THEROM Enthalpy of Nesrralizacion 185 Post-Laboratory Questions (uUse the spaces
ID: 968190 • Letter: T
Question
THEROM Enthalpy of Nesrralizacion 185 Post-Laboratory Questions (uUse the spaces procided for the ansuers and additional paper if necosary) 1. Obtain from your classmates or from your laboratory instructor, the molar enthalpies of neutralization of hydrochloric acid, sulfuric acid, and phosphoric acid when each reacts with sedium hydroxide (1) Write a chemical equation for each of these neutralization reactions (2) Compare the molar enthalpies of neutralization of hydrochloric acid, sulfuric acid, and phosphoric acid. Discuss the similarities and differences of these data 2. When a neutralization reaction was carried out using 100.0 mL of 07890M NH, water and 1000 ml of 0.7940M acetic acid,AT was found to be4% C The specific heat of the reaction mixture was 410 g'K- and its density was 1,03 g ml.. The calorimeter constant was 336J K (1) Calculate , utrn for the reaction of NH, and acetic acid. (2) At the end of the experiment, it was discovered that the thermometer had not been calibrated 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 Aentn calculated in (2) aboveExplanation / Answer
ANSWER
1.
1. HCl + NaOH --------> NaCl + H2O
H2SO4 + 2NaOH --------> Na2SO4 + 2H2O
H3PO4 + 3NaOH ---------> Na3PO4 + 3H2O
2. The enthalpy of neutralisation is measured as per the number of moles of water formed. When one mole of water is formed the enthalpy of neutralisation is - 57.7KJ. In case of HCl, H2SO4 and H3PO4 one two and three moles of water are formed respectivelly. Hence enthalpy of neutralisation will be -57.7KJ, 2 X -57.7KJ and 3 X -57.7KJ repectivelly.
2.
1. Mass of the solution = total volume X density = 200mL X 1.03g/mL = 206.0g
Heat released = Mass X specific heat capacity X change in temperature
Heat released = 206.0 X 4.104 X 4.76 =4024.2J = 40.24KJ
2. The uncalibrated thermometer reads 0.5C lower than actual, therefore the actual temperature is
4.76 - 0.5 = 4.26oC
Heat released = 206.0 X 4.104 X 4.26 = 3603.3J =36.03KJ
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