Question

When $1$ -pentyne $(A)$ is treated with $4\, N$ 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} =-R T \ln K_{1} $
$=-8.314 \times 448 \times 2.303 \log \frac{13}{952}=16 kJ $
$\Delta G_{2}^{\circ} =-R T \ln K_{2} $
$=-8.314 \times 448 \times 2.303 \log \frac{35}{952} $
$=12.3 \,kJ $