Well, this is why I think there's a little ambiguity here. We do know the two solvents will form a solution, so they can't be overwhelmingly unhappy to mix together, BUT if you just look at the dielectric constants (or dipole moments), you'll see that acetone is considerably more polar than carbon tetrachloride. From this I would propose that the intermolecular forces are weaker in the mixture than they are in the separated solvents. This would lead to a conclusion that mixing the two solvents together is endothermic.
Honestly with nothing else to go on, it's a bit of a judgment call. Thankfully we do have the benefit of knowing the answer beforehand, but if you had told me it was exothermic, I'd have probably been able to convince myself that this is correct ("well, acetone is relatively polar but it does have some methyl groups, so..."). Much easier to compare two mixtures to each other than to predict a single mixture in an absolute sense.
Here's the thing to remember: In general if you're mixing two solvents that have similar polarity, you are more likely to have an exothermic process, and if you are mixing two solvents with very different polarity, you are more likely to have an endothermic process. That can be rather vague because at what point do the polarities of two mixtures become "similar"? Good question, and that's why it's ambiguous.
Let's just push onward and see if I can't help you understand why this is the case. I couldn't find any real data for carbon tetrachloride mixtures with a cursory search of the literature, but enthalpies of mixing for water-alcohol mixtures are readily available. Take a look at this plot:
(In the event that doesn't show up, the direct source is here, look at just the top part of the figure:
http://www.nature.com/srep/2014/140314/srep04377/fig_tab/srep04377_F2.html)
(EDIT: I just looked at it on my phone, and I see it is way oversize. Sorry about that. Maybe I can fix it later. I can't access file sharing sites from here.)
What you're looking at is the enthalpy of mixing, ΔHe, for several different water-alcohol mixtures as a function of the mole fraction of alcohol, measured at 298 K. The black data is water-methanol, the red is water-ethanol, the blue is water propanol and the purple is water-isopropanol. Let's just ignore the isopropanol data and focus on the linear alcohols, because the comparison here is the most straightforward. I think it should go without saying that as you go from methanol to ethanol to n-propanol, the degree of polarity of the alcohol is decreasing, and therefore the favorability of its interactions with water should decrease, because water is the most polar alcohol or of the bunch. In case that's not obvious, the dielectric constants for these solvents at 298 K are as follows: water (79), MeOH (33), EtOH (25), PrOH (20.1).*
You'll notice two things from these plots. First is that the enthalpy of mixing can change quite substantially based on what the relative mole fractions of the two solvents are. In the case of propanol, the enthalpy of mixing can actually change sign - when the majority of the mixture is water, the enthalpy of mixing is negative, but when the majority of the mixture is alcohol, the enthalpy of mixing is positive. I point this out because it highlights another point of ambiguity in the original question you post: the relative mole fraction of carbon tetrachloride to acetone is not specified. It may not matter in this specific system, but still, it's an important consideration that is kind of ignored.
The more important thing you'll notice from these plots is that as the number of carbons on the alcohol increases, the enthalpy of mixing becomes more positive, for most mole fractions. So we can make a general conclusion that as the alcohol becomes less polar, mixing becomes less exothermic, and therefore, less thermodynamically favorable. In the case of propanol at around 70% mole fraction, the mixing is actually endothermic.
Now, a final thing to notice is that, positive or negative, the enthalpy of mixing absolute value is always rather small, below 1 kJ/mol. This should agree with practical experience: when you mix water and alcohol together, rarely do you feel them heat up (or cool down). So while these solutions are, strictly speaking, not ideal at most mole fractions, mixing is still primarily an entropically driven process.
Consider, for example, the case of propanol and water. No matter what the mole fractions are of water and propanol, they will ALWAYS mix together at room temperature, even at mole fractions where the enthalpy of mixing is positive (at 70% propanol and 30% water, say). We can roughly calculate why.
At this mole fraction, the ΔH
e is roughly +200 J/mol. The Gibbs energy of mixing is ΔH
e-TΔS
e. So what is ΔS
e?
For an solution with random mixing, which we'll approximate this to be, ΔS
e per mole is given by:
[tex]\frac {\Delta S_e}{n} = -R(\chi_w \ln \chi_w + \chi_a \ln \chi_a)[/tex]
Where χ
w and χ
a are the mole fractions of water and alcohol, respectively.
For a 70% alcohol in water mixture, you can determine ΔS
e/n to be ~ 5.1 J/mol K. At 298 K and ΔH
e = +200 J/mol, ΔG
e = -1314 J/mol. Mixing is still very spontaneous despite the endothermic enthalpy change. In fact, if you set ΔG
e to zero and solve for T, you can determine that the mixture won't separate until you cool it down to about 39 K - of course, they wouldn't be liquid then, and this is based on a whole lot of assumptions, but I hope you get the basic point.
Anyway, all this long-windedness is to say that the enthalpy of mixing depends in large part on the strength of intermolecular interactions in the mixture compared to those in the isolated components. This comparison will be more likely to be favorable when the two solvents share similarities in their polarity, and more likely to be unfavorable when the two solvents are very different. Do keep in mind that thinking about polarity is still a simplification of the problem and can sometimes lead you astray if you take it too much at face value.**
For carbon tetrachloride and acetone, the solvents are different in terms of polarity (dielectric constant of carbon tet is ~2, for acetone it's 21), so it seems reasonable to conclude that the enthalpy of mixing could plausibly be positive, though not positive enough to overcome the entropic driving force for mixing. Whether or not I would have come up with that without knowing the answer ahead of time is another story, though. This is why I think it's not a particularly good question - to ask how a mixture will perform in any absolute way is a bit of a stretch in my opinion, particularly because in many cases it depends on how much of A there is in the mixture and how much of B. So - don't feel too bad that you didn't really know how to answer it.
Ok, that leaves us with the final question. Entropