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Thermodynamics

# 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, $95.2 \%\: 2-$ pentyne(B) and $3.5 \%$ of $1,2-$ pentadiene(C). The equilibrium was maintained at $175^{\circ}C$.  Calculate $\Delta G^{\circ}$ for the equilibrium  $B \leftrightharpoons A$

$\begin {array} {1 1} (A)\;-11.45\: kJ & \quad (B)\; -15.997\: kJ \\ (C)\;15.997\: kJ & \quad (D)\;11.45 \: kJ \end {array}$

$1-Pentyne(A) \leftrightharpoons 2-Pentyne(B) \leftrightharpoons 1,2-Pentadiene(C)$
$\: \: \: \: \: 1.3 \% \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: 95,2 \% \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3.5 \%$
$\Delta G_1^{\circ} = -2.303RT\: log \large\frac{|A|}{|B|}$$= -2.303 \times 8.314 \times 448 \times log \bigg (\large\frac{1.3}{95.2} \bigg)$$= 15997.8\: J = 15.997 \: kJ$
Ans : (C)