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Topic: Using Gibbs to find Vapor Pressure  (Read 5744 times)

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Offline CKabes

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Using Gibbs to find Vapor Pressure
« on: February 10, 2013, 06:28:23 PM »
For the process CH3OH(l) :rarrow: CH3OH(g)
ΔG°= 4.3 kJ/mol at 25°C. What is the vapor pressure of CH3OH(l) at 25°C in mm Hg?

Can someone please provide an equation or a start?
I thought I would find entropy, but I would require ΔH, or I could find ΔH if I could first find ΔS, and then possibly use the Clausius-Clapeyron equation to find vapor pressure?

Offline UG

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Re: Using Gibbs to find Vapor Pressure
« Reply #1 on: February 10, 2013, 10:20:59 PM »
Can someone please provide an equation or a start?
Yup, there is an equation which relates ΔG° to the equilibrium constant K

Offline CKabes

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Re: Using Gibbs to find Vapor Pressure
« Reply #2 on: February 10, 2013, 10:59:27 PM »
ΔG=-RT ln(K) ?

How do I procede after I find K?

Offline UG

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Re: Using Gibbs to find Vapor Pressure
« Reply #3 on: February 10, 2013, 11:01:20 PM »
How do I procede after I find K?

Well, what is the expression for K for the reaction CH3OH(l)  :rarrow: CH3OH(g) and what does it express?

Offline CKabes

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Re: Using Gibbs to find Vapor Pressure
« Reply #4 on: February 10, 2013, 11:41:15 PM »
K is the ratio of product to reactant at equilibrium where the forward reaction rate is equal to the reverse. The equilibrium expression for this reaction would be:

CH3OH(g) / CH3OH(l), however because CH3OH(l) is a pure substance, the Equilibrium Expression would simply be K = CH3OH(g)

Is that correct?

Offline UG

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Re: Using Gibbs to find Vapor Pressure
« Reply #5 on: February 11, 2013, 12:21:09 AM »
CH3OH(g) / CH3OH(l), however because CH3OH(l) is a pure substance, the Equilibrium Expression would simply be K = CH3OH(g)
Ok, so you have K is equal to some value. What does this value signify, is it a concentration, a pressure, a ratio perhaps?

Offline CKabes

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Re: Using Gibbs to find Vapor Pressure
« Reply #6 on: February 11, 2013, 08:14:50 AM »
Here's what I said earlier:
Quote
K is the ratio of product to reactant at equilibrium where the forward reaction rate is equal to the reverse

Offline curiouscat

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Re: Using Gibbs to find Vapor Pressure
« Reply #7 on: February 11, 2013, 08:54:34 AM »
Here's what I said earlier:
Quote
K is the ratio of product to reactant at equilibrium where the forward reaction rate is equal to the reverse

" ratio of product to reactant" doesn't mean anything much quantitatively.

e.g. What's "ratio of you to me"? Our weights? Our heights? Our money? Our ages?

Get the point?

Offline CKabes

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Re: Using Gibbs to find Vapor Pressure
« Reply #8 on: February 11, 2013, 08:49:00 PM »
Alright, let me try again,

K is the ratio of the concentration of products to reactants in moles at a certain temperature

Offline curiouscat

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Re: Using Gibbs to find Vapor Pressure
« Reply #9 on: February 11, 2013, 11:40:29 PM »
Alright, let me try again,

K is the ratio of the concentration of products to reactants in moles at a certain temperature

That's a specific case. It'd be wrong in general.

e.g. It's not even consistent with this that you wrote earlier:

Quote
because CH3OH(l) is a pure substance, the Equilibrium Expression would simply be K = CH3OH(g)


Offline Stovn0611

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Re: Using Gibbs to find Vapor Pressure
« Reply #10 on: February 14, 2013, 08:39:06 PM »
For the equation ΔG=-RT ln(K), K represents the Kp equilibrium constant - think about what Kp is a ratio of

Offline CKabes

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Re: Using Gibbs to find Vapor Pressure
« Reply #11 on: February 14, 2013, 09:00:17 PM »
Kp is the partial pressure of the products devided by the partial pressure of the reactants?

Offline Stovn0611

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Re: Using Gibbs to find Vapor Pressure
« Reply #12 on: February 14, 2013, 09:12:32 PM »
Yes exactly(and it works the exact same way as for concentrations and you have to take the partial pressure to the power of whatever the coefficient of that substance is in the reaction)

For example, in this reaction:

N2 + 3H2 -> 2NH3

Kp = PNH32/(PH23PN2)

So in this reaction, CH3OH(l) :rarrow: CH3OH(g)

Kp = PCH3OH since the liquid form will not be able to exert any pressure

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