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)?

At 333.33 K we expect 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,

Putting the given values into the above formula as follows.

      T =

         = 333.33 K

Thus, we can conclude that at 333.33 K we expect to boil.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

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