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Calculate difference, ΔH-ΔE=Δ(PV) for the combustion reaction of 1 mole of heptane. (Assume standard state conditions and 298 K for all reactants and products.) The ideal gas law must be used. The incremental change in volume due to liquids is neglected (cf the volume of 1 mole of H2O(l) is 18 mL, whereas the volume of 1 mol of a gas at 298 K is about 24,450 mL. Be careful to account for the total change in number of moles of gas molecules (be sure the signs and units are correct too). Note that certain alkanes are gases in the standard state.

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Answer:

ΔPV = -9911 J

Explanation:

The combustion of 1 mole of heptane is:

C₇H₁₆(l) + 11O₂(g) → 7CO₂(g) + 8H₂O(l)

The change in number of moles of gas molecules is:

Δn = 7 moles products - 11 moles reactants = -4 moles

Using ideal gas ΔPV is:

ΔPV = ΔnRT

Where:

Δn is -4mol

R is 8,314472 J/molK

T is 298K

Replacing:

ΔPV = -9911 J

I hope it helps!

The difference for the combustion reaction of 1 mole of heptane is

ΔPV = -9911 J

The combustion of 1 mole of heptane is:

Chemical reaction:

C₇H₁₆(l) + 11O₂(g) → 7CO₂(g) + 8H₂O(l)

The change in number of moles of gas molecules is:

Δn = 7 moles products - 11 moles reactants = -4 moles

Ideal gas Law:

The pressure of a gas times its volume equals the number of moles of the gas times a constant (R) times the temperature of the gas. It is given by:

[tex]\triangle PV = \triangle n RT[/tex]

where,

Δn = -4mol

R = 8.314472 J/molK

T = 298K

On substituting the values in the above formula:

[tex]\triangle PV = \triangle n RT\\\\\triangle PV = -4*8.314*298\\\\\triangle PV =-9911J[/tex]

Thus, the difference for the combustion reaction of 1 mole of heptane is

ΔPV = -9911 J.

Find more information about Ideal gas here:

brainly.com/question/25290815

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