a) How this experiment is an illustration of Hess's Law?b) If you were to repeat this experiment in the “real world”, what are some of the major sources of error that could occur? PROCEDUREHEAT OF REACTION (simulation)Reaction 1In this experimental simulation you will determine and compare the quantity of heat energyreleased in three exothermic chemical reactions.a. In the glassware menu, take out a 50 mL graduated cylinder and a foam cup. From toolsmenu, take out the scale. From the chemical stockroom, move the distilled water and solidNaOH onto the workbench.Reaction 1: solid sodium hydroxide dissolves in water to form an aqueous solution of ionsb. Transfer 50.0 mL of water to the foam cup. To do this, drag the carboy of water onto thegraduated cylinder. Before you release the mouse button, the cursor will show a "plus sign" toindicate that it is the recipient. A transfer text bar will appear, enter "50.0" (mL) and click onpour. You will notice that the graduated cylinder reeds 50.0 mLNaOH(s) → Na*(aq) + OH (aq) + X¡ kJ (DH9olution)Reaction 2: Sodium hydroxide reacts with an aqueous solution of hydrogen chloride to formwater and an aqueous solution of sodium chloride.c. Weigh about 1 gram of solid sodium hydroxide pellets, NaOH (s), directly into the foam Cupand record its mass to the nearest 0.01 gram. To do this place the foam Cup on the balance so itregisters a mass, then click the "Tare" button. Drag the NaOH bottle onto the foam Cup. Whenyou release the mouse, the bottle will be tipped to show that it is in pour mode. Next type "1.00"(grams) into the transfer bar and click pour. Note that the balance now reads the mass of thetransferred NaOH. Record this mass in your table. You may not take the Cup off the scale andremove the scale from the workbench.NaOH(s) + H*(aq) + Cl'(aq) → H2O(1) + Na*(aq) + CI(aq) + X, kJ (DHreaction)Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution ofhydrogen chlorideform waand an aqueous solution of sodium chloride.Na" (aq) + OH (aq) + H*(aq) + Cl'(aq) → H2O(1) + Na*(aq) + Cl'(aq) + X3 kJ (DHneutralization)In order to accurately measure the heat released in each reaction, we will be using a calorimeter.For this experiment a Styrofoam Cup will act as a calorimeter. The change in temperature thatoccurs for each reaction will be used to calculate the energy released in kilo joules per mole ofsodium hydroxide used (kJ/mol). We can assume for our calculations that any heat transferred tothe Styrofoam cup and surrounding air will be negligible. We can also assume that the specificheat of water is 4.18 J/g°Cd. Click on the graduated cylinder, record its temperature and then drag it onto the foam Cup.When you release the mouse, the graduated cylinder will be tipped to show that it is in pourmode. Enter “50.0" (mL) in the transfer bar and then click pour. Record the highest temperature.Reaction 2a. Take 0.5 M HCl from the strong acid cabinet and foam Cup, and a 50mL graduated cylinderfrom the glassware menu and place them on the workbench. The procedure for reaction 2 isthe same as for reaction 1 except that 50.0 mL of 0.50 M hydrochloric acid solution is used inplace of the water. After measuring 50.0 mL of HCl solution into the graduated cylinder,proceed as before with steps (b) through to (d) of the procedure.Reaction 3a. Take out another graduated cylinder, a fresh foam Cup, the 1.0 M HCl and the 1.0M MNaOH. (If you are running out of room, you may remove the previous chemicals fromworkbench.) Measure 25.0 mL of 1.0 M HCl solution into a foam Cup. Pour an equalvolume of 1.0 M NAOH in the clean graduated cylinder.b. Record the temperature of each solution to the nearest 0.1°C. Pour the sodium hydroxidesolution into the foam Cup and record the highest temperature obtained during thereaction. ResultsReaction123Mass of NaOH1.00 g1.00 gVolume of NaOH +50.0 mL50.0 mL25.0 mLHCl or H2OMass of solution51.0 g51.0 g50.0 gInitial TemperatureFinal Temperature25.00 •C25.00 •C25.00 •C30.30 •C36.97 •C31.7 °CΔΤ30.30 – 25.0036.97 – 25.031.7 – 25.00= 5.30 •C11.97 C= 6.7 °C%3D51.0 g * 4.18 * 5.30= 1129 j/ 1.129 kJ51.0 g * 4.18 * 11.97= -2500 j/ -2.5 kJ50.0 g * 4.18 * 6.7= -1400.3 j / -1.4 kJ0.025 molHeat of reactionMol NaOH0.025 mol0.0250 molEnthalpy, kJ/mol1.129/0.025-2.5/0.025-1.4/0.025-45 kJ/mol= -100.0 kJ/mol= -56.0 kJ/mol

Given reactions with their heat of reactions are: Reaction 1: Solid NaOH dissolves in water to form…

Answered: a) How this experiment is an…

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...