Question

Given the following thermodynamic data, determine the temperature above which CH3OH "boils" according to the reaction below.

CH2OH(l) CH3OH(g)

DELTA Hvap = +38 kJ and DELTA SvapP = +113 J/K

A.) 3.00*103 K

B.) 3.00 K

C.) 150 K

D.) 0.34 K

E.) 336 K

Answer 1

given

CH3OH (l) ----> CH3OH (g)

while boiling

the above system will be at equilibrium

we know that

at equilibirum

dG = 0

now

dG = dH - TdS

so

dH - TdS = 0

dH = TdS

T = dH / dS

using given values

we get

T = 38 x 1000 / 113

T = 336.28 K

so

the temperature is 336 K

so

the answer is E) 336 K