Question

The enthalpy of vaporization of water is $386$ ${\text{KJ}}$. What is the change in entropy of water on boiling point?
A)0.5KJ
B) 1.02KJ
C)1.5KJ
D)22.05KJ

Answer 1

Hint: To find the entropy change using the formula for Gibbs free energy change-
$\Delta {\text{G = }}\Delta {\text{H - T}}\Delta {\text{S}}$ where $\Delta {\text{G}}$ is change in free energy, $\Delta {\text{S}}$ is entropy change, $\Delta {\text{H}}$ is enthalpy change and ${\text{T}}$ is the temperature in Kelvin. Put $\Delta {\text{G}}$$ = 0$ for boiling water and given values to get the answer.

Complete step by step answer:
Given, the enthalpy of vaporization of water $\Delta {\text{H}}$=$386$ ${\text{KJ}}$
We have to find the entropy change of boiling water $\Delta {\text{S}}$. We know that the formula for Gibbs free energy change is-$\Delta {\text{G = }}\Delta {\text{H - T}}\Delta {\text{S}}$ where $\Delta {\text{G}}$ is change in free energy, $\Delta {\text{S}}$ is entropy change, $\Delta {\text{H}}$ is enthalpy change and ${\text{T}}$is the temperature in Kelvin.
The liquid and vapour phase during evaporation of water is in equilibrium which makes free energy change $\Delta {\text{G}}$$ = 0$
Since the temperature of boiling water=${100^ \circ }{\text{C}}$ and we know that the standard temperature at Kelvin is $273{\text{K}}$ so to make the temperature of boiling water at Kelvin we will add the standard value to the temperature of boiling water.
$ \Rightarrow {\text{T = }}100 + 273 = 373{\text{ K}}$
Then, the formula becomes-
$ \Rightarrow 0 = \Delta {\text{H - T}}\Delta {\text{S}} \Rightarrow \Delta {\text{S}} = \dfrac{{\Delta {\text{H}}}}{{\text{T}}}$
On putting the values in the formula, we get-
$ \Rightarrow \Delta {\text{S}} = \dfrac{{386}}{{373}} = 1.02{\text{KJ}}$

Hence, the correct answer is ‘B’.

Note:
The direction of a chemical reaction is indicated by the sign of $\Delta G$ and it also tells –
If $\Delta G < 0$ then, the reaction is spontaneous in the direction it is written(backward or forward).This means that no external energy is needed for the reaction to occur.
If $\Delta G = 0$then, the system is at equilibrium. No change occurs in either forward or backward direction.
If$\Delta G > 0$ then the reaction is not spontaneous. This means the reaction needs input of free energy to make the reaction go forward instead of backward.