Exp. 4 - Heat of Reaction

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University of Nebraska, Lincoln *

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Chemistry

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Dec 6, 2023

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Experiment 4 Heat of Reaction Total Points: 40 Student’s Name: Will DePooter Lab Section: 501 REPORT Your report should include the following: Data Tables 4-1 and 4-2 Example calculations Answers to questions NOTE: You must show your work for all calculations; no work, no credit. DATA TABLE 4-1. (8 pts) Trial Mass of KCl (g) Initial Temperature (°C) Final Temperature (°C) 1 .501 22.3 19.3 2 1.00 22.3 17.5 3 2.00 22.5 13.5 4 3.00 22.5 10.9 Trial Mass of CaCl 2 •2 H 2 O (g) Initial Temperature (°C) Final Temperature (°C) 1 .500 22.4 25.0 2 1.00 22.5 27.8 3 2.00 22.5 33.9 4 3.00 22.5 37.3 DATA TABLE 4-2. (10 pts) KCl Heat of Water (J) Heat of Reaction of the Salt (kJ/mol) Trial 1 -125.52 18.600 Trial 2 -200.832 14.972 Trial 3 -376.56 14.036 Trial 4 -485.344 12.060 Average Heat of Reaction for KCl = 14.917 CaCl 2 •2 H 2 O Heat of Water (J) Heat of Reaction of the Salt (kJ/mol) Trial 1 108.784 -31.915 Trial 2 221.752 -32.6 Trial 3 476.976 -35.060 Trial 4 619.232 -30.345 Average Heat of Reaction for CaCl 2 •2 H 2 O = -32.48 Example Calculations for Heat of Water (2 pts) and Heat of Reaction of the Salt (2 pts) ( for the first trial of each salt only):
Experiment 4 Heat of Reaction Total Points: 40 qsalt= - qwater =10x4.184 (19.3-22.3) = -125.52J = (1.25/.501)(MMKCl/1mol)(10^-3) Heat of reaction = 18.600 qwater=-qsalt = 10x4.184(25.0-22.4) =108.784 qwater 108.784 qsalt= -138.072 Heat of reaction= -31.915 QUESTIONS 1. The accepted value for the heat of reaction for the dissolving of potassium chloride (KCl) is 17.2 kJ/mol of KCl. Determine the percent error in your measured value. (2 pts) Percent error = 14.917-17.2/ 17.2 x 100 Percent error = 13.27% 2. According to the Calcium Chloride Handbook printed by Dow Chemical Company in 2003, the accepted value for the heat of reaction for the dissolving of calcium chloride dihydrate (CaCl 2 •2 H 2 O) is -44.8 kJ/mol of CaCl 2 •2 H 2 O. Determine the percent error in your measured value. (2 pts) Percent Error = | experimental accepted | accepted × ( 100 ) Percent error = -32.48-(-44.8)/ -44.8 Percent error = 27.5% 3. Give at least one possible source of error in your experiment that might explain why your values for KCl and CaCl 2 •2 H 2 O differ from these accepted values. (2 pts) One possible error that we observed while preforming the lab is that the salts would stick to the little cups that we were using to get the initial weight and we couldn’t find an effective way to get all the salt out without creating a big mess or getting some of it outside the styrofoam cup the reaction was taking place in. 4. Medical cold packs can use various chemicals in order to produce a cold compress to help reduce inflammation caused by injuries. Based on the accepted value in Question 1, how many grams of potassium chloride would have to be dissolved into 750 g of 20.0°C water to produce a 4.0°C solution? (Remember: specific heat of water is 4.184 J/g•°C) (6 pts) qwater = 750x4.184x(4-20) = -50208J
Experiment 4 Heat of Reaction Total Points: 40 qsalt= 50208J = 50.208KJ Mols of KCL= 2.919 Grams of KCL= 217.62 You would need to have 217.62 grams of KCl that would dissolve into 750 grams of water to get the desired temperature of 4C. 5. The solubility limit of CaCl 2 •2 H 2 O in boiling water is 152 g CaCl 2 •2 H 2 O in 100 g water. If you dissolved enough CaCl 2 •2 H 2 O in water to reach this solubility limit, would the water boil if the water was initially 20°C? (6 pts) Moles of CaCl2 x H2O = 152g/ 147.012(g/mol) KJ produce by CaCl2 = -46.32KJ = -46320J 46320J= 100x4.184x(Tf-20) Tf= 130.7075C The water would be boiling if 152 grams of CaCl2xH20 dissolved into100g of water. We know this because water boils at 100C and we found a temperature of 130.7075C which is. more that 100C.
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