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Finger [1]
2 years ago
8

Determine the enthalpy of the products of ideal combustion, i.e., no dissociation, resulting from the combustion of an isooctane

-air mixture
Chemistry
1 answer:
jeka942 years ago
8 0
The combustion of isooctane (C₈H₁₈) is written below:

2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (l)

The formula for heat of combustion is:

ΔHc = (∑Stoichiometric coefficient×ΔHf of products) - (∑Stoichiometric coefficient×ΔHf of reactants), where ΔHf is heat of formation.

ΔHf of isooctane = -259.2 kJ/mol
ΔHf of O₂ = 0 kJ/mol
ΔHf of CO₂ = -393.5 kJ/mol 
ΔHf of H₂O = <span>-285.8 kJ/mol
</span>
ΔHc = [(16 mol×-393.5 kJ/mol )+(18 mol×-285.8 kJ/mol)] - [(2 mol×-259.2 kJ/mol) + (25 mol*0 kJ/mol)]
ΔHc = -10,922 kJ
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Explanation:

Given -

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To Find -

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Now,

Let A be the organic compound.

Then,

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Now,

Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.

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And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).

Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.

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We know that 1° alcohol on oxidation gives aldehyde.

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Here 2 Carbon and 1° alcohol is used.

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Its cleared that Compound A is Ethanol.

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