(a) The standard free-energy difference between 2,2 -dimethylpropane and pentane is $6.86 \mathrm{~kJ} \mathrm{~mol}^{-1}$ (1.64 kcal mol $^{-1}$ ); 2,2 -dimethylpropane is the more stable compound. If the two were present in an equilibrium mixture, what would be the percentage of each in the mixture at $25^{\circ} \mathrm{C} ?$
(b) The energy difference between anti-butane and either one of the gauche-butane conformations is $2.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\left(0.67 \mathrm{kcal} \mathrm{mol}^{-1}\right)$ (Fig. $2.5, \mathrm{p} .53$ ). Treating this difference as a standard free energy, calculate the amounts of gauche-and anti-butane present in equilibrium in one mole of butane at $25^{\circ} \mathrm{C}$. (Remember that there are two gauche conformations.)