Question

If enthalpy of vaporization of water is 186.5kj/mol,entropy of its vaporatoon is

Answer 1

Answer : The entropy of vaporization is $0.5KJ/mole/k$.

Solution : Given,

$\Delta H_{vap}=186.5KJ/mole$

The boiling point of water = $100^oC=273+100=373k$      $(0^oC=273k)$

Vaporization means the phase changes from liquid to vapor at boiling point.

Formula used :

$\Delta S_{vap}=\frac{\Delta H_{vap}}{T_b}$

where,

$\Delta S_{vap}$ = entropy of vaporization

$\Delta H_{vap}$ = enthalpy of vaporization

$T_b$ = boiling point temperature

Now put all the given values in this formula, we get

$\Delta S_{vap}=\frac{186.5KJ/mole}{373k}=0.5KJ/mole/k$

Therefore, the entropy of vaporization is $0.5KJ/mole/k$.

Answer 2

See the image attached

Hope it helps

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