Question

Chloroform, formerly used as an anaesthetic and now believed to be a carcinogen, has a heat of vaporization ΔHvaporization = 31.4 kJ mol-1. The change, CHCl3(l) CHCl3(g) has ΔSº = 94.2 J mol-1 K-1. At what temperature do we expect CHCl3 to boil (i.e., at what temperature will liquid and vapor be in equilibrium at 1 atm pressure)?

Answer 1

Answer: At 333.33 K we expect $CHCl_{3}$ to boil.

Explanation:

We know that at the boiling point both the liquid and gaseous state are in equilibrium with each other. Hence, free energy change for this reaction will be equal to zero.

Also,

       $\Delta G^{o} = \Delta H^{o} - T \Delta S^{o}$

Putting the given values into the above formula as follows.

      $\Delta G^{o} = \Delta H^{o} - T \Delta S^{o}$

     $0 = 31400 J/mol - T \times 94.2 J/mol K$

      T = $\frac{31400 J/mol}{94.2 J/mol K}$

         = 333.33 K

Thus, we can conclude that at 333.33 K we expect $CHCl_{3}$ to boil.

Answer 2

Answer:

Chloroform is expected to boil at 333 K (60 $^{0}\textrm{C}$).

Explanation:

For liquid-vapor equilibrium at 1 atm, $\Delta G^{0}$ = 0.

We know, $\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$ , where T is temperature in kelvin scale.

Here both $\Delta H^{0}$ and $\Delta S^{0}$ are corresponding to vaporization process therefore T represents boiling point of chloroform.

So, $0=(31.4\times 10^{3}\frac{J}{mol})-[T\times (94.2\frac{J}{mol.K})]$

or, T = 333 K

So, at 333 K (60 $^{0}\textrm{C}$) , chloroform is expected to boil.