contestada

Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. 390. k 412 k 368 k 466 k 255 k

Respuesta :

Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c
pstnonsonjoku

The normal boiling point is 412 K.

The following information was provided in the question;

P1 = 55.1 mm Hg

T1 = 35°C + 273 = 308 K

P2 = 1 atm = 760 mmHg

T2 = Boiling point

ΔH vap = 32 .1 kJ/mol

Using the Clausius -Clapeyron equation;

ln(P2/P1) =  -ΔH vap/R(1/T2 - 1/T1)

ln(760 mmHg/55.1 mm Hg ) = -32 .1  × 10^3/8.314 × (1/T2 - 1/308)

2.6 =  -3860.96 × (308 - T2/308 T2)

2.6 = -1189175.68 + 3860.96T2/308 T2

985.6 T2=  -1189175.68 + 3860.96T2

985.6 T2 - 3860.96T2 = -1189175.68

-2875.36T2 = -1189175.68

T2 = -1189175.68 /-2875.36

T2 = 412 K

Learn more: https://brainly.com/question/4699407

Otras preguntas