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 $-690 \mathrm{~kcal} / \mathrm{mol}$ at $25^{\circ} \mathrm{C},$ then calculate the enthalpy of combustion of butane if all the butane molecules are in gauche conformation. The potential energy of gauche form is $2 \mathrm{~kcal} / \mathrm{mol}$ higher than anti form but it is $3 \mathrm{~kcal} / \mathrm{mol}$ less than eclipsed form and $5.5 \mathrm{~kcal} / \mathrm{mol}$ less than fully eclipsed form.

$-689 \mathrm{~kcal} / \mathrm{mol}$

$-692 \mathrm{~kcal} / \mathrm{mol}$

JEE CHEMISTRY MCQ: THERMODYNAMICS

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Analyzing +4 | -1

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 \mathrm{~kcal} / \mathrm{mol}

25^{\circ} \mathrm{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 \mathrm{~kcal} / \mathrm{mol}

higher than anti form but it is

3 \mathrm{~kcal} / \mathrm{mol}

less than eclipsed form and

5.5 \mathrm{~kcal} / \mathrm{mol}

less than fully eclipsed form.

View answer and explanation

Correct option: C

-691 \mathrm{~kcal} / \mathrm{mol}

Option C (-691 kcal/mol) is correct because it calculates the enthalpy of combustion for the gauche conformation based on the given probabilities and energy differences.

JEE CHEMISTRY MCQ: THERMODYNAMICS

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