Thermodynamics and Thermochemistry - Result Question 69
69. 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 $(A), 95.2 % \quad 2$-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.
$ \begin{array}{lll} B \rightleftharpoons A, \Delta G^{\circ}{ } _1=\text { ? } \\ B \rightleftharpoons C, \Delta G^{\circ}{ } _2=\text { ? } \end{array} $
From the calculated value of $\Delta G^{\circ}{ } _1$ and $\Delta G^{\circ}{ } _2$ indicate the order of stability of $(A),(B)$ and $(C)$. Write a reasonable reaction mechanism showing all intermediates leading to $(A),(B)$ and $(C)$.
(2001, 10M)
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Answer:
Correct Answer: 69. ($12.3$ kJ)
Solution:
- At equilibrium :
$\Rightarrow \quad \Delta G_1^{\circ} =-R T \ln K_1 $
$ =-8.314 \times 448 \times 2.303 \log \frac{13}{952}=16 \mathrm{~kJ} $
$\Delta G_2^{\circ} =-R T \ln K_2 $
$ =-8.314 \times 448 \times 2.303 \log \frac{35}{952} $
$ =12.3 \mathrm{~kJ}$
BETA
