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Topic: How do I calculate enthalpy and entropy of this really complicated problem?  (Read 22771 times)

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Offline webassignbuddy

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Given:
K1 = 1.00 x 10-14
T1 = 298.15 K
K2 = 3.12 x 10-15
T2 = 283.15 K
ΔH° = ?
ΔS° = ?

Reaction containing variables needed:

ΔG = ΔH - TΔS
(-RT ln K) =  ΔH - TΔS

What do I do though??? I have 2 K's! And the equation
ln(K2/K1) = (ΔH/R)(1/T2 - 1/T2) doesn't give me the right ΔH value!
The answer for ΔH° is supposed to +54.4.
« Last Edit: October 23, 2013, 04:37:26 PM by Borek »

Offline webassignbuddy

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Anyone willing to help?

Offline magician4

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think about "two equations with two unknowns" , with [itex]\Delta[/itex] G(T) = - RT lnK(T) giving you access to named equations of the type [itex]\Delta[/itex] G(T) = [itex]\Delta[/itex]H - T [itex]\Delta[/itex]S, and K(T1), K(T2) , T1 , T2 as given values

regards

Ingo

p.s.:"really complicated" ? you ain't seen nothin' yet...
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Offline Corribus

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Your Van't Hoff equation is wrong.  Using the right one, I get pretty close to your right answer.  Also, don't forget to include the units.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline webassignbuddy

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think about "two equations with two unknowns" , with [itex]\Delta[/itex] G(T) = - RT lnK(T) giving you access to named equations of the type [itex]\Delta[/itex] G(T) = [itex]\Delta[/itex]H - T [itex]\Delta[/itex]S, and K(T1), K(T2) , T1 , T2 as given values

regards

Ingo

p.s.:"really complicated" ? you ain't seen nothin' yet...

Wait but how did you derive [itex]\Delta[/itex] G(T) = - RT lnK(T)? and what do you use it to solve?
It isn't on the equation sheet my teacher gave our class :(

Offline Corribus

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Van't Hoff is derived from the ΔG expression, not the other way around.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline webassignbuddy

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think about "two equations with two unknowns" , with [itex]\Delta[/itex] G(T) = - RT lnK(T) giving you access to named equations of the type [itex]\Delta[/itex] G(T) = [itex]\Delta[/itex]H - T [itex]\Delta[/itex]S, and K(T1), K(T2) , T1 , T2 as given values

regards

Ingo

p.s.:"really complicated" ? you ain't seen nothin' yet...

Wait but...you mentioned 2 equations and 2 unknowns, right?
ΔG = ΔH - TΔS
-RT ln K = ΔH - TΔS

So,

    -(0.08314)(298.15)ln(1.00 x 10-14) = ΔH - (298.15)ΔS
+[-(0.08314)(283.15)ln(3.12 x 10-15)] = ΔH - (283.15)ΔS
––––––––––––––––––––––––––––––––––––––––––––––––

Nothing on the right side of the equation cancels so that I can solve for ΔH first  ???

Offline webassignbuddy

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Van't Hoff is derived from the ΔG expression, not the other way around.

Ok I'm more confused...can you please show me the equation I'm supposed to use? Because the main ΔG expression I think I'm supposed to use is ΔG = ΔH - TΔS
which is basically -RT ln K = ΔH - TΔS.

But I don't know where to go from there because that equation has the 2 unknowns that I'm looking for, and I can't get TΔS to cancel in the equation i posted above this post because the T's have 2 different values.

Offline magician4

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Wait but how did you derive [itex]\Delta[/itex] G(T) = - RT lnK(T)?
It isn't on the equation sheet my teacher gave our class :(
take a look at your own equations:
Reaction containing variables needed:
ΔG = ΔH - TΔS
(-RT ln K) =  ΔH - TΔS
now: try to combine those two equations
shouldn't be too hard to do...


and what do you use it to solve

with K(T) given, calculate [itex]\Delta[/itex] G (T) belonging to , for example:
K298 K = 10-14 :rarrow: [itex]\Delta G[/itex]298 K = - RT lnK = - R * 298 * ln (10-14)

so, you calculate [itex]\Delta G[/itex]298 K and [itex]\Delta G[/itex]283 K

... and put up your secondary equaitions of the general structure ΔG = ΔH - TΔS
for example: [itex]\Delta G[/itex]283 K = ΔH - 283 K* ΔS

 :rarrow: two equations with two unknowns resulting thereof

regards

Ingo
There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another theory which states that this has already happened.
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Offline webassignbuddy

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solve like this??

