Reaction Order and Rate Law Name Data‚ Calculations‚ and Questions A. Calculate the initial and final concentrations as needed to complete Tables 1 and 2. Data Table 1: Varying the Concentration of 1.0 M HCl | | | | |Concentrations | | |# Drops |# Drops |# Drops |Initial
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Lab 3: Reaction Order and Rate Laws Name: Student Lab Partners: None Date of Experiment: 16 March 2014 Location: My House Course Number: CHE112 Abstract: When combining compounds‚ there are always two roles considered in the reaction. These are reactants and products. Reactants are the initial compounds before mixing. The products are synonymous with the term byproduct‚ or what remains when the reaction occurs. For example‚ Sodium (Na+) and Chlorine (Cl-)
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Top of Form 1. What is the pH of a buffer that results when 33 g NaHCO3 is mixed with 100.0 mL of 2.50 M NaOH and diluted with water to 250 mL? (Ka for HCO3- = 4.8 10-11) (Points : 10) 8.20 10.07 10.12 10.32 10.56 2. What is the pH of an aqueous solution of 0.30 M HF and 0.15 M F-? (Ka of HF = 7.2 10-4) (Points : 10) 1.83 2.84 3.14 3.44 10.86 3. Which of the following mathematical expressions
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The reaction order‚ based on the rate law‚ was first order with respect to crystal violet and second order with respect to OH-. The rate law was as follows: Rate law = k [CV]1[OH-]1 where k equaled 2.61. In order to determine the reaction order with respect to crystal violet‚ the graph that described the relationship between ln[CV] and time (seconds) was Figure 2. Not only did Figure 2 generate a more linear relationship‚ but it had the highest R2 value of 0.992 than ([CV] versus time) and ([1/CV]
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volume). 3. [10 points] The following reaction is found to be first order in A: A −→ B + C If half of the starting quantity of A is used up after 56 seconds‚ calculate the fraction that will be used up after 6.0 minutes. 4. [15 points] The rate law for the decomposition of ozone to molecular oxygen 3 O3 (g) −→ 3 O2 (g) is rate = k [O3 ]2 . [O2 ] The mechanism for this process is k1 O3 O + O2 k−1 k 2 O + O3 −→ 2 O2 Derive a rate law from these elementary steps. Clearly
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Determination of Rate and Order of a Reaction Results This experiment used a spectrometer to find the wavelength with maximum absorbance in a green food coloring solution. For this particular solution the wavelength was 629.7 nm. The system was then calibrated to that and was set to measure the food coloring and bleach solution. The measured visible light absorbance of the mixed solution was collected over a time of 200 seconds and eight points were then selected and placed into the Absorbance
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Effect of Temperature on Solubility of a Salt Ahmed Mohammed 17-nov-2013 Abstract In this experiment‚ you will study the effect of changing temperature on the amount of solutet will dissolve in a given amount of water. In this experiment‚ you will completely dissolve different quantities of potassium nitrate‚ KNO3‚ in the same volume of water at a high temperature. As each solution cools‚ you will monitor temperature using a computer-interfaced Temperature Probe and observe
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Determining Order from Rates of Reactions AP Chemistry Purpose The focus of this experiment is to recognize that when aqueous solutions of potassium iodate ion (KIO3-) and bisulfite ion (HSO3-) are mixed‚ a series of reactions will occur‚ and the final reaction is signaled by the appearance of a dark blue color. My partners and I investigate how the concentration of the reactants affects the rate of reaction. The purpose of this lab is to find
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RATE OF REACTIONS. The reaction rate (rate of reaction) or speed of reaction for a reactant or product in a particular reaction is intuitively defined as how fast or slow a reaction takes place. For example‚ the oxidative rusting of iron under the atmosphere is a slow reaction that can take many years‚ but the combustion of cellulose in a fire is a reaction that takes place in fractions of a second (right). Chemical kinetics is the part of physical chemistry that studies reaction rates. The concepts
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Aim: To determine the rate law of chemical reactions Introduction: The aqueous solution of potassium peroxydisulphate can oxidize potassium iodide as follow: Equation: K S 0 (aq) + 2KI (aq)->2K S0 (aq)+I (aq) 2 2 8 2 4 2 KI(aq) +I (aq) -> KI (aq) 2 3 _________________________________________________ K S 0 (aq) +3KI(aq) -> 2K S0 (aq)+ KI (aq) 2 2 8 2 4 2 The rate law of this reaction can be represented as follow: Rate=k[S208 2-]^a [I-]^b When the concentration of peroxydisulphate ions
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