Question

Acetic acid forms dimer in vapour phase. The dimer is held together by two hydrogen bonds with a total strength of $66.5$ kJ per mole of dimer. If at ${25}^o$C, the equilibrium constant for the dimerization is $1.3\times {10}^3$. $\Delta S^o$ for the reaction $2C{H}_{3}COOH\rightleftharpoons (C{H}_{3}COOH{)}_2$ will be ($log1.3=0.1139$) :

• A. $-0.163$ kJ
• B. $-0.232$ kJ
• C. $-0.342$ kJ
• D. $-0.456$ kJ

Answer 1

The correct option is A $-0.163$ kJ

$2C{H}_{3}COOH\left(g\right)\to \left(C{H}_{3}COOH{)}_2\right(g)$

$\Delta G^o=-2.303RTlogK$

$\Delta G^o=-2.303\times 8.314\times 298\times log(1.3\times {10}^3)=-17767.69J=-17.77kJ$

Again,

$\Delta G^o=\Delta H^o-T\Delta S^o$

$-17.77=-66.5-298\times \Delta S^o$

$\Delta S^o=\frac{(-66.5+17.77)}{298}$

$\Delta S^o=\frac{-48.73}{298}=-0.163kJ$

Therefore, $\Delta S^o=-0.163kJ$