Question

The ΔS for the vaporisation of 1 mol of water is 88.3J/molK. The value of ΔS for the condensation of 1 mol of vapour will be:
A. 88.3 J/molK
B. \[{\left( {88.3} \right)^2}\] J/molK
C. -88.3 J/molK
D.\[\dfrac{1}{{88.3}}\] J/molK

Answer 1

Hint:Recall the mathematical relation between change in entropy (\[\Delta S\] ) and change in enthalpy (\[\Delta H\] ). Keep in mind that condensation is the reverse process of vaporisation. Use both these concepts together to answer this question.

Formula used:
\[\Delta S = \dfrac{{\Delta H}}{T}\] … (1)
\[\Delta S\] = entropy change (Units: \[Jmo{l^{ - 1}}{K^{ - 1}}\] )
\[\Delta H\] = change in enthalpy (Units: \[Jmo{l^{ - 1}}\] )
 T = temperature (Units: K)

Complete step-by-step answer:
The enthalpy of a system (H) is a thermodynamic parameter that is, in essence, a measure of the energy content of the system. During a chemical reaction, as reactants convert into products, the enthalpy associated with the reaction changes. This change in the enthalpy (\[\Delta H\]) is given by the difference in the enthalpies of the products and the reactants.
\[\Delta H = {H_{products}} - {H_{reactants}}\]
For the vaporisation of water, the change in enthalpy is the difference between enthalpies of water in its liquid state and vapour state.
\[{H_2}O(l) \to {H_2}O(g)\]
\[\Delta H = {H_{{H_2}O(l)}} - {H_{{H_2}O(g)}} = {\Delta _{vap}}H\] … (2)
We know that condensation is the process in which a substance gets converted from its gaseous/vapour state to its liquid state. Thus, we can say that condensation and vaporisation are reverse processes. In the case of water, the condensation process is represented as
\[{H_2}O(g) \to {H_2}O(l)\]
and the change in enthalpy is \[\Delta H' = {H_{{H_2}O(g)}} - {H_{{H_2}O(l)}} = {\Delta _{condensation}}H\] … (3)
Comparing equations (2) and (3), we can conclude that \[{\Delta _{condensation}}H = - {\Delta _{vap}}H\] … (4)
For the vaporisation process, \[{\Delta _{vap}}S = \dfrac{{{\Delta _{vap}}H}}{T}\] [from equation (1)] … (5)
Similarly for the condensation process, \[{\Delta _{condensation}}S = \dfrac{{{\Delta _{condensation}}H}}{T}\] … (6)
\[ \Rightarrow {\Delta _{condensation}}S = \dfrac{{ - {\Delta _{vap}}H}}{T}\]
\[ \Rightarrow {\Delta _{condensation}}S = - {\Delta _{vap}}S\] [using equation (5)]
\[ \Rightarrow {\Delta _{condensation}}S = - 88.3J/molK\]

Thus, option C is correct.

Note:Notice that in equations (5) and (6), T remains the same for both condensation and vaporisation processes. This is because both condensation and vaporisation are surface phenomena i.e., they occur at the surface of the water where the liquid water interfaces with the air around it. At the surface, some water molecules go from the liquid state into the vapour state (vaporisation) while others go into the liquid state from the vapour state (condensation) creating an equilibrium between the two states \[{H_2}O(l) \rightleftharpoons {H_2}O(g)\].