Question

The enthalpy of vaporization of trichloromethane (chloroform, CHCl3) is 29.4 kJ mol−1 at its normal boiling point of 334.88 K. Calculate (i) the entropy of vaporization of trichloromethane at this temperature and (ii) the entropy change of the surroundings.

Answer 1

a) the entropy of vaporization is ΔS = 0.08779 kJ/(mol*K)

b) the entropy of the surroundins is ΔS surr ≥ -0.08779 kJ/(mol*K )

Step-by-step explanation:

from the second law of thermodynamics

ΔS ≥ ∫dQ /T

where ΔS= change in entropy of trichloromethane , Q= heat exchanged with the surroundings, T= absolute temperature

since the boiling of trichloromethane is at constant temperature if it is in a pure state ( with no other substances), then

ΔS ≥ ∫dQ /T = 1/T ∫dQ = ΔH/T

ΔS ≥ ΔH/T

since the process is also reversible

ΔS = ΔH/T = 29.4 kJ mol−1/ 334.88 K = 0.08779 kJ/(mol*K)

for the surroundings

ΔS universe = ΔS + ΔS surr ≥ 0 (the entropy of the universe always increases)

ΔS surr ≥ -ΔS = -0.08779 kJ/(mol*K)

ΔS surr ≥ -0.08779 kJ/(mol*K )