Use standard enthalpies of formation to calculate delta H for the following reac

Question

Use standard enthalpies of formation to calculate delta H for the following reaction 2NaHCO3 (s)rightarrow Na2O(5) + 2CO2(g) + H2O(g) Express your answer using three significant figures. ?Use standard enthalpies of formation to calculate delta H for the following reaction CO(g) + H2O(g) rightarrow H2(g) + CO2(g) Express your answer using three significant figures. Use standard enthalpies of formation to calculate delta H for the following reaction 3NO2(g) +H@O(l) rightarrow 2HNO3(aq) + NO(g) Express your answer using three significant figures. Use standard enthalpies of formation to calculate delta H for the following reaction C6H12O6(s) + 6O2(g) rightarrow 6CO2(g) + 6H2O(g) Express your answer using three significant figures.

Explanation / Answer

1)

2NaHCO3 --> Na2O + H2O + 2CO2
you need the heat of formation of NaHCO3, Na2O, H2O, CO2

NaHCO3 (s) ____ Delta Hf = ( - 226.5 ) ; Delta Gf = ( - 203.6 )
Na2O (s) ____ Delta Hf = ( - 99.4 ) ; Delta Gf = ( - 90.0 )
CO2 (g) ____ Delta Hf = ( - 94.05 ) ; Delta Gf = ( - 94.26 ) ** per Mole **
H2O (g) ____ Delta Hf = ( - 57.796 ) ; Delta Gf = ( 54.634 )
For Delta H
Enthalpy of Reaction = Sum [ Enthalpy of Formation of Product(s) ] - Sum [ Enthalpy of Formation of Reactant(s) ]

= (-99.4-(2*94.05)-(57.796)) - (-2*226.5) = 105.704

Part B

CO + H2O ? CO2 + H2

delta H = delta Hf(CO2) + delta Hf (H2) - [ delta Hf(CO) + delta Hf(H2O) ]

delta Hf for elements in their standard state is zero so delta Hf(H2) = 0

putting the values...

delta H = -393.5 + 0 - [ -110.5 + -241.8 ] = -393.5 + 352.3 = -41.2 kJ

Part C

3NO2(g) + H2O(l) --> 2HNO3(aq) + NO(g)

standard enthalpy of reaction for this reaction using these values:
NO2(g) = 34kJ/mol
H2O(l) = -286kJ/mol
HNO3(aq) = -207kJ/mol
NO(g) = 90.0kJ/mol

delta H = (sum of product enthalpies) - (sum of reactant enthalpies)

= (2 mol x -207 kJ/mol + 1 mol x 90 kJ/mol) - (3 mol x 34kJ/mol + 1 mol x -286 kJ/mol)

= (-414 + 90) - (102 - 286)

= -140 kJ
The unit for enthalpy is kJ/mol, but this reaction is produces 140 kJ per 3 moles of NO2... it would be enough to say -140 kJ for the reaction -- that's more than likely the answer they'll expect on a test. Besides, in the math we're multiply moles with kJ/mol, which should yield a kJ answer only.

part D

C6H12O6(s) + 6O2(g)  -->6CO2(g) + 6H2O(l)

dHf CO2(g) = ?393.52 kJ/mole
dHf H2O(l) = -285.8 kJ/mole
dHf glucose = -1250 kJ/mole

dH reaction = sum(dH products) - sum(dH reactants)
dH reaction =6*(?393.52 kJ/mole + -285.8 kJ/mole) - (-1250 kJ/mole) = 1396.92

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