Question

Trouton’s rule states that the entropy of boiling at the normal point is 85 J/mol * K. (a) Does the data from Example 3.2 support Trouton’s rule? (b) H2O has a heat of vaporization of 40.7 kJ/mol. Does the Delta vapS for H2O at its normal boiling point support Trouton’s rule? Can you explain any deviation? (c) Predict the boiling point of cyclohexane, C6H12, if its Delta vapH is 30.1 kJ/mol. Compare your answer to the measured normal boiling point of 80.7°C.

Ex 3.2

What is the change in entropy when 1.00 g of benzene, C6H6, boils reversibly at its

boiling point of 80.1°C and a constant pressure of 1.00 atm? The heat of vaporization

of benzene is 395 J/g.

Because the process occurs at constant pressure, Delta vapH for the process equals the heat,

q, for the process. Because vaporization is an endothermic (that is, energy-in) process,

the value for the heat is positive. Finally, 80.1°C equals 353.2 K. Using equation 3.14:

Delta S = +395J/353.2K =+1.12 J/K

for 1 g of benzene. Because this represents the entropy change for 1 g of benzene,

we can also write this Delta S as +1.12 J/g*K. The entropy of the system—the benzene—

is increasing in this example.

Answer 1

(a) entropy of boiling from data of 32a.

S = 395J/gm/353.1 T = 1.12 J/g

Convert J/mol to J/gm

85J/mol = 85J/1mol * 78g/mol =1.089 J/g ~1.1 J/g

Two data are same

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(b) Troton's rule will be accurate for non-associated liquid. Deviation will occur in case of water as it is Hydrogen bonded.

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H vap = (85 J/K/mol ) * Tb

30.1 *10^3 J/mol = (85 J/K/mol ) * Tb

or, Tb = 354.12 K = 81.12oC

Calculated and actual value matches closely.