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Question: The enthalpy of vaporization of water is \(386\) \({\text{KJ}}\). What is the change in entropy of w...

The enthalpy of vaporization of water is 386386 KJ{\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

Explanation

Solution

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

Complete step by step answer:
Given, the enthalpy of vaporization of water ΔH\Delta {\text{H}}=386386 KJ{\text{KJ}}
We have to find the entropy change of boiling water ΔS\Delta {\text{S}}. We know that the formula for Gibbs free energy change is-ΔG = ΔH - TΔS\Delta {\text{G = }}\Delta {\text{H - T}}\Delta {\text{S}} where ΔG\Delta {\text{G}} is change in free energy, ΔS\Delta {\text{S}} is entropy change, ΔH\Delta {\text{H}} is enthalpy change and T{\text{T}}is the temperature in Kelvin.
The liquid and vapour phase during evaporation of water is in equilibrium which makes free energy change ΔG\Delta {\text{G}} =0 = 0
Since the temperature of boiling water=100C{100^ \circ }{\text{C}} and we know that the standard temperature at Kelvin is 273K273{\text{K}} so to make the temperature of boiling water at Kelvin we will add the standard value to the temperature of boiling water.
T = 100+273=373 K\Rightarrow {\text{T = }}100 + 273 = 373{\text{ K}}
Then, the formula becomes-
0=ΔH - TΔSΔS=ΔHT\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-
ΔS=386373=1.02KJ\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 ΔG\Delta G and it also tells –
If ΔG<0\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 ΔG=0\Delta G = 0then, the system is at equilibrium. No change occurs in either forward or backward direction.
IfΔG>0\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.