recitation week 1 (rw1) chem 2510 dr. chang\n8. for the given cyclohexane chair:\ndraw a \two - circle\…

Explanation:

Step1: Calculate $\Delta G$

From the A - values, assume the relevant groups contribute to $\Delta G$. Here, if we consider the difference in A - values for the groups in the axial and equatorial positions as a measure of $\Delta G$ (simplified approach), and given the formula $\Delta G=- 2.303RT\log K_{eq}$. First, find $\Delta G$ from the A - values. Let's assume the difference in A - values gives $\Delta G$ in kcal/mol. So $\Delta G = 1.2$ kcal/mol. Convert $T = 25^{\circ}C=298K$ and $R = 0.001987$ kcal/(mol·K).

Step2: Solve for $K_{eq}$

We have $\Delta G=-2.303RT\log K_{eq}$. Rearranging for $\log K_{eq}$, we get $\log K_{eq}=-\frac{\Delta G}{2.303RT}$. Substitute $\Delta G = 1.2$ kcal/mol, $R = 0.001987$ kcal/(mol·K) and $T = 298K$ into the formula. [ \begin{align*} \log K_{eq}&=-\frac{1.2}{2.303\times0.001987\times298}\ &=-\frac{1.2}{2.303\times0.592126}\ &=-\frac{1.2}{1.36366}\ &\approx - 0.88 \end{align*} ] Then $K_{eq}=10^{- 0.88}\approx0.13$.

Step3: Calculate the distribution of axial to equatorial

Let the fraction of the equatorial form be $x$ and the fraction of the axial form be $1 - x$. We know that $K_{eq}=\frac{x}{1 - x}$. Solving for $x$: [ \begin{align*} 0.13&=\frac{x}{1 - x}\ 0.13(1 - x)&=x\ 0.13-0.13x&=x\ 0.13&=x + 0.13x\ 0.13&=1.13x\ x&=\frac{0.13}{1.13}\approx0.115 \end{align*} ] The fraction of the axial form is $1 - x=1 - 0.115 = 0.885$. The percentage of axial is $88.5%$ and the percentage of equatorial is $11.5%$.

Answer:

$K_{eq}\approx0.13$, Axial: $88.5%$, Equatorial: $11.5%$