SA 3 - scientific argument on identifying enthalpy of reaction using calorimetry experiment PDF

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scientific argument on identifying enthalpy of reaction using calorimetry experiment...

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Experiment 3 Determination of enthalpy for a reaction using calorimetry Experiment performed on 3/29/2021 Chemistry 1215L section 12

1. Question: What is the enthalpy of dissociation of HSO4-?

2.Claim: From the low variation of my standard deviations of the 2 data sets(reaction between NaOH and NaHSO4 and between NaOH and HCl) 0.14 and 0.42 respectively and the respective means for these two reactions 12.1 and 12.4, I can say that my values/data has low variation of difference and are close to each other, thus my precision is high. Speaking of accuracy, the calculated %error is 106% which means that my accuracy is very low .

3.Evidence: Balanced equation of NaOH and NaHSO4(equation 1): NaOH(aq) + NaHSO4(aq) = H2O(l) + Na2SO4(aq) Net ionic equation: OH- + H+ = H2O

When NaOH and NaHSO4 react, two things happen to form the products H+ is removed from HSO4- and H2O is formed 1. HSO4-(aq) + OH-(aq) = H+(aq) + OH-(aq) + SO42-(aq) 2. H+(aq) + OH-(aq) + SO42-(aq) = H2O(l) + SO42-(aq) The difference between these two reactions is that H+ is removed from HSO4 Balanced equation of NaOH and HCl(equation 2): NaOH(aq) + HCl(aq) = H2O(l) + NaCl(aq) Net ionic equation: H+ + OH- = H2O

Intermediate equation 2(delta H2): HSO4-(aq) = H+(aq) + SO42-(aq) (reaction we are trying to measure using Hess law)

First I collected my data(values of temperature) and calculated delta T for each trial of each equation(2 trials for reactions between NaOH and HCl as well NaOH and NaHSO4), then I calculated the heat/energy of the calorimeter by multiplying the mass of the solution (NaOH and NaHSO4 for the first reaction as well as the second reaction between NaOH and HCl) by the specific heat capacity by average Delta T And after that I got the value of the heat of each reaction by reversing the sign of the value of

the heat of the calorimeter. After that, I obtained Delta H for both reactions by dividing the heat of the reaction by the number of moles of the limiting reactant (NaHSO4 and HCl respectively since they have less volume in both reactions than NaOH) 20ml vs 21ml. Then I applied Hess law to obtain Delta H of reaction 2 (the intermediate reaction) using the following formula. Delta H1 = Delta H2 + Delta H3 Finally, I found the enthalpy of the dissociation of HSO4-. Using the calculated values that I have Experiment 1: NaOH and NaHSO4, Experiment 2: NaOH and HCl

Delta T of trial 1 of experi ment 1 12

Delta T of trial 2 experime nt 1 12.2

Mean(avera ge) Of experiment 1 12.1

Standar d deviati on

Delta T of trial 1 experimen t2

Delta T of trial 2 experim ent 2

0.14

12.1

12.7

Formula of mean: Mean = sum of terms/number of terms Formula of standard deviation

Values of Delta T of reaction 1 (NaOH and NaHSO4): Trial 1, Delta T = 12(Tfinal – Tinitial = 32.3 – 20.3) Trial 2, Delta T = 12.2(Tfinal – Tinitial = 32.4 – 20.2)

Mean(a SD veravera ge of experim ent 1) 12.4 0.42

Mean (average) = (12 + 12.2) / 2 = 12.1 Standard deviation: 0.14(high precision)

Values of Delta T of reaction 2 (NaOH and HCl): Trial 1, Delta T = 12.1( Tfinal – Tinitial = 31.3 – 19.2) Trial 2, Delta T = 12.7(Tfinal – Tinitial = 32 – 19.3) Mean (average) = (12.1 + 12.7) / 2 = 12.4 Standard deviation: 0.42(high precision) Percent error: absolute value of :experimental value – true value/true value*100

Referring to the table in Appendix B, the true value of is -21.76 KJ/mol. My experimental value is 1.28 KJ/mol % Error = (1.28+21.76) / 21.76) * 100 %Error = 106%

4.Justification: The low value of standard deviation that I got from both reaction was low which shows that my data is precise. Finding Delta H2 using Hess law then finding the enthalpy of dissociation of HSO4- accordingly. To check if my data and experiment succeeded I found the %error. The %error is 106% which shows that my accuracy is bad, leading me to say the experiment did not succeeded...