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Chem lab (3).docx

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Department
Chemistry
Course
CHEM 120L
Professor
Sue Stathopulos
Semester
Fall

Description
Heat of Neutralization with Strong and Weak Wlectrolytes By: Anas Jumale Student ID: 20417117 Partner: Cassandra Downs Instructor: Sue Stathopulos TeachingAssistant:Adrianna Richards CHEM 120L Date Performed: November 9, 2012 Date Submitted: November 23, 2012 Introduction Introduction All chemical reactions involve energy change. The study of energy change is an important part of chemistry. Fundamental to the thermo-chemistry is the law of conservation of energy, which states that energy is neither created nor destroyed, but can be converted from one form to another.An exothermic reaction is one in which stored chemical energy is converted to heat energy (heat is released to the surroundings), conversely an endothermic reaction is one in which heat energy is converted to chemical energy (heat is absorbed into the system). The enthalpy (H) of a substance, sometimes called its heat content, is an indication of its total energy content. The equation for calculating the enthalpy change (heat change) is given byΔH = 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 (Petrucci et al, 2010). 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) (Lab Manual). Reactants can fall into two categories, strong and weak electrolytes. Strong electrolytes are substances that completely, or almost completely, ionize and dissociate in solution. Some examples include solutions of HCl, KOH, NaOH and KCl . Weak electrolytes, on the other hand, only partially dissociate into ions . These include acetic acid, phenol and ammonia. With the neutralization of weak electrolytes, the heat of reaction can be either smaller or larger than the value of the amount of heat produced with strong electrolytes (Kenneth,2009) In this experiment, the enthalpy change for a specific pair of acid-base neutralization will be determined., the molarity of an HCl solution of unknown concentration will also be determined(Lab manual) 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] Trials Temperature (˚C) Initial Final A 2.061M HCl 2.121M 1 23.8 36.3 2 23.3 36.1 B 2.061M HNO 3 2.133M 1 23.4 36.0 2 23.5 35.4 C 2.061M 0.5115M 1 23.8 27.4 2 23.7 27.3 D 2.061M Unknown HCl#4 ? 1 23.7 37.1 2 22.9 35.9 Results & Calculations Summary Table of the Calculations Part Trial [NaOH] [Acid] ΔT (˚C) Moles of q (kJ) q / mol (M) (M) H 2 (mol) (kJ/mol) A 1 2.008 1.994 12.5 0.07976 -4.71 -59.01 2 2.008 1.994 12.8 0.07976 -4.82 -60.4 Average partA: 2.121 12.7 0.07976 -5.06 -59.7 B 1 2.008 2.035 12.6 0.0814 -4.74 -58.2 2 2.008 2.035 11.9 0.0814 -4.48 -55.0 Average part B: 2.133 12.3 0.0814 -4.59 -56.6 C 1 2.008 0.5115 3.6 0.0256 0.941 36.8 2 2.008 0.5115 3.6 0.0256 0.941 36.8 Average part C: 0.5115 3.6 0.0256 0.941 36.8 D 1 2.008 2.90 13.4 0.0939 - -59.7 2 2.008 3.03 13.0 0.0911 - -59.7 Average part D: 2.008 2.97 13.2 0.0925 - -59.7 Sample Calculations PartA 1. # of moles of water produced HCl (aq)NaOH (aq) NaCl (aq) H 2 (l) From HCl: From NaOH n = CV n = CV = 1.994 mol/L * 0.040L = 2.008 mol/L * 0.050L = 0.07976 mol of HCl = 0.1004 mol of NaOH = 0.07976 mol of H O2 = 0.1004 mol of H O2 *Therefore, HCl was the limiting agent, thus 0.07976 mol of water was produced. 2. Enthalpy per mole of water Enthalpy/mole =ΔH/n ΔH = CmΔT = (CmΔT) / n m = 90g = (4.184 J/(˚C*g) * 90g * 12.5˚C) / 0.07976 mol = 59014J/mol = 59.014 kJ/mol *Heat of reaction = - heat absorbed by water = - 59.014 kJ/mol Part B 1. # of moles of water produced HNO 3(aq)NaOH (aq) NaNO 3(aq) H 2 (l) From HNO : 3 From NaOH n = CV n = CV = 2.035 mol/L * 0.040L = 2.008 mol/L * 0.050L = 0.0814 mol of HNO 3
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