ThermochemistryHard

Question

Butane exists in various conformations in nature. At any given instant, the probability that a given butane molecule is in anti, gauche, eclipsed and fully eclipsed conformation is 0.7, 0.2, 0.06 and 0.04, respectively. If the molar enthalpy of combustion of natural butane is −690 kcal/mol at 25o C, then calculate the enthalpy of combustion of butane if all the butane molecules are in gauche conformation. The potential energy of gauche form is 2 kcal/mol higher than anti form but it is 3 kcal/mol less than eclipsed form and 5.5 kcal/mol less than fully eclipsed form.

Options

A.−690 kcal/mol
B.−689 kcal/mol
C.−691 kcal/mol
D.−692 kcal/mol

Solution

Let the enthalpy of combustion of gauche form be –x kcal/mol.

Now, $690 = 0.7 \times (x - 2) + 0.2 \times x + 0.06 \times (x + 3) + 0.04 \times (x \times 5.5) $$$\therefore x = 691$$

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