Use tabulated heats of formation to determine the standard heats of the followin

Question

Use tabulated heats of formation to determine the standard heats of the following reactions in kJ, letting the stoichiometric coefficent of the first reactant in each reaction equal one. Check stoichlometry. Ntrogen (N2) and oxygen (02) react to form nitrogen tetraokide the tolerance is +/-2% By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Oxidation of Alkanes Gaseous isopentane +oxygen react to form carbon monoxide+ liquid the tolerance is +/-2% By accessing this Question Assistance, you will learm while you earn points based on the Point Potential Policy set by your instructor Oxidation of n-Alkanes Liquid n-hexane + oxygen react to to form carbon dioxide + water vapor the tolerance is +/-2% By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Reduction of Sodium Sulfate Liquid sodium suifate reacts with carbon (solid) to form liquid sodium sulfide and carbon dioxide (g)- the tolerance is +/-2% By accessing this Question Assistance, you will learn while you earn points base

Explanation / Answer

Ans

Part a - oxidation of nitrogen

The balanced reaction

N2(g) + 2O2(g) = N2O4(g)

Heat of reaction = sum of heat of formation of products - sum of heat of formation of reactants

H = Hf(N2O4) - 2*Hf(O2) - Hf(N2)

= 10 - 2*0 - 0

= 10 kJ

Part b - oxidation of alkanes

The balanced reaction

C5H12(g) + 8O2(g) = 5CO(g) + 6H2O (l)

Heat of reaction = sum of heat of formation of products - sum of heat of formation of reactants

H = 6*Hf(H2O) + 5*Hf(CO) - 8*Hf(O2) - Hf(C5H12)

= 6*(-285.8) + 5*(-110.5) - 8*0 - (-153.7)

= - 2113.6 kJ

Part c -

oxidation of n-alkanes

The balanced reaction

C6H14(l) + (19/2)O2(g) = 6CO2(g) + 7H2O (g)

Heat of reaction = sum of heat of formation of products - sum of heat of formation of reactants

H = 7*Hf(H2O) + 6*Hf(CO2) - (19/2)*Hf(O2) - Hf(C6H14)

= 7*(-241.8) + 6*(-393.5) - (19*0/2) - (-198.7)

= - 3854.9 kJ

Part d - reduction of sodium sulfate

The balanced reaction

Na2SO4(l) + 2C(s) = Na2S(l) + 2CO2(g)

Heat of reaction = sum of heat of formation of products - sum of heat of formation of reactants

H = Hf(Na2S) + 2*Hf(CO2) - (2)*Hf(C) - Hf(Na2SO4)

= (-323.94) + 2*(-393.5) - 2*(0) - (-1356.38)

= 245.44 kJ

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