Question

Acetic acid forms dimer in vapour phase. The dimer is held together by two hydrogen bonds with a total strength of $66.5{\text{ kJ}}$ per mole of dimer. If at ${25^o}{\text{C}}$ , the equilibrium constant of the dimerization is $1.3 \times {10^3}$ . $\Delta {{\text{S}}^o}$ for the reaction ${\text{2C}}{{\text{H}}_3}{\text{COOH}} \to {\left( {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right)_2}$ will be : $\left( {\log 1.3 = 0.1139} \right)$
A.$ - 0.163{\text{ kJ}}$
B.$ - 0.232{\text{ kJ}}$
C.$ - 0.342{\text{ kJ}}$
D.$ - 0.456{\text{ kJ}}$

Answer 1

Hint: We shall require knowledge about Gibbs free energy and its relation with the equilibrium constant and entropy (formula given). We shall first calculate the Gibbs free energy of this reaction using the equilibrium constant. Then, we shall use the Gibbs free energy to calculate entropy of the reaction.
Formula used:
$\Delta {\text{G = }} - 2.303{\text{RT}} \times {\text{logK}}$ where $\Delta {\text{G}}$ is the Gibbs free energy, R is a constant $\left( {8.314{\text{ J }}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}} \right)$ , T is the temperature and K is the equilibrium constant. (Eq. 1)
$\Delta {\text{G = }}\Delta {\text{H - T}}\Delta {\text{S}}$ where $\Delta {\text{H}}$ is the enthalpy change and $\Delta {\text{S}}$ is the entropy change (Eq. 2)

Complete step by step answer:
When a compound converts into a dimer, new bonds are formed between two molecules. The strength of those bonds depends upon the stability of the dimer. As the number of moles decrease, so there is also a slight decrease in the entropy of the system. We shall first calculate the Gibbs free energy of the system with the help of the equilibrium constant. We shall substitute the appropriate values in Eq. 1 with the appropriate units:
${\text{T = 2}}{{\text{5}}^o}{\text{C = 298K}}$ , ${\text{K = }}1.3 \times {10^3}$
\[ \Rightarrow \Delta {\text{G = }} - 2.303 \times 8.314 \times 298 \times {\text{log}}\left( {1.3 \times {{10}^3}} \right)\]
Solving this for $\Delta {\text{G}}$ , we get:
$\Delta {\text{G}} = - 17767.69{\text{ J = }} - 17.77{\text{ kJ}}$
Now, we shall use this value to calculate $\Delta {\text{S}}$ of the reaction:
$ \Rightarrow - 17.77{\text{ = }} - 66.5{\text{ }} - {\text{ }}\left( {{\text{298}} \times \Delta {\text{S}}} \right)$
Solving this for $\Delta {\text{S}}$ , we get:
$\Delta {\text{S}} = \dfrac{{ - 66.5 + 17.77}}{{298}}$
$ \Rightarrow \Delta {\text{S}} = - 0.163{\text{ kJ}}$

$\therefore $ The $\Delta {\text{S}}$ of the given reaction will be $ - 0.163{\text{ kJ}}$ , i.e. option A.

Note:
In the question as $\Delta {\text{H}}$ is given as the energy of the bonds, i.e. the bond dissociation energy so, in this case we shall use $\Delta {\text{H}}$ with a negative sign. This is because we are calculating the change in entropy for the bond formation of dimer and not the dissociation.