April 03, 2020, 08:16:41 PM
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### Topic: Chemical potential  (Read 135 times)

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#### jere8184

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##### Chemical potential
« on: March 22, 2020, 01:46:28 PM »
Hi just signed up and as I have been struggling with understanding two pieces of information which seem conflict with each other and its been bugging me.

The problem is I have been told
(1) The equation for the chemical potential of a solution is:

μ = μ° + R*T*lnX

Where x is the mole fraction of the solvent.

This means that the chemical potential of the pure phase (μ°) is higher than the chemical potential of the solution (μ) by R*T*lnX amount,

As the R*T*lnX is negative.

(2) I have also been told that raising the temperature is needed to allow for a pure liquid to decrease its chemical potential to a point where it is equal to (and therefore in equilibrium with) the vapour phase of a solution which is in equilibrium with said solution.

But the equation above shows that all that increasing temperature would do is decrease the chemical potential of the solution. Causing the chemical potential of the pure liquid and solution to become even more different and further away from equilibrium.

Any help would be appreciated.

#### mjc123

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##### Re: Chemical potential
« Reply #1 on: March 23, 2020, 07:38:56 PM »
I don't know where you heard (2), but it's rubbish. At any temperature, the pure liquid has a higher vapour pressure than the solution. That means you have to increase the temperature of the solution to get its vapour pressure to equal that of the pure liquid. That's why a solution has a higher boiling point than the pure solvent.

#### demoninatutu

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##### Re: Chemical potential
« Reply #2 on: March 24, 2020, 04:05:52 AM »
Or, to put mjc123's reply another way, you need to lower the temperature of a pure solvent to get the same vapour pressure as a corresponding solution.

I suggest it's very easy to mis-remember these things, mis-hear them (and write them down wrong), or mis-state them (when giving a lecture).