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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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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$

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

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$1-Pentyne(A) \leftrightharpoons 2-Pentyne(B) \leftrightharpoons 1,2-Pentadiene(C)$
$\: \: \: \: \: 1.3 \% \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: 95,2 \% \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3.5 \%$
$ \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)
answered Mar 20, 2014 by thanvigandhi_1
 

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