Butane exist in various conformations in nature like in staggard and in eclipsed form. At any given instant, the probability that a given butane molecule is in anti, gauche, eclipsed-1, ( E 1 ) and

Question

Butane exist in various conformations in nature like in staggard and in eclipsed form. At any given instant, the probability that a given butane molecule is in anti, gauche, eclipsed-1, $(E_{1})$ and in eclipsed-2 $(E_{2})$ conformation is $0.7, 0.2, 0.06$ and 0.04 respectively. If the molar enthalpy of combustion of natural butane is -690 kcal/mole at 25°C.

Calculate magnitude of enthalpy of combustion (kcal/mol) of butane if all the butane molecule are in gauche conformation.

TestHub · Accepted Answer

Energy Relation: Since the combustion products are identical, a higher potential energy ( $E$ ) corresponds to a more exothermic (higher magnitude) enthalpy of combustion:

$∣Δ H_{co mb} ∣ = ∣Δ H_{co mb, an t i} ∣ + Δ E$

Average Enthalpy: The total enthalpy is the weighted average of all conformations:

$∣Δ H_{co mb, a vg} ∣ = \sum P_{i} ∣Δ H_{co mb, i} ∣ = ∣Δ H_{co mb, an t i} ∣ + \sum P_{i} Δ E_{i}$

Relative Potential Energies: From the potential energy diagram (taking anti as $0$ ): $Δ E_{an t i} = 0$ , $Δ E_{g a u c h e} = 2 k c a l / m o l$ , $Δ E_{E_{2}} = 5 k c a l / m o l$ , and $Δ E_{E_{1}} = 5.5 k c a l / m o l$ .

Calculation: Substituting the values gives $690 = ∣Δ H_{co mb, an t i} ∣ + [0.7 (0) + 0.2 (2) + 0.06 (5.5) + 0.04 (5)]$ , which simplifies to $∣Δ H_{co mb, an t i} ∣ = 689.07 k c a l / m o l$ .

Final Answer: For the pure gauche conformation ( $Δ E = 2 k c a l / m o l$ ), the magnitude is:

$∣Δ H_{co mb, g a u c h e} ∣ = 689.07 + 2 \approx 691.00 k c a l / m o l$

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