logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

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 \[\] $B \leftrightharpoons C$

$\begin {array} {1 1} (A)\;-12.3\: kJ & \quad (B)\;12.3\: kJ \\ (C)\;123.007\: J & \quad (D)\;-123007\: J \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
$ \Delta G_2^{\circ} = -2.303\: RT+log \large\frac{|C|}{|B|}$$ = -2.303 \times 8.314 \times 448 \times log \bigg (\large\frac{3.5}{95.2} \bigg)$$ = 12300.7\: J = 12.3\: kJ$
Ans : (B)
answered Mar 20, 2014 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...