Question

What's the boiling point water (in celsius) at the top of Mt Everest (atmospheric pressure =0.54 atm)? I've tried about everything and I keep getting wrong answers. At 25C Substance delta H (kj/mol) delta S (j/mol*k) delta G (kj/mol) H2O (l) -285.8 70.0 -237.1 H2O (g) -241.8 188.8 -228.6 Find Tboiling point in celsius.

Answer 1

H2O (l) <------------------> H2O (g)

Kp = P_H2O = 0.54 atm

delta G = -R T ln Kp

delta G = -RT ln P

= - 8.314 x 10^-3 x 298 x ln 0.54

delta G  = 1.53

delta H = delta H (H2O g) - delta H (H2O l )

= -241.8 + 285.8 = 44 kJ /mol

delta S = products - reactants = H2O gas - H2O liquid

= 188.8 -70 = 118.8 J / K

= 0.1188 kJ

now gibbs helmholtz eqaution:

deltG = deltaH - T delta S

1.53 = 44 - T (0.1188 )

T =357.5 K

T = 84.5 oC -------------------> answer boiling point at tot of Mt everest

note: can you please give me reply once you get the answer . i need your reply please