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.

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 )

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 )

answer:

a

explanation: