Calculation of ÄH rxn for Part I 0 1. Assume the heat capacity of the final solu

Question

Calculation of ÄH rxn for Part I 0

1. Assume the heat capacity of the final solution is 4.184 J K-1 g-1. Using the final mass of the
solution in the calorimeter, calculate qcontents from equation (3.9). Calculate qsystem from
equation (3.8).

2. Calculate the moles of H2O (aq.) produced, nrxn. Be sure to identify the limiting reagent in
the reaction.

3. Calculate the standard molar enthalpy change, ÄH rxn, for reaction (3.1). 0

4. CHEM 1020: Calculate the average ÄH rxn from the two runs. 0

Calculation of ÄH rxn for Part II 0

1. Again, assume the heat capacity of the final solution is 4.184 J K g . Calculate qcontents and -1 -1
qsystem.

2. Calculate the number of moles of NaCH3COO(aq.) produced, nrxn. Again, identify the
limiting reagent.

3. Calculate the standard molar enthalpy change for reaction (3.2) for each of the two runs.

4. CHEM 1020: Calculate the average ÄH rxn from the two runs. 0

Questions:

1. Why should ÄH rxn for reaction (3.2) be less than ÄH rxn for reaction (3.1)?

2. Using Hess’s Law and your results from parts I and II, calculate ÄH rxn, for reaction (3.3). 0

3. Calculate the ÄH f of the OH ion given that the ÄH f (H2O) = -285.8 kJ mol . 0 - 0 -1

4. Imagine an endothermic reaction was studied (i.e. ÄH rxn > 0). If the calorimeter were not 0
perfect and qcalorimeter < 0, would the measured ÄH rxn be more or less endothermic than the 0
true ÄH rxn? 0

5. Use tabulated standard molar enthalpies of formation (from the textbook) to calculate ÄH rxn 0
for the following (unbalanced) reactions:
a. Al(s) + Fe2O3(s) ÿ Fe(s) + Al2O3(s)
b. PbO(s) + C(s) ÿ CO(g) + Pb(s)
c. CaCO3(s) ÿ CO2(g) + CaO(s)

6. The decomposition of baking soda (sodium bicarbonate) to sodium carbonate, water and
carbon dioxide is endothermic. If the decomposition of 10.0 g of sodium bicarbonate
requires 5.06 kJ, what is the ÄH f of sodium bicarbonate?

UOIT Chem. 1800U, SIS: Exp. 3-48 Results, Discussion and Questions NOTE: This is a very long report to prepare. Do not leave it to the last minute! Part I: NaOH+HC Run I mass of empty calorimeter /g mass of calorimeter + solution (end of run)/g230 mass of solution /g l2 112.30 meter + solution (end ofrun) / g Time /s Temp. / Time/s Temp. / 60 23 80 2.3 23 26 12 o S 2t 29 2% 2.3 2 8 Ao 26 2 8 28 2% 8 103

Explanation / Answer

Part I

1. q = 99.142 x 4.184 x 2 = 829.620 J

2. Moles of water = 99.142 / 18 = 5.51 mols

3.delta H = 829.62 J (I)

delta H = 834.16 J (II)

4. Average delta H = 831.89 J

Questions.

3. Missing delta Hrxn.

4. For an endothermic reaction with ÄH rxn > 0. If the calorimeter were not 0
perfect and qcalorimeter < 0, the measured ÄH rxn be less endothermic than the 0
true ÄH rxn.

5. Calculation of delta Hrxn = sum of delta Hf products - sum of delta Hf reactans

a. Al(s) + Fe2O3(s) ÿ Fe(s) + Al2O3(s)

delat Hrxn = (-1669.8) -(-826) = -843.8 kJ/mol

b. PbO(s) + C(s) ÿ CO(g) + Pb(s)

delta Hrxn = (-110.525) - (-217.9) = 107.38 kJ/mol

c. CaCO3(s) ÿ CO2(g) + CaO(s)

delta Hrxn = (-393.09 - 635.09) - (-1206.9) = 178.72 kJ/mol

6. 2NaHCO3 = Na2CO3 + H2O + CO2 delat Hrxn = 5.06 kJ

For 1 mole NaHCO3 decomposition = 5.06/2 = 2.53 kJ/mol

moles of NaHCO3 calculated for 10 g = 10/84.0066 = 0.119 mols

So delta H for 10 g NaHCO3 decomposition = 2.53 x 0.119 = 0.30107 kJ

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