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Question: Acetic acid \(C{H_3}COOH\) can form a dimer \({(C{H_3}COOH)_2}\) in the gas phase. The dimer is held...

Acetic acid CH3COOHC{H_3}COOH can form a dimer (CH3COOH)2{(C{H_3}COOH)_2} in the gas phase. The dimer is held together by two by two H-bonds with a total strength of 66.5kJ66.5kJ per mole of the dimer. If at 25∘C25^\circ C, the equilibrium constant for the dimerization is 1.3×1031.3 \times {10^3}. Calculate ΔS∘\Delta S^\circ for the reaction:
2CH3COOH(g)⇌(CH3COOH)2(g)2C{H_3}COOH(g) \rightleftharpoons {(C{H_3}COOH)_2}(g)

Explanation

Solution

In this question, to get value of (ΔS∘\Delta S^\circ ) first we need to find the value of Gibbs free energy(ΔG∘\Delta G^\circ) and then to put the value of Gibbs free energy in the standard equation for Gibbs free energy that is the difference of enthalpy change and product between entropy change and temperature.

Complete step by step answer:
In the question, we are given an equilibrium constant so we can find Gibbs free energy (ΔG∘\Delta G^\circ ) from it. It is the free energy associated with a chemical reaction that can be used to do work. It has following relation with equilibrium constant:
ΔG∘=−2.303RT×log⁡K\Delta G^\circ = - 2.303RT \times \log K
Where, R=8.314JK−1mol−1R = 8.314J{K^{ - 1}}mo{l^{ - 1}} (Gas constant)
T=25+273=298KT = 25 + 273 = 298K (here we change temperature units from celsius degree to kelvin)
K=1.3×103=K = 1.3 \times {10^3} = equilibrium constant (given)
Now, ΔG∘=−2.303×8.314×298×log⁡(1.3×103) ΔG∘=−17.767kJ  \Delta G^\circ = - 2.303 \times 8.314 \times 298 \times \log (1.3 \times {10^3}) \\\ \Delta G^\circ = - 17.767kJ \\\
Now, to find entropy change (ΔS∘\Delta S^\circ ) which is the measure of randomness or uncertainty, we will use the Gibbs free energy equation which can be given as :
ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ −(1) - (1)
Where, ΔH∘=\Delta H^\circ = Enthalpy change == 66.5kJ66.5kJ(It is the amount of heat evolved or absorbed in a chemical reaction at constant pressure)
T=298K=T = 298K = Temperature
ΔS∘=\Delta S^\circ = entropy change
Now, by putting all values in equation −(1) - (1) we get:
\-17.767=−66.5−298×ΔS∘ ΔS∘=−66.5−17.767298 ΔS∘=−0.163kJ  \- 17.767 = - 66.5 - 298 \times \Delta S^\circ \\\ \Delta S^\circ = \dfrac{{ - 66.5 - 17.767}}{{298}} \\\ \Delta S^\circ = - 0.163kJ \\\
Hence, the required value of entropy (ΔS∘\Delta S^\circ ) is −0.163kJ - 0.163kJ.

Note:
In this question, by the negative entropy we mean that the randomness or disorder is decreasing and if the entropy has positive value then it means that the randomness or disorder is increasing.