effect of temperature on the rate of reaction. To investigate the order of reaction with respect to hydrogen peroxide and ethanoic acid (acetic acid) by the use of an oscillating clock reaction. To determine the activation enthalpy with and without catalytic ions and use this to compare the effectiveness. To investigate the rate equation‚ rate constant and possible mechanism for this reaction. Background research: The Briggs-Rauscher reaction The Briggs-Rauscher reaction also known as the oscillating
Premium Chemical kinetics Reaction rate Rate equation
CHEM 104: General Chemistry II Spring 2014 Exam 1 Version A Instructions: 1. Do NOT open this exam until you are told to do so. Opening the exam before the start of the exam is considered an act of Academic Dishonesty. 2. All answers must be bubbled into the scantron answer sheet. 3. No credit will be given for answers written on this exam. 4. You must use a #2 pencil to bubble in the answers on your scantron. 5. This is a multiple choice exam consisting of 30 questions (7 pages). Each question
Premium Chemical kinetics Rate equation Reaction rate
graphs‚ and calculate rate constants. By determining the number of drops of sodium hypochlorite necessary to make a diluted food dye change to colorless in less than three minutes helps calculate the absorbance which can then be analyzed to find the pseudo rate constant and eventually leading to the rate constant and the rate law. The results show that the m and n are both 1st order which makes the overall order of the blue dye 2nd order. Results: In this experiment‚ rate laws and graphs help calculate
Premium Chemistry Chemical reaction Reaction rate
Chapter 36 Rate of chemical reaction Page Number Class practice 1 Think about 5 Chapter exercise 5 Chapter 37 Factors affecting rate of reaction Class practice 11 Think about 14 Writing practice 14 Chapter exercise 14 Chapter 38 Molar volume of gases at room temperature and pressure (r.t.p.) Class practice 21 Self-test 24 Think about 26 Chapter exercise 27 Part exercise 34 Chapter 39 Dynamic equilibrium Class practice 44 Think about 45 Chapter exercise 46 Chapter 40 Equilibrium constant Class practice
Premium Mole Carbon dioxide Hydrochloric acid
concentrations. The calculated Km constant was found to be 0.22 for all reactions. The Vmax values for each inhibition ion were 0.00986 for the phosphate ion and 0.00436 for the fluoride ion. The inhibitor constant‚ Ki‚ was determined to be 0.0967 for the phosphate ion. The inhibitor constant for the fluoride ion was determined to be 0.137. Both ion inhibitors resembled competitive inhibition. Introduction Enzyme kinetics is a classification of chemically catalyzed reactions that take place in the presence
Premium Enzyme Chemical reaction Chemistry
Enzymes are a type of proteins that are formed by Amino acids and help speed up metabolic reactions. They are able to do this by interacting with the substrate . The substrate is what is being breaking down in the reaction. The substrate comes in to contact with the enzyme by binding to the enzyme’s unique active site . This is called the enzymesubstrate complex . The complex may cause a reduction in the activation energy for the reaction. the rates of reaction can be affected based on concentration‚ pH‚ Temperature
Free Chemical reaction Enzyme Catalysis
kinetics‚ which can also be explained as reaction kinetics‚ studies the rates of chemical processes and reactions. A lot of this has to do with the speed at which a chemical reaction occurs for instance the rate at which reactants and products appear and disappear. In this case chemical kinetics goes further in depth to understand what conditions affect the rate of a given reaction. This is important because it allows scientists to determine what type of reaction can occur‚ and how they can alter the
Premium Reaction rate Chemical reaction Chemical kinetics
mid-term). This is taken from an earlier final exam. 1. (4 pts) Permanganate ion oxidizes sulfide ion to elemental sulfur (which you should represent as a single S atom) and is reduced to MnO2 in basic solution. In the balanced equation for this reaction‚ which you should determine in the work area‚ how many water molecules and hydroxide ions are needed in the final balanced equation? Work area: # of H2O ____4______ # of OH- ___8_______ MnO4- + S2- ---> S + MnO2 x3 Oxidation: S2- ----->
Premium Chemical kinetics Reaction rate Electrochemistry
(M) | Initial rate (sec -1) | 1 | .01 | .01 | .002246 | 2 | .01 | .005 | .001348 | 3 | .005 | .01 | .001627 | 4 | .0075 | .005 | .001126 | 5 | .005 | .0075 | .001267 | Order of Reaction: n=1;m=1 R1/R2=.002246/.001348=k[.01][.01]/k[.01][.005] 1.666172=2n n=1 Rate law Expression: Rate=k[FeCl3][KI] Rate Law constant: Trial 1: .002246=k[.01][.01] k=22.46 Trial 2: k=26.96 Trial 3: k=32.54 Trial 4: k=30.027 Trail 5: k=33.787 Average Rate Constant: Trial1+2+3+4+5=145
Premium Chemical reaction Rate equation Reaction rate
in rate is only due to the change in concentration of A (Propanone). The concentration of propanone in experiment 2 is the double of it in experiment 1. Something similar happens with the rate: In experiment 2 the rate is almost the double than the rate of experiment 1. This means that the order of [C3H6O] is 1. 2) In experiment 4 and 5 [Solution B] is the only one which changes. The concentration of iodine in experiment 5 is the half than in experiment 4. It is noticeable that the rate didn’t
Premium Chemistry Chemical reaction Oxygen