Chemistry - Grade 12+ - yTeach Publisher - Class activity - Learning object - Self-study resource - Chemical equilibrium and equilibrium constant. Kinetics%20of%20Crystal%20Violet%20Fading. Rate Law Determination of the Crystal Violet Reaction > Experiment #30 from Chemistry with Vernier. Recommended for grades 9–12. Loaded: 0% Progress: 0% Introduction In this experiment, you will observe the reaction between crystal violet and sodium hydroxide. One objective is to study the relationship between concentration of crystal violet and the time elapsed during the reaction. A simplified version of the equation is: The rate law for this reaction is in the form: rate = k[CV+]m[OH–]n, where k is the rate constant for the reaction, m is the order with respect to crystal violet (CV+), and n is the order with respect to the hydroxide ion.
As the reaction proceeds, a violet-colored reactant will be slowly changing to a colorless product. Absorbance vs. time: A linear plot indicates a zero order reaction (k = –slope). ln Absorbance vs. time: A linear plot indicates a first order reaction (k = –slope). 1/Absorbance vs. time: A linear plot indicates a second order reaction (k = slope).
Objectives In this experiment, you will Observe the reaction between crystal violet and sodium hydroxide. Kinetic Molecular Theory: Pressure. Java Security Settings: This web page employs Java, which requires specific security settings for correct operation. If the applets on this page do not run correctly, consult the Virtual Chemistry ExperimentsFAQ or the Physlet Physicsweb site for establishing the correct security settings. Concepts What is the origin of the pressure exerted by a gas? As is evident from the previous exercises, the Kinetic Molecular Theory proposes that a gas is composed of a large number of particles in rapid motion. Each particle occasionally collides with a wall of the container. As a result of that collision, the particle exerts a force on the wall.
What does the Kinetic Molecular Theory predict about the pressure of a gas? Effect of Particle Mass The simulation below depicts a system containing helium atoms (red, FW 4.00) and krypton atoms (blue, FW 83.80). 1. 2. 3. 4. Factors affecting the Pressure In the simulation below you may select the number of molecules and the molecular mass of the molecules. 1. Chemistry Online @ UTSC. Solubility Theory Solubility plays a critical role in organic chemistry. Its applications are extensive, ranging from purification (extraction) to identification of unknown compounds. It is the latter that will be discussed. Using solubility as a means of identification takes advantage of the properties associated with certain functional groups.
While it does not determine the exact chemical structure, as spectroscopy techniques do, it provides a rough idea of key functional groups present and the degree of hydrocarbon character. What is solubility? Solubility can be defined as the maximum amount of solute that can dissolve in a fixed amount of solvent at a specific temperature. What is considered “dissolved”?
A solute is considered “dissolved” when a homogenous solution is formed, with no suspended particles. What takes place at a molecular level? The formation of a solution is a 3 step process that can be represented by an enthalpy diagram: Example: Acetone being dissolved in water WHY? Ch03_11. 3.11 Solutions and Solution Stoichiometry Many chemical reactions are rapid and reproducible when carried out in the liquid state. As a result, chemists often dissolve solids in a liquid and carry out reactions in a liquid medium. Most of the reactions in our bodies occur in aqueous (water) solutions. In Chapter 1, we noted that a solution is a homogeneous mixture of two or more substances. A solution of sugar in water does not consist of tiny particles of solid sugar dispersed among droplets of liquid water. Rather, individual sugar molecules are randomly distributed among water molecules in a uniform liquid medium. A solution is homogeneous right down to the molecular level.
The components of a solution are the solute(s)—the substance(s) being dissolved—and the solvent—the substance doing the dissolving. Components of a Solution activity The concentration of a solution refers to the quantity of solute in a given quantity of either solvent or solution. Molar Concentration Suppose we let. How Buffers Work. How does a mixture of a weak acid and its conjugate base help buffer a solution against pH changes? If we mix a weak acid (HA) with its conjugate base (A-), both the acid and base components remain present in the solution. This is because they do not undergo any reactions that significantly alter their concentrations. The acid and conjugate base may react with one another, HA + A- → A- + HA, but when they do so, they simply trade places and the concentrations [HA] and [A-] do not change. In addition, HA and A- only rarely react with water.
By definition, a weak acid is one that only rarely dissociates in water (that is, only rarely will the acid lose its proton H+ to water). Likewise, since the conjugate base A- is a weak base, it rarely steals a proton H+ from water. So, the weak acid and weak base remain in the solution with high concentrations since they only rarely react with the water. See the next page for a tutor that leads you through the calculations involved in making a buffer. Heat of Reaction for the Formation of Magnesium Oxide Lab Answers. Planning A: Refer to lab handout entitled, Heat of Reaction for the Formation of Magnesium Oxide. Planning B: Refer to lab handout entitled, of Reaction for the Formation of Magnesium Oxide. Data Collection: Quantitative Table I: Qualitative Table I: Data Analysis: Therefore, the average temperature high for the MgO and HCl solution in trial 1 was 30.0⁰C.
Therefore, the average temperature high for the MgO and HCl solution in trial 1 was 46.0⁰C. Therefore, the average temperature high for the MgO and HCl solution in trial 2 was 29.0⁰C. Therefore, the average temperature high for the Mg and HCl solution in trial 2 was 44.5⁰C. Data Processing: Discussion: This investigation was conducted in order to determine the enthalpy of formation for magnesium oxide by manipulation of the three equations given. . -593.3KJ/mol and that the thermo chemical equation (target equation) for the combustion of magnesium is (see right) . ChemLab - Chemistry 3/5 - The Enthalpy of Formation of MgO - Chemistry. The Adiabatic Calorimeter In most systems, without special precautions, the system not only exchanges enthalpy with its surroundings but also suffers a change in its temperature.
In order to avoid such a complicated situation, thermochemists use the adiabatic calorimeter, a device so well insulated thermally that the transfer of energy (and enthalpy) as heat across the boundary between the calorimeter and its surroundings is effectively zero during the experiment. In practice, perfect insulation or matching of temperatures between system and surroundings is never achieved, but, if the energy flow is small and steady, the necessary corrections can be made. The use of an adiabatic calorimeter can be understood with the help of the following diagram of the thermodynamic states. The reaction path observed in the laboratory is adiabatic, along the path labeled 3: A (at T1) + B (at T1) C(at T2) + D(at T2) q3 = Hpath 3 = 0 Hpath 3 = Hpath 1 + Hpath 2 = 0 Hpath 3 = 0 , we have Hpath 1 = - Hpath 2. Theoretical Yield. The theoretical yield of a reaction is the amount of product that would be formed if the reaction went to completion.
It is based on the stoichiometry of the reaction and ideal conditions in which starting material is completely consumed, undesired side reactions do not occur, the reverse reaction does not occur, and there no losses in the work-up procedure. Theoretical yield calculations are carried out in the same way as they were in general chemistry: the moles of limiting reactant determines the moles of product. To calculate theoretical yield: Balance the reaction and determine the stoichiometry or ratios of reactants to products. Find the number of moles of each starting material used. Determine which reagent is limiting; remember that catalysts, solvents, or any compounds that are not part of the actual chemical reaction cannot be the limiting reagent.
Calculate the moles of product expected if the yield were 100% based on the limiting reagent. Example 4: mass = volume x density.
Kent's Chemical Demonstrations Movies Page 1. Elemental Analysis Tutorial and Quiz. Using the above measurements and the power of math, we will come up with a formula for glucose. As usual we can't count atoms, but we can measure their mass. Here we have measured the mass of water absorbed by measuring the before and after weights of the MgSO4 canister. Then we can convert that to a count of hydrogen atoms: The increase of mass in MgSO4 canister is just from H2O. However, the NaOH canister's increase in weight is from CO2 absoption but some mass from the original NaOH is lost as water vapor. 2NaOH + CO2 --> Na2CO3 + H2O The canister's start weight was from the canister itself plus an unknown mass of the NaOH inside. Let N be the starting grams of NaOH. Canister after glucose burned: 241.8 g = C + N + S - M Canister with NaOH start weight: 155.2 g = C + N Subtracting the the start weight from the end weight: 86.6 g=(C+N+S-M) - (C+N) = S - M This shows that the difference in weight is the grams of Na2CO3 created minus the grams of NaOH converted.