   -(0.08314)(283.15)ln(3.12 x 10-15)] = ΔH - (283.15)ΔS
-[-(0.08314)(298.15)ln(1.00 x 10-14) = ΔH - (298.15)ΔS]

Which becomes:

   -23.54ln(3.12 x 10-15)] = ΔH - (283.15)ΔS
-[-24.04ln(1.00 x 10-14) = ΔH - (298.15)ΔS]

Which becomes:

 (786.25) = ΔH - (283.15)ΔS
-(774.95) = ΔH - (298.15)ΔS
________________________
(786.25)-(774.95) = 0 - (283.15)ΔS + (298.15)ΔS

11.3 = 15ΔS
11.3/15 = ΔS
0.753 = ΔS

^ That doesn't make any sense

Offline Corribus

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There are multiple, essentially equivalent ways to solve this problem.  Here's how I did it.

First use the Van't Hoff equation to solve for ΔH°. 

Then because ΔG° equals both -RT ln K AND ΔH° - TΔS°, therefore -RT ln K = ΔH° - TΔS°.  You can insert your ΔH° value into this and use one of your K and T data pairs.  Solve for ΔS°. Done.  (Doesn't matter what K/T pair you use; ΔH° and ΔS° are approximated as roughly independent of temperature, which you'll see if you put in the numbers.)

Or you can solve directly from the -RT ln K = ΔH° - TΔS° equation, two equations with two unknowns.  Both are essentially the same because the Van't Hoff equation is basically derived by integrating -RT ln K = ΔH° - TΔS°. 

I got ΔH° = 54,400 J/mol and ΔS° = ~-85.6 J/mol K
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline webassignbuddy

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There are multiple, essentially equivalent ways to solve this problem.  Here's how I did it.

First use the Van't Hoff equation to solve for ΔH°. 

Then because ΔG° equals both -RT ln K AND ΔH° - TΔS°, therefore -RT ln K = ΔH° - TΔS°.  You can insert your ΔH° value into this and use one of your K and T data pairs.  Solve for ΔS°. Done.  (Doesn't matter what K/T pair you use; ΔH° and ΔS° are approximated as roughly independent of temperature, which you'll see if you put in the numbers.)

Or you can solve directly from the -RT ln K = ΔH° - TΔS° equation, two equations with two unknowns.  Both are essentially the same because the Van't Hoff equation is basically derived by integrating -RT ln K = ΔH° - TΔS°. 

I got ΔH° = 54,400 J/mol and ΔS° = ~-85.6 J/mol K

Yeah but THAT'S the problem though....i have absolutely no idea how to go about solving for ΔH FIRST because I'm having trouble manipulating a specific formula that I'm supposed to be using... :(

Offline webassignbuddy

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Wait but how did you derive [itex]\Delta[/itex] G(T) = - RT lnK(T)?
It isn't on the equation sheet my teacher gave our class :(
take a look at your own equations:
Reaction containing variables needed:
ΔG = ΔH - TΔS
(-RT ln K) =  ΔH - TΔS
now: try to combine those two equations
shouldn't be too hard to do...


and what do you use it to solve

with K(T) given, calculate [itex]\Delta[/itex] G (T) belonging to , for example:
K298 K = 10-14 :rarrow: [itex]\Delta G[/itex]298 K = - RT lnK = - R * 298 * ln (10-14)

so, you calculate [itex]\Delta G[/itex]298 K and [itex]\Delta G[/itex]283 K

... and put up your secondary equaitions of the general structure ΔG = ΔH - TΔS
for example: [itex]\Delta G[/itex]283 K = ΔH - 283 K* ΔS

 :rarrow: two equations with two unknowns resulting thereof

regards

Ingo

I know that much but do I set TΔS = to 0 for both??
I did this:

ΔG298.15 K = 774.95
ΔG283.15 K = 786.25

I know how to do that no problem.
But what I DON'T know is how to set up both equations in such a way that I can find ONE of the variables I'm looking for, especially when none of them cancel out...

 (786.25) = ΔH - (283.15)ΔS
-(774.95) = ΔH - (298.15)ΔS

^ when I do this, I get an extremely small number for ΔS which is completely 100% wrong.

 (786.25) = ΔH - (283.15)ΔS
+(774.95) = ΔH - (298.15)ΔS

When I do this, i go nowhere...because I end up with 1561.12 = 2 ΔH - 581.3 ΔS. I STILL have two unknown variables.

Offline Corribus

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This is just plug and chug, so there's really not much I can do to help you.

You know

[tex]\ln \frac{k_1}{k_2}=\frac{-\Delta H^o}{RT} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]

Put in your values for k1, k2, T2, T1, solve for ΔH°.  If you're not getting the right answer, you're making a basic arithmetic or algebra mistake. 

EDIT: In the post above, try subtracting the equations rather than adding them, if you want to do the direct method.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline webassignbuddy

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This is just plug and chug, so there's really not much I can do to help you.

You know

[tex]\ln \frac{k_1}{k_2}=\frac{-\Delta H^o}{RT} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)[/tex]

Put in your values for k1, k2, T2, T1, solve for H°.  If you're not getting the right answer, you're making a basic arithmetic or algebra mistake.

I tried that originally but for some reason didn't get the right answer on paper and then I erased it. I'm gonna try it again.
Also,
R = 0.08314 kJ/mol Kthough, right? not 8.314 J/mol K?

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