Here is my data and results referenced in question 1: The temperature change was
ID: 509016 • Letter: H
Question
Here is my data and results referenced in question 1:The temperature change was recorded for two trials of the reactions between HCl+NaOH and NaHSO4+NaOH each. Each solution used was 2.0M and had 25mL of the acid and 26mL of the base per reaction. The average temperature change was calculated between the two trials.
I am having a really difficult time calculating the enthalpy of each reaction and the heat of dissociation. 1. Data analysis: Show calculations of the enthalpy of each reaction you did and your determination of the heat of dissociation using Hess's law. 2. Summary of Experimental results: Present your results in a clear and concise way. You should report the experimentally determined values of the following thermodynamic quantities: a. AT b. q AH(2). 3. Discussion Questions: a. What are the theoretical values of AH(1), AH(2), and AH(3) calculated using enthalpies of formation? You can find the enthalpies of different formations in the Appendix B of this manual. b. How are your experimentally determined enthalpies compared against the theoretical values you calculated? Calculate the percent differences. c. If the final temperature of a measurement is mistakenly determined to be lower than the correct value, will the enthalpy you calculated be higher or lower than the correct one? Explain
Explanation / Answer
total volume of solution = 25ml+26ml= 51ml, assuming the density of water as 1g/ml, mass of water = density* Volume= 1*51= 51 gm
enthalpy change= mass*specific heat* temperature difference
for trial-1
Assuming specific heat = 4.184 J/g.deg.c ( same as that of water) , enthalpy change= 51*4.184*12.9=2752.654Joules
the reaction between acid and base is HCl + NaOH------>NaCl+ H2O
moles of acid = Molarity* Volume(L)= 2*25/1000 =0.05, moles of base= 2*26/1000=0.052
all the HCl gets consumed. molar enthalpy = 2752.654/0.05 joules/mole =55053 J/mole= 55.053 Kj/mole
Trial-2 : molar enthalpy = 51*1*4.184*17.3/0.05=73831J/mole= 73.831Kj/mole
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