Jor
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diluting a solution: enthalpy change
I soon have a test on physical chemistry.
I know everything quite well, but in an exam of 1 year ago, they ask for the enthalpy change when diluting a glucose solution with an equal volume of
water.
Now I am not sure what the answer is, and I cannot find it in my textbook.
When you mix equal amounts of for example benzene and toluene, enthalpy and internal energy change is supposed to be zero. So I figure that diluting a
glucose solution should also give a zero enthalpy change?
This is ofcourse not the case when dissolving things like NaOH or CaCl2 (negative entropy change) or NaCl (positive entropy change), but I suppose
this is due to the fact that the substance is ionised when dissolved, wich glucose doesn't.
Am I wrong?
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woelen
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The enthalpy does change in such a case, not due to a chemical reaction, but due to equalization of pressure.
Suppose you have two compartments separated by a membrane, which only allows water molecules to pass through. On one side you have the solution of
glucose, on the other side you have pure water. What happens is that there is a certain osmotic pressure over the membrane. This pressure can be used
to do some work (i.e. there is some potential energy in the system). Now, if you mix the two solutions, then the pressure difference has gone to zero
and no work can be done anymore.
You can compare this with two compartments, in one there is a gas, in the other there is a vacuum. The pressure difference can be used to do some
work. With your glucose solution there also is a form of 'pressure'-difference.
Please look up your textbooks for the math behind this stuff, but now you at least know where to look.
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DJF90
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I believe Woelen has hit the nail on the head; the enthalpy change will occur due to the osmotic pressure of the system, which then causes a change in
the internal energy, U, of the system, and hence the enthalpy, H. Quantifying it should be possible, but you'll have to look at a chapter on "Liquids
and solutions" or "solution thermodynamics" to work it out.
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Magpie
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Quote: Originally posted by Jor ![](images/xpblue/lastpost.gif) |
I know everything quite well, but in an exam of 1 year ago, they ask for the enthalpy change when diluting a glucose solution with an equal volume of
water.
When you mix equal amounts of for example benzene and toluene, enthalpy and internal energy change is supposed to be zero. So I figure that diluting a
glucose solution should also give a zero enthalpy change?
This is ofcourse not the case when dissolving things like NaOH or CaCl2 (negative entropy change) or NaCl (positive entropy change), but I suppose
this is due to the fact that the substance is ionised when dissolved, wich glucose doesn't.
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This is a problem normally described under "heat of solution."
The heat of solution varies with the starting and ending concentrations of the particular case you're given.
Toluene and benzene are very close analogs so mixing them likely shows no heat of solution, as you say. This is not true of water and glucose,
however.
Glucose does have a heat (or enthalpy) of solution in water as indicated here:
http://pubs.acs.org/doi/abs/10.1021/ja01316a029
Other factors, such as solvation, affect the heat of solution besides ionization.
Do you mean "enthalpy" when you've said entropy? These have very different meanings, as you likely know.
[Edited on 21-4-2010 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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Jor
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Thanks for all the replies. Yes I understand it now, although I am not completely sure how to calculate the exact enthalpy change.
Yes Magpie, I did not mean entropy ![:)](./images/smilies/smile.gif)
Are you sure about the benzene and toluene. You seem to say that the change in enthalpy when mixing is very small but my nextbook says (delta)H=0 for
mixing these, the same with Hexane and heptane!
Yes woelen, we also had to learn the things you describe, e.g when you have 2 solutions (1 pure water) seperated by a membrame. It is amazing that
even a 1g/L glucose solution can do a lot of work!
I had the exam today, and I think I passed it. There were no questions on osmotic pressures this time (wich was not the case the past years). This is
a shame, as me I was quite good at making those exercises. Lots of questions on entropy, for example why is (delta)S = nR Ln(V2/V1) equal to (delta)S=
nR Ln(P1/P2). I am really not good at math, but I managed to do most things right.
Now there is still to pass quantum mechanics and math (wich I both find very difficult!) and then I can fully concentrate on the subjects I like
(organic, inorganic, biochemical). If ofcourse i indeed passed this exam ![:(](./images/smilies/sad.gif)
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woelen
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Magpie, I don't think that the heat of solution is the issue in Jor's question. Only at high concentration or when going from solid to solution this
is something which matters. In this question I belive that the glucose already is completely solvated and mixing it with more water does not add extra
solvation.
