Question

Butane exists in various conformations in nature like in staggard and in eclipse. At any given instant the probability that a given butane molecule is in anti, guache,eclipsed-1$\left(E_1\right)$ and in eclipsed-2$\left(E_2\right)$ conformation is $0.7,0.2,0.06$ and $0.04$ respectively
if the molar enthalpy of combustion of natural butane is $-690Kcal/mol$ at ${25}^{\circ }$C, Suppose the magnitude of enthalpy of combustion (kcal/mol) of butane is $691x$ if all the butane molecules are in gauche conformation. Then $x$ is:

Answer 1

The anti form is more stable than the gauche form by about 0.9 kcal/mol.
Hence, when the proportion of the gauche form increases from 0.2 to 1, the enthalpy of combustion of butane will be $690kcal/mol+0.8\times 0.9kcal/mol\simeq 691kcal/mol$.