28S Post-Laboratory Questions use the spaces procided for the ansovers and addit
ID: 968344 • Letter: 2
Question
28S Post-Laboratory Questions use the spaces procided for the ansovers and additional paper if necessary 1. Obtain from your classmates or from your laboratory instructor instructor, the molar enthalpies of of hydrochloric acid, sulfuric acid, and phosphoric acid when each reacts with sodium hydroxide. () Write a chemical equation for each of these neutralization reactions of Hel + NaOH NaCl + H2O (2) Compare the molar enthalpies of neutralization of hydrochloric acid, sulfuric acid, and phosphoric acid. Discuss the similarities and differences of these data. Thw enthaiPs of neutro Lieation is measured as Per the amel Hasou one, two,cunel theee me les cf uater are formel Hence, enthaiP of neutalitation will be 2. When a neutralization reaction was carried omel HrSau one, two, tes fechvelt and reaction was carried out using 100.0 mL of 0.7890M NiH, water and 1000 ml of ON40M acetic acid. AT was found to be4%C. The speific heat of the reaction mixture was 4104 g , K , and its density was 1.03 g mL , Thecalonmeter constant was 336 l (D Calculate AHeuten for the reaction of NH, and acetic acid Heat releasels masssPec.fic luat atac.td x cha.. sPee.fic heat afac.t. Iump Heat releasedfnass 40.2 kJ 46 24 k3 answer the end of the experiment, it was discovered that the thermometer had not been calibrated. At When it was calibrated, it was found that the thermometer read 0.50 'C low. What effect would this thermometer reading have on the reported Afiton Calculated in (2) above? The uncali bratel thurmandlay reodso actual, therefor·fu ached temperature is 4-76-o.5-4.26 Heetrelees ed : 206.014-1a44.26 ! 3663.3 J 6.63 KExplanation / Answer
3) considering delta T = 4.70 oC + 273 = 277.7 K
heat released q = m. c. delta T
= 206.0 g x 4.104 J.g-1.K-1 x 277.7 K
= 234774.24 J
= 234.77 KJ
4) percent error = (( Experimental value - Actual value ) / Actual value) x 100
= ( ( 234.83 - 234.77) / 234.77) x 100
= 0.025 %
3. 1) Glass is not a good insulator of heat as a styrofoam cup would be . Glass has a lower specicific heat capacity than the styrofoam which means it takes less energy for glass to accumulate or lose heat than it does styrofoam . Therefore, a glass calorimeter will have a lower calorimeter constant due to heat escaping.
2.) Delta H would be lower in value in case of glass ,I think .
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