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Chem lab (1).docx

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School
University of Waterloo
Department
Chemistry
Course
CHEM 120L
Professor
Sue Stathopulos
Semester
Fall

Description
Osmosis and Its Impact on Structures ofAnimal and Plant Cells By: Seung-HunAnthonyYu Student ID: 20419514 Partner: Jeffrey Xiang Instructor: Sue Stathopulos Teaching Assistant: Joshua Zimmerman CHEM 120L Date Performed: November 6, 2012 Date Submitted: November 20, 2012 Introduction In a chemical reaction, change in energy of a system is often depicted by change in heat (q). The First Law of Thermodynamic states that energy is conserved; energy can be neither lost nor destroyed. That is, if a system was to gain or lose heat, it must interact with the source of heat; the surroundings of the system. This derives the equation: q = - q . This essential system surrounding means that in a perfect system, all the heat lost from the system would be transferred to the surrounding, and vice versa.Areaction can be either endothermic or exothermic. In an endothermic reaction, a net absorption of heat occurs. In this case, q value would be positive for the system while it is negative for the surrounding, as in an endothermic reaction, the system would absorb heat from the surrounding. Conversely, an exothermic reaction occurs when a net release of heat occurs. In other words, heat will be transferred from the system to the environment, causing q value of the system to be negative, while it is positive for the surrounding environment. The purpose of this experiment is to determine the heat of the reaction (q) of various neutralization reactions through calorimetry. In addition, in the last part of the experiment, the concentration of unknown hydrochloric acid will be determined using calorimetry and thermochemical properties. Caorimetry is common technique used to determine the heat of reaction. It uses calorimeter, a device that can effectively isolate a chemical reaction, so that the temperature of the system is unaffected by its surrounding. However, calorimeter cannot isolate a reaction in terms of pressure. This means that the pressure of the reaction will be same as the pressure of the surrounding. Hence, the pressure is assumed to be constant for the purpose of this experiment (as air pressure of the room would have a pressure change that is negligible). The q value at a constant pressure is more commonly known as enthalpy (ΔH). This derives the equation; ΔH = q = C * m *ΔT whereΔH is enthalpy, in joules, m is the mass of the system in grams,ΔT is a change in temperature in degrees (or kelvin), and C is the specific heat capacity of the reaction, in joules per degree per grams. In this experiment, the specific heat capacity of each solution will be assumed to equal to that of water, at 4.184 J / (deg * g). In partAand part B of this experiment, neutralization reactions between strong electrolytes occur. The reactants of these two parts – sodium hydroxide, hydrochloric acid, and nitric acid) would ionize completely as they are strong acid and bases, and the resultant reaction would be H (from acid) + OH (from base) → H O, a2 other reactants would simply dissociate and cancel each other out at the aqueous state.As a result, partAand part B are expected to yield + similar results, and are expected to result inΔH close to -55.90kJ per mol of H . The negative value of ΔH indicates that reaction between these strong electrolytes should give us an exothermic reaction. Part C of the experiment will study a reaction between sodium hydroxide and phenol; a neutralization reaction between a strong base and weak acid. Unlike the reactions between two strong electrolytes, weak acids tend to undergo dissociation reaction prior to participating in the reaction for forming water. This causes enthalpy to change as the dissociation of weak acid would also change enthalpy, and final enthalpy would be the sum of the two reactions. Hence, neutralization involving weak electrolytes generally have varying enthalpy. This part is expected to yield a positive enthalpy ofΔH = 25.3kJ / mol, and is expected to result in an endothermic reaction. Lastly, for part D, reaction between hydrochloric acid and sodium hydroxide would be studied. However, unlike partA, the concentration of hydrochloric acid is unknown, and the goal of this exercise is to find the unknown, assuming thatΔH = -55.90kJ/mol Experimental Procedure The experimental procedure used for this experiment was outlined in CHEM 120L lab manual, Experiment #4 (pages 52-54).All steps were followed without deviation. Experimental Observations Table of Concentration and Temperature of Solutions Part [NaOH] Acid [Acid] A 2.061M HCl 2.121M B 2.061M HNO 3 2.133M C 2.061M 0.5115M D 2.061M Unknown HCl#4 ? Part A Part B Initial 60s Trial 1 (˚C) Trial 2 (˚C) 30s for 5 min Trial 1 (˚C) Trial 2 (˚C) Initial 60s 10s 37.0 36.0 30s 36.7 36.0 10s 20s 36.9 36.0 60s 36.5 36.0 20s 30s 36.9 35.9 90s 36.2 36.0 30s 40s 36.9 35.9 120s 36.2 35.9 40s 50s 36.9 36.0 150s 36.1 35.9 50s 60s 37.0 35.9 180s 36.1 35.8 60s 210s 36.0 35.8 240s 36.0 35.8 270s 36.0 35.7 300s 35.9 35.7 Table for Temperature Change in PartAand B Part C Part D Initial Trial Trial 30s Trial Trial Initial Trial Trial 30s Trial Trial 60s 1 (˚C) 2 (˚C) for 5 1 (˚C) 2 (˚C) 60s 1 (˚C) 2 (˚C) for 5 1 (˚C) 2 (˚C) min min 10s 24.9 25.9 30s 25.0 25.8 10s 39.5 40.0 30s 39.0 39.6 20s 24.9 25.9 60s 25.0 25.8 20s 39.5 40.0 60s 39.0 39.5 30s 24.9 25.9 90s 25.0 25.8 30s 39.4 40.0 90s 39.0 39.2 40s 25.0 25.9 120s 25.0 25.8 40s 39.2 40.0 120s 38.9 39.1 50s 25.0 25.8 150s 25.0 25.8 50s 39.2 40.0 150s 38.9 39.0 60s 25.0 25.9 180s 25.0 25.8 60s 39.2 39.8 180s 38.9 39.0 210s 25.0 25.8 210s 38.9 39.0 240s 25.0 25.8 240s 38.7 39.0 270s 25.0 25.8 270s 38.5 39.0 300s 25.0 25.8 300s 38.5 38.9 Table for Temperature Change in Part C and D Results & Calculations Summary Table of the Calculations Part Trial [NaOH] [Acid] ΔT (˚C) Moles of q (kJ) q / mol (M) (M) H2O (mol) (kJ/mol) A 1 2.061 2.121 13.9 0.0848 -5.23 -61.7 2 2.061 2.121 13.0 0.0848 -4.88 -57.5 Average partA: 2.121 13.5 0.0848 -5.06 -59.6 B 1 2.061 2.133 22.9 0.0853 -4.59 -53.8 2 2.061 2.133 22.9 0.0853 -4.59 -53.8 Average part B: 2.13
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