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Magpie
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Quote: Originally posted by woelen ![](images/xpblue/lastpost.gif) | Magpie, I don't think that the heat of solution is the issue in Jor's question. Only at high concentration or when going from solid to solution this
is something which matters. In this question I belive that the glucose already is completely solvated and mixing it with more water does not add extra
solvation. |
Woelen, Jor does not give a beginning concentration of glucose and therefore we do not know how concentrated (or dilute) the original solution was. A
standard way of giving heats of solution is the "heat of solution at infinite dilution." This is the value per mole of solute when adding more water
produces no noticeable change in enthalpy. This usually occurs at 100 to 200 moles of water per mole of solute.
If you take a fairly concentrated solution and then add to it an equal volume of water you will indeed get a noticeable change in enthalpy for many
solutes. My textbook shows this graphically very well. I'll see if I can find such a graph and post it.
It does seem that glucose has a very small, but measureable, heat of solution according to the link I posted above.
Jor, it seems that the mixing of benzene and toluene, and heptane with hexane, produce no noticeable heats of solution. This is, I suppose, expected,
due to their close chemical similarity.
I have never seen how heats of solution can be calculated without a look-up table, graph, or empirical formula. Ie, this information comes from
experimental data.
The single most important condition for a successful synthesis is good mixing - Nicodem
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JohnWW
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Any change in enthalpy on the mixing of two pure liquids (or other substances) under adiabatic conditions would be detectable by either an increase or
decrease in temperature of the mixture, depending on whether there is an increase or decrease in enthalpy and also the specific heat of the mixture.
Because the production of such a mixture from pure substances is an increase in disorder or randomness, there would always be an accompanying increase
in ENTROPY; - except possibly if two reagents can be mixed in stoichiometric proportions to create a completely new pure substance, with no byproducts
produced in the same phase nor any of the original reagents remaining in the same phase to create another mixture.
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entropy51
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Heat of Dilution
F = RT ln (C1/C2)
C2 = C1 / 2
R = 8.314 J/K-mol
T = 25 C = 298 K
F = 1681 J per mol of glucose (e.g. for starting with 1 Liter of 1 M glucose) which is 401.6 calories
For extra credit, what is the increase in temperature?
dT = dQ / ( Mass x heat capacity)
dT = 401.6 cal / ( 1 cal/gm-C x 2000 g) = 0.2 C ![:P](./images/smilies/tongue.gif)
[Edited on 21-4-2010 by entropy51]
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Magpie
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Quote: Originally posted by Magpie ![](images/xpblue/lastpost.gif) |
I have never seen how heats of solution can be calculated without a look-up table, graph, or empirical formula. Ie, this information comes from
experimental data.
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Entropy, that formula you gave looks vaguely familiar. Is F the enthalpy of formation? Is the formula derived from that of osmotic pressure vs
concentration? I could not find it in my Daniels & Alberty "Physical Chemistry" text. My Chem E text says explicitly that this data must be
determined experimentally. It's likely the formula is for ideal solutions only. But that's true for most all formulae, for that matter.
The single most important condition for a successful synthesis is good mixing - Nicodem
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entropy51
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Of course it's for ideal solutions only, like most everything else in P Chem. I think it only applies to non-electrolytes, such as glucose.
I didn't look up the derivation, my recollection is that it's derived using Gibbs potential. It is an enthalpy.
And I'm shocked that you didn't come up with first!
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Magpie
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Quote: Originally posted by Jor ![](images/xpblue/lastpost.gif) |
I know everything quite well, but in an exam of 1 year ago, they ask for the enthalpy change when diluting a glucose solution with an equal volume of
water.
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According to this source (Eqn 4-12):
http://www.chem1.com/acad/webtext/thermeq/TE4B.html
The equation,
G(dilution) = RTln(C2/C1)
is valid only for ideal solutions, and in that case delta H = 0.
The single most important condition for a successful synthesis is good mixing - Nicodem
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