Question

When $1$ -pentyne $(A)$ is treated with $4N$ alcoholic $KOH$ at $175^{\circ }C$, it is converted slowly into an equilibrium mixture of $1.3\%$ 1-pentyne $(A),95.2\%2$-pentyne $(B)$ and $3.5\%$ of 1,2 -pentadiene $(C)$. The equilibrium was maintained at $175^{\circ }C$. Calculate $\Delta G^{\circ }$ for the following equilibria.
$B\rightleftharpoons A,\Delta G_1^{\circ }=?$
$B\rightleftharpoons C,\Delta G_2^{\circ }=?$
From the calculated value of $\Delta G^{\circ }$, and $\Delta G_2^{\circ }$ indicate the order of stability of $(A),(B)$ and $(C)$. Write a reasonable reaction mechanism showing all intermediates leading to $(A),(B)$ and $(C)$

Answer 1

At equilibrium : $\underset{95.2\%}{B}\rightleftharpoons \underset{1.3\%}{A}$
$\Rightarrow K_1=\frac{13}{952}$
$\underset{95.2\%}{B}\rightleftharpoons \underset{3.5\%}{C}$
$\Rightarrow K_2=\frac{35}{952}$
$\Rightarrow \Delta G_1^{\circ }=-RT\ln K_1$
$=-8.314\times 448\times 2.303\log \frac{13}{952}=16kJ$
$\Delta G_2^{\circ }=-RT\ln K_2$
$=-8.314\times 448\times 2.303\log \frac{35}{952}$
$=12.3kJ$