Calculate the standard heat of reaction (Delta H compositefunction) in kJ/mol fo

Question

Calculate the standard heat of reaction (Delta H compositefunction) in kJ/mol for the combustion of methane gas (CH_4) to carbon dioxide and water vapor (H_2 O gas) using only the bond energies (BE) and the standard heats of formation (Delta H compositefunction _f) provided below: CH_4 (g) + 2O_2 (g) rightarrow CO-2 (g) + 2H_2 O(g) Delta H compositefunction = ? C(s) + O_2 (g) rightarrow CO_2 (g) Delta H compositefunction _f = - 385.8 kJ/mol C(s) + 2 H_2 (g) rightarrow CH_4 (g) Delta H compositefunction _f = - 74.8 kJ/mol BE_H-H = 432 kJ/mol BE_O=O = 495 kJ/mol BE_O-H = 467 kJ/mol

Explanation / Answer

We are given three reactions.

(1) C (g) + O2 (g) ------> CO2 (g); Hf0 = -385.8 kJ/mol.

Use the BEO=O to calculate BEC=O in CO2. O2 has 1 O=O bond and CO2 has 2 C=O bonds. Use Hess’s law to find out BEC=O.

1*(495 kJ/mol) = Hf0 + BEC=O

===> 495 kJ/mol = (-385.8 kJ/mol) + BEC=O

===> BEC=O = ½*(495 + 385.8) kJ/mol = 440.4 kJ/mol.

(2) C (s) + 2 H2 (g) -------> CH4 (g); Hf0 = -74.8 kJ/mol

Use the BEH-H to calculate BEC-H in CH4. H2 has 1H-H bond and we have 2 moles of H2. CH4 has four C-H bonds. Use Hess’s law.

2*(432 kJ/mol) = Hf0 + 4*BEC-H

===> 864 kJ/mol = -74.8 kJ/mol + 4*BEC-H

===> BEC-H = ¼*(864 + 74.8) kJ/mol = 234.7 kJ/mol.

Use the bond energies to calculate the enthalpy change for the reaction

CH4 (g) + 2 O2 (g) -------> CO2 (g) + 2 H2O (g)

The enthalpy of the reaction is

Hf0 = 4*BEC-H + 2*BEO=O – 2*BEC=O– 4*BEO-H = (4*234.7 kJ/mol) + 2*(495 kJ/mol) – 2*(440.4 kJ/mol) – 4*(467 kJ/mol) = -820 kJ/mol (ans).

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