Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 216.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid.
Calculate the normal boiling point of this liquid.

Answer 1

Use the Arrhenius equation.

Answer 2

ln(216/92)=(DeltaH vapor/8.3145)(1/296- 1/3180

2.348= 0.00234 DeltaHV/8.3145

DeltaHv=0.000281

Im kinda confused and don't know where to go from there?

Answer 3

i meant (1/296 - 1/318)

Answer 4

I meant Clausius-Clapeyron equation. You have it right and have substituted correctly. Give me a second.

Answer 5

Thanks! for the first part, my answer is supposed to be kj/mol

Answer 6

0.8535 = x(2.337E-4)/8.314
0.8535*8.314 = 0.0002337x
Solve for x but confirm what I've done.

Answer 7

x will be in J/mol and you must convert to kJ/mol.

Answer 8

for x I got 30380.8 but I'm not sure where you got 0.8535?

And i don't know how to answer the second part (answer in degrees celsius)

Answer 9

For the second part use the same equation but I would change the p2:T2 pair to make them the p1:T1 pair, then use the new p2 as 760 (that's the definition of boiling point; i.e., when the vapor pressure of the liquid = atmospheric pressure which of course is 760 mm), then T2 is the unknown which is the boiling point (in Kelvin).

Answer 10

0.8535 is ln (216/92) if I punched in the numbers correctly.
216/92 = 2.348 and ln of that number is
0.85349 which I rounded to 0.8535.

Answer 11

1n(216/760)=(30.38kj/8.3145) (1/318-1/T)

-1.258= 3.735 (0.00314-1/T)
-1.258= 0.0117-1/T
-1.2697= 1/T
T =-1.2697


I feel like i did this wrong

Answer 12

Why did you plug kJ/mol back into the equation. When you solve it delta H came out in J/mol and it must go back as J/mol. The ln(216/760) is ok but that's all I checked since dHvap wasn't substituted properly.
Good luck. It's past my bed time.