In redox reaction, one element or compound is reduced and gains electrons, while on the other hand, the other element or compound is oxidized and loses electrons. For this lab, through the given oxidation and reduction numbers from balancing the equation and the electrons, it was shown that Manganese was reduced while Iron was oxidized. Also from the equation, the movement of electrons can be noticed, as it was added or subtracted from the substances. In order to balance an equation, there must be same number of elements on both sides, with the exception of hydrogen and oxygen. From there, in order to balance oxygen, water molecules are enumerated to the opposite of the equation/reaction.
Catalysts operate by decreasing the value of the activation energy for the reaction. [1] The iodide ions lower the activation energy, making it easier for the reactants to convert to the products and vice versa, hence speeding up the reaction. We use an adaptation of the Arrhenius equation ln k’ = -Ea/RT + ln A . where k’ is the rate constant, Ea is the activation energy, R is the universal gas constant and T is temperature. [2] to determine the Activation Energy.
Investigating One Factor That Affects How The pH Changes During a Titration Introduction: In this experiment I will be investigating how a certain factor changes the pH during a titration. pH changes when acid is added to alkali, or vice versa, as the ions in each are able to neutralise each other. The H+ ions, found in all acids, bond with the OH- ions, found in all alkalis, to form the completely neutral molecule water, or H2O, as shown by this symbol equation: H+(aq) + OH- (aq) ( H2O (l) The pH of a solution of acid would start at 1 or two, and be deep red in colour when tested with universal indicator, but when alkali begins to be added, therefore starting to neutralise some of the H+ ions, the pH will start to increase. When there is a complete balance between the amount of H+ in the acid and OH- in the alkali, this would mean the solution is completely neutral, and at this point it would have a pH of 7, and would be green in colour through testing with universal indicator. Once more alkali is added to this, the solution will begin to increase again in pH, going up to 12 – 14, and being blue/purple in colour through testing with indicator.
Of course the saturated solutions take time this experiment is done in a two day lab setting part one should be completed during the first day and ready for use on day two. Remember because Le Châtelier’s principle if you don’t filter the solution will rise. The removal of OH- from the UN filtered solution will cause more Ca(OH)2 to be dissolved for the equilibrium to restore. The second part of this lab takes knowledge from past experiments done in this laboratory. Using the Primary method used to standardize the concentration of HCl, this will bring much lower percent error in the Ksp For this standardization sodium carbonate should be used for determining the amount of hydroxide in the solution.
Oladapo Adaramola Lab # 6 Chemistry equilibrium: Formation constant of a complex ion system Aim: Find the equilibrium concentrations of FeSCN+2 from a graph of absorbance vs. known FeSCN+2 concentrations. Calculate the equilibrium constant (K) for the reaction. Procedure part 1: In order to know the relation between the absorbance of a solution and its concentration, it is necessary for the experimenter to prepare a calibration graph of the molar concentration of FeSCN2+ vs. absorbance. The problem associated with this is that since the reaction is an equilibrium reaction, it has not gone to completion, and the concentration of FeSCN2+ in solution is difficult to determine. Thus, the experiment should “force” the reaction to go almost to completion by the addition of an excess (over 200 times of the amount needed) of Fe3+ ions to a relative small quantity of HSCN.
Which way will the following equilibrium shift if the total pressure on the system is decreased? 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g) Answer________________ 4. Hydrogen peroxide is decomposed as follows: H2O2(l) → H2(g) + O2(g) H = +187 kJ Predict the direction of equilibrium shift by each of the following imposed changes: a) Increase the [H2] ........................................Answer ______________________ b) Decrease the [O2] .......................................Answer ______________________ c) Decrease the total pressure ........................Answer ______________________ d) Increase the temperature............................Answer ______________________ e) Add MnO2 as a catalyst.............................. Answer ______________________ 5. Consider the following reaction at equilibrium: H2(g) + I2(g) → 2HI(g) a) Addition of more H2 gas to the container will do what to the rate of the forward reaction? Answer ________________________ b) If, for a while, the rate of the forward reaction is greater than the rate of the reverse
Determination of Freezing Point and Verification of Freezing Point Depression for a Mixture by Linah Richer Partners: Maike Blakely CHM317 Preformed: Nov. 8th, 2013 Report: Nov. 19th, 2013 Abstract: The purpose of this experiment is to determine the freezing point of the solvent, biphenyl, and verify the freezing point depression equation ΔTf=-ikfnsolutemsolvent. The addition of solute will lower the vapor pressure of the solvent/solute mixture resulting in the lowering of the freezing point for the mixture solution. The experiment resulted in the experimental difference in the two trial freezing points to be 0.08 K leading to a calculated change in temperature freezing to be -1.92 E -5 K, and an overall decrease in temperature freezing of the solute/solvent solution compared to the pure solvent solution. Introduction: Experimentally it is understood that the addition of a nonvolatile solute to a solvent will lower the vapor pressure, raise the boiling point and lower the freezing point. In this experiment, the freezing point for the solvent biphenyl will be determined theoretically and experimentally, as well as the verification of the freezing point depression equation for a solvent/solute mixture.
One of the major sources of error in this experiment is that the volume of metal ball will rise while the surrounding water’s temperature rise. The metal ball has fixed amount of gas, so with higher volume the inner pressure of the metal ball will be lower. In order to reduce the error and improve this experiment, we can use a metal with lower coefficient of volume expansion. Objective: The objective of the second lab is to determine the relationship between the pressure and volume of gas with fixed temperature. Based on the Bayle’s Law, the pressure of the gas is inversely proportional to the volume with fixed temperature.
INTRODUCTION The purpose of this experiment was to determine the pKa of an acid-base indicator (methyl red). All indicators are weak acids, and as such have a dissociation constant, or Ka. The subscript a signifies that the constant is for an acid. The lower case p indicates that the value is the negative log (base 10) of the constant. The Ka expression, Ka=H+In-HIn [1] (1) is the equilibrium for the dissociation of the weak acid, where HIn is the acid and In- is the conjugate base.
For instance, the neutralization of HCl by NaOH is written as: HCI + NaOH -----------> NaCl + HOH Equations for neutralization reaction are balanced so that the amount of H+ will be equal to the amount of OH-. The addition of a specific amount of base required to neutralize an acid in a sample involves a titration. An indicator in a sample will change color when all the H+ from an acid has been neutralized. The addition of a base should be stopped when the indicator changes color which therefore determines the endpoint. At this point the volume of base used to neutralized the acid can be determined.