Final answer:
To calculate the equilibrium percentages of n-butane conformers, the energy difference given can be used in the Boltzmann distribution equation. At room temperature, the gauche conformation makes up approximately 43.6%, while the anti conformation makes up about 56.4%.
Step-by-step explanation:
The energy difference between the gauche and anti conformations of n-butane is provided as 0.9 kcal/mol. To find the percentages of each of these conformers at equilibrium, we must use this energy difference in the Boltzmann distribution equation. This equation gives the probability (P) of a particular energy state relative to another energy state using the formula:
P = e-(ΔE / RT)
Where ΔE is the energy difference between the states, R is the gas constant (which has a value of 8.314 J/mol·K or 1.987 cal/mol·K), T is the temperature in Kelvin, and e is the base of the natural logarithm. Assuming room temperature of 298K, the energy difference of 0.9 kcal/mol is equivalent to 3768 J/mol. The probability that a molecule is in the gauche conformation relative to the anti conformation is:
Pgauche = e-(3768 J/mol / (8.314 J/mol·K * 298 K)) ≈ e-1.52 ≈ 0.218
This means that the gauche conformation is present at about 21.8% relative to the anti conformation. Assuming two gauche conformations for butane, the total percentage for the gauche conformer would be roughly 43.6%, and the remaining percentage would be the anti conformation at about 56.4%.
This calculation assumes that no other conformations are significantly populated at room temperature, which is a reasonable approximation.