Question

this is a preexam for this tuesday. Need help solving and understand please.

Benzene has a heat of vaporization of 30.72 kJ/mol, and a normal boiling point of 80.1 deg. Celsius. At what temperature does benzene boil when the external pressure is 745 torr?

well since I have a pressure, and i have what seems to be 2 temperatures, 1 i gotta find. I have to use the ln (p2/p1)= - Delta H vap / R (1/t2 - 1/t1) right?

@aaronq Im back I think I fixed it

Answer 1

okay, so this: \(ln(\dfrac{P_2}{P_1})=\dfrac{-\Delta H_{vap}}{R}(\dfrac{1}{T_2}-\dfrac{1}{T_1})\)

is called the "clausius-clapeyron equation", although it's name is not really important.

you basically need to plug in the values given, and solve.

Just keep track of what P2 and T2 are and P1 and T1 are.

you have: heat of vaporization of 30.72 kJ/mol
T= 80.1 deg. P=normal conditions (1 atm)

they ask at what T? will it boil if the P= 745 torr

can you try to set it up?

Answer 2

okay. well this is what I got, from information given.

Answer 3

\[\Delta H=30.72 \frac{ kj }{ mol }\]
\[T_{1} 80.1 C +273=353.1 Kelvin\]
\[P_{1} = 1atm_ normal boilig point = 760 torr\]
\[P_{2} = 745 torr\]

Answer 4

perfect

Answer 5

now all you do it plug it in and solve for T2

Answer 6

you don't have to write it here, i'll do it also and we'll compare answers, hopefully we have the same one.

Answer 7

don't forget to convert dHvap to match the units of R

Answer 8

YAY! im back on haha sorry. im back

Answer 9

dHvap? how hold on let me just try and ill show u what i have :)

Answer 10

okay, imma put what i have @aaronq

Answer 11

\[\ln \frac{ 760 }{ 745 } = \frac{ -30.72 \frac{ kJ }{ mol } }{ .082 \frac{ L * atm }{ mol * K } } (\frac{ 1 }{ T } - \frac{ 1 }{ 353 K })\]

Answer 12

@aaronq

Answer 13

you put the 353 in the wrong place