Question

The enthalpy of evaporation of liquid diethyl ether ${({C_2}{H_5})_2}O$ is $26kJ\;mo{l^{ - 1}}$ at it boiling point ${35^\circ }C$. Calculate $\Delta {S^\circ }$ for conversion of:
(a) Liquid to vapour
(b) Vapour to liquid at ${35^\circ }C$

Answer 1

As we know that the enthalpy of evaporation is the amount of energy that is added to a liquid substance which helps in vaporisation or conversion of that liquid substance into a gaseous state. And the entropy is the randomness of a system whereas the enthalpy of condensation is the amount of energy required to convert a gaseous substance into liquid state.

Complete solution: As we know that entropy of a system is defined as the randomness or disorderness of the system and it is given as the ratio of change in enthalpy to the temperature of its boiling point. We can show this ratio as:
$\Delta S = \dfrac{{\Delta H}}{{{T_{bp}}}}$, Where the change in enthalpy is the enthalpy of evaporation.
Now, we are given that the enthalpy of evaporation is $26kJ\;mo{l^{ - 1}}$ which should be first converted to joules because the change in entropy is always given in joules per kelvin temperature per mole, thus the temperature will be equivalent to the given temperature plus the standard temperature which is $273K$.
Hence, $\Delta {H_{vaporisation}} = 26 kJ\;mo{l^{ - 1}} = 26000J{K^{ - 1}}mo{l^{ - 1}}$
${T_{bp}} = 35 + 273 = 308K$
Thus putting both these values in the formula of change in entropy we will get:
$\Delta S = \dfrac{{26000}}{{308}} \\\ \Rightarrow 84.42J{K^{ - 1}}mo{l^{ - 1}} \\\$
Therefore, the change in entropy for conversion of liquid into vapour is $84.42J{K^{ - 1}}mo{l^{ - 1}}$.
Now we also know that the conversion of vapour phase into liquid state is called the process of condensation. So, the entropy of condensation is basically the opposite of the entropy of vaporization. We can show this as below:
$\Delta {S_{condensation}} = - \Delta {S_{vaporisation}}$
Therefore we can say that the entropy of condensation is equivalent to $\Delta {S_{condensation}} = - 84.42J{K^{ - 1}}mo{l^{ - 1}}$
Thus, the change in entropy for conversion of vapour into liquid at ${35^\circ }C$ is $- 84.42J{K^{ - 1}}mo{l^{ - 1}}$.

Note: Always remember that change in entropy takes place when one mole of liquid is converted to vapour at its boiling point and entropy of condensation takes place when one mole of vapour is converted back to its liquid state and entropy is a function of temperature and change in entropy is inversely related to temperature. Also entropy increases when temperature increases and the number of gaseous products increases.