also be involved in chemical kinetics. Reaction Rates and Stoichiometry In acidic solutions, hydrogen peroxide and iodide ion react according to the equation: H2O2 + 2H+ + 3I- = 2 H2O + I3-. In this reaction, the reaction Rate can be expressed as decreasing Rate of H2O2, - d[H2O2]/dt decreasing Rate of H+, - d[H+]/dt decreasing Rate of I-, - d[I-]/dt increasing Rate of H2O, + d[H2O] /dt increasing Rate of I3-, d[I3-]/dt However, from the stoichiometry, you can easily see the following relationship: d[H2O2] 1 d[H+] 1 d[I-] 1 d[H2O] d[I3-] - ------- = - - ------ = - - ------- = - ------ = -------- ...
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...Second Order Reaction Kinetics Abstract The objective of this experiment was to determine if the specific rate constant of the reduction reaction of hexacyanoferrate (III) ion with ascorbic acid (C6H8O6) is affected by the ionic strength of the solution and the charges of the ion species within the solution. A Cary Bio 50 Spectrophotometer and its associated software was used to measure the absorbance of the solution at a wavelength of 418 nm. Analysis of the data collected supports the conclusion that the ionic strength of the solution and the charges of ions on the activated complex have a direct relation to the rate constant. The experimental value of 2.237 for the ionic strength of 0.025 M when compared to the literature value of 1.72 in a similar experiment by Nobrega and Rocha, has a percent error of 30%. For the ionic strength of 0.05 M, the percent error between the experimental value 2.851 and the literature value of 2.58 was 11%. The ionic strength of 0.1 had an error of -1% between the literature value of 3.97 and the experimental value 3.924. For the ionic strength of 0.2 there was a -23% error between the experimental value of 5.301 and the literature value of 6.89 (Nobrega, 1997). The experimental value of the product of the charges of the ions in the activated complex, 2.1, was lower than the expected value 3, which could be due to the shielding of the charges on the reactants due to the ions witihin the solution. Introduction The reduction of the hexacyanoferrate...
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...STK 2023 PHYSICAL CHEMISTRY 2 ASSIGNMENT 1 Question 1 (a) Rate of formation = 13d[O2]dt Rate of disappearance = - 12d[O3]dt (b) Rate of formation = 12d[HF]dt = d[O2]dt Rate of disappearance = - 12d[HOF]dt (c) Rate of formation = 12d[BrNO]dt Rate of disappearance = - 12d[NO]dt= - d[Br2]dt Question 2 2NO(g) + H2(g) N2O(g) + H2O(g) (a) Rate = k[NO]2[H2] (b) 0.005 = k(0.075)2(0.400) 0.005 = k(2.25 x 10-3) k = 2.22 M-2 s-1 Question 3 Temperature (K) | Rate constant (M/s) | 1/T (K-1) | ln k | 573 | 2.91 x 10-6 | 0.00175 | -12.75 | 673 | 8.38 x 10-4 | 0.00149 | -7.08 | 773 | 7.65 x 10-2 | 0.00130 | -2.57 | Plot of ln k against 1/T (K-1) Slope of graph, m = -7.08-(-12.75)0.00149-0.00175 = 5.67- 0.00026 = 21807.69 Using Arrhenius equation, m = - Ea8.314 J/Kmol 21807.69 = - Ea8.314 J/Kmol Ea = 181 kJ/mol Question 4 (a) Rate of reaction of NO is second order. Rate of reaction of H2 is first order. Total rate of reaction = third order. (b) If [NO] is doubled, the rate of reaction quadruples. (c) If [H2] is halved, the rate of reaction is halved. Question 5 Initial concentration = Ao Final concentration = A Time, t = 65s – 45s = 20s Use ln A = -kt + ln Ao ln (8.70 x 10-3M) = - (k)(20s) + ln (7.30 x 10-2M) -4.744 = - (k)(20s) + (-2.617) 2.127 = (k)(20s) k = 0.106 s-1 Question 6 The rate law is determined based on the slow reaction since the fast reactions are...
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