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pav-90 [236]
2 years ago
7

g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole

of methane under these conditions
Chemistry
1 answer:
Anna [14]2 years ago
3 0

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

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Answer: Option (B) is the correct answer.

Explanation:

Degree of randomness of the molecules of a substance is known as entropy. More is the kinetic energy between the molecules of a substance more will be the degree of randomness.

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On the other hand, in solid state molecules are much more closer to each other as they arr held by strong intermolecular forces of attraction. Therefore, they have very less entropy.

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Since, 1 mole is producing 2 moles. This means that degree of randomness is increasing as both the molecules are present in gaseous form.

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3 years ago
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