Question

Mercury boils at 357°C and its average enthalpy of vaporization is 60.9 kJ/mol.

Estimate the equilibrium vapour pressure of mercury at 25°C.

Answer 1

Hey @ Jadedry, I don't know if you're working on this or not but I appreciate the help. I know this is a Clausius-Clapeyron Equation type question and I've set it up as such but its the inital pressure I can't figure out what to put for it, or maybe I don't put anything since its the "equilibrium vapour pressure"

Answer 2

I'm typing it out! just bear with me. c:

Answer 3

If you're just worried about what to put for initial pressure you should try 760 torr.

Answer 4

alright, i know its not easy typing these questions

Answer 5

Which is the pressure at 1 atmosphere, it seems like the best option if they don't tell you what pressure they're working at.

(Yeah, it does take a while, sorry! ;n;)

Answer 6

You have to use the Clausius-Clapeyron Equation.

which is:

\[\ln(P _{1} / P _{2}) = -(\Delta H/R) (\frac{ 1 }{ T_{1} } - \frac{ 1 }{ T _{2}})\]

Where P1 is the equilibrium vapour pressure of mercury at 25°C, and P2 is 760 torr (this is the value of torr at 1 atmosphere; since the pressure is not stated, you use 1 atmosphere.)

Delta H is average enthalpy of vaporization (in joules!) and R is the gas constant, which equals: 8.31447 joules/kelvin/mole

T1 and T2 = your given temperatures.

You should (I think) end up with this:

\[\ln(P _{1}/760) = - (\frac{ 60900 }{ 8.31447 })(\frac{ 1 }{ 298.15 }-\frac{ 1 }{ 630.15 })\]

Answer 7

And then solve for P1.

(I converted the temps to kelvin.)

Answer 8

Ya this was my exact set-up except i didnt use torr

Answer 9

if you want atmosphere you could just use 1 then.

Answer 10

in place of 760.

Answer 11

This was more of a second opinion question aha. I have another if you have the time i'll close this one so you can gain another medal

Answer 12

Thanks! I'll have a look at the next one you post and see if I can help. ;u;