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Sergeeva-Olga [200]
3 years ago
11

The enthalpies of formation of the compounds in the combustion of methane,

Chemistry
2 answers:
Naya [18.7K]3 years ago
5 0

Answer:

The amount of heat released by the combustion of 2 moles of methane = 1605.1 kJ

Option 3 is correct.

ΔH°(combustion of 2 moles of methane)

= -1,605.1 kJ

Explanation:

The balanced equation for the combustion of methane is presented as

CH₄ + 2O₂ → CO₂ + 2H₂O

The heat of formation for the reactants and products are

CH₄ (g): ΔHf = –74.6 kJ/mol;

CO₂ (g): ΔHf = –393.5 kJ/mol;

and H₂O(g): ΔHf = –241.82 kJ/mol.

ΔHf for O₂ = 0 kJ/mol

ΔH°rxn for the combustion of methane is given as

ΔH°rxn = ΣnH°(products) - ΣnH°(reactants)

ΣnH°(products) = (1×-393.5) + (2×-241.82)

= -877.14 kJ/mol

ΣnH°(reactants) = (1×-74.6) + (2×0)

= -74.6 kJ/mol

ΔH°rxn = -877.14 - (-74.6) = -802.54 kJ/mol

For 2 moles of methane, the heat of combustion = 2 moles × -802.54 kJ/mol

= -1,605.08 kJ = -1,605.1 kJ

Hope this Helps!!!

inn [45]3 years ago
4 0

Answer:

C, -1,605.1 kJ

Explanation:

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Answer:

1-Pentene

Explanation:

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Answer:

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Explanation:

Let's consider the following thermochemical equation.

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Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.

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