2. Using Hess’s Law and your results from parts I and II, calculate ÄH rxn, for

Question

2. Using Hess’s Law and your results from parts I and II, calculate ÄH rxn, for reaction CH3COOH(aq) ->CH3COO (aq) + H (aq.

3. Calculate the ÄH f of the OH ion given that the ÄH f (H2O) = -285.8 kJ mol^ -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 true ÄH rxn?

5. Use tabulated standard molar enthalpy changes of formation (from the textbook) to calculate ÄH rxn for the following (unbalanced) reactions:

a. C2H2(g) + O2(g) ->CO2(g) + H2O(g)

b. SO2 (g) + O2(g) ->SO3(g)

c. Fe2O3(s) + Al(s)(g) -> Fe(s) + Al2O3 (s)

6. The human body “burns” glucose (C6H12O6) for energy.

a. Write a balanced chemical reaction for the complete combustion of glucose to CO2(g) and H2O(l) such that the stoichiometric coefficient for glucose is 1.

b. When 1.00 g of glucose are completely burned at 298.15 K and 1 bar of pressure, 15.7 kJ are released. Calculate the ÄH f of glucose.

Explanation / Answer

3. H+ + OH- --> H2O            dH = -55.8 kJ/mol (from literature)

dH = dH(products) - dH(reactants)

-55.8 = (-285.8) - (0 + dH(OH-))

dH[OH-] = -230 kJ/mol

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4. For an endothermic reaction, If the calorimeter is not perfect zero.

qcal < 0. The measured dHrxn would be less endothermic than the true value.

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5. dHrxn

a. dHrxn = dH(products) - dH(reactant)

               = (-393.51 - 241.82) - (226.7) = -862.03 kJ/mol

b. dHrxn = (-395.7) - (-296.84)

               = -98.86 kJ/mol

c. dHrxn = (-1669.8) - (-824.2)

              = -845.6 kJ/mol

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6. Glucose

a. C6H12O6 + 6O2 --> 6CO2 + 6H2O

b. dH = 15.7 kJ x 180.156 g/mol/1 g

          = 2828.45 kJ/mol

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