Predict the normal boiling point of triethylborane (C6H15B) using the following data:?

Question

Predict the normal boiling point of triethylborane (C6H15B) using the following data:
ΔH°f (25°C) ΔG°f (25°C)
C6H15B(l) –194.6 kJ/mol 9.4 kJ/mol
C6H15B(g) –157.7 kJ/mol 16.1 kJ/mol


A)90°C
B) –21°C
C) 365°C
D) 256°C

Answer 1

The enthalpy and free energy changes at 25 C for the vaporization reaction are:

dH = -157.7 kJ/mol -(-194.6 kJ/mol) = 36.9 kJ/mol
dG = (16.1 - 9.4) kJ/mol = 6.7 kJ/mol

dG = dH - T*dS, so at 25 C (298.15 K) the entropy change for the reaction is:

dS = -(dG-dH)/T = -((6.7 - 36.9) kJ/mol)/298.15 K
dS = 101.291 J/(mol*K)

If one assumes that the difference in heat capacities between triethylborane gas and liquid is the same at 25 C and at the boiling point (i.e., that dH and dS of the reaction are constant as a function of temperature), then we can set dG equal to zero, and find the temperature at which this occurs:

dG = 0 = 36.97 kJ/mol - T*101.291 J/mol*K

(36.9 kJ)/(101.291 J/K) = 364.3 K = 91.1 C (answer A is the closest)

The actual boiling point of triethylborane is 95 C (see source).

Answer 2

A)90°C

Answer 3

A) 90

Answer 4

Triethylborane

Answer 5

Pisces.