Question

Calculate the enthalpy of vaporization per mole for ethanol. Given, $\Delta S = 109.8J{K^{ - 1}}mo{l^{ - 1}}$ and boiling point of ethanol is ${78.5^o}C$ .

Answer 1

Hint: The entropy associated with this process increases as it involves the change of the state of ethanol from a liquid to a gas. Therefore, the entropy will be positive. The enthalpy however will also be positive as a liquid requires external heating to change its state.

Formula used: $\Delta {S_{vapourisation}} = \dfrac{{H_{vapourisation}}}{{\Delta {T_{boiling\; point} }}}$
$\Delta S$ denotes the change in entropy of ethanol during vaporization, $H$ for enthalpy of the ethanol for vaporization per mole, $T$ is the boiling point of ethanol.

Complete step by step answer:
Enthalpy of ethanol can be defined as the amount of heat that is involved in the process where ethanol is converted from liquid to gas. It is required to be found in this reaction
The change in entropy is the increase in disorderliness of the state of ethanol as it goes from one state to another.
Finally using the temperature at which ethanol changes from liquid to gas, we can find the enthalpy of the reaction using the above reaction. The step by step derivation of the answer is given below.
$\Delta {S_{vapourisation}} = \dfrac{{H_{vapourisation}}}{{\Delta {T_{boiling\; point} }}}$
Here we know that, $\Delta {S_{vapourisation}} = 109.8J{K^{ - 1}}mo{l^{ - 1}}$and the boiling point is ${78.5^o}C$ .
It is required to convert ${78.5^o}C$ into the kelvin scale. This can be done by following the steps below.
$T = {78.5^o}C$
To convert into the kelvin scale we have to carry out the step shown below,
$T = 273 + 78.5$
$ \Rightarrow 351.5K$
$\therefore T = 351.5K$
Now, we by plugging the values in the above equation, we get,
$109.8 = \dfrac{{{H_{vaopourisation}}}}{{351.5}}$
Shifting the denominator to the numerator we get,
$109.8 \times 351.5 = H$
$38594.7 = H$

Therefore, we can conclude that the enthalpy of vaporization of the process is \[38594.7J\] which can also be written as \[38.6KJ\] .

Note: This \[38.6KJ\] value which is the enthalpy of vaporization denotes that this amount of energy is involved with the process of vaporization of ethanol.
Remember that for vaporization reactions, Boiling point is considered as the temperature in the above formula that is used. For fusion reactions, the temperature considered is the melting point of the reaction.
Also keep in mind that the temperature should always be in Kelvin scale. And for conversion from Celsius to kelvin the temperature given must be added with $273K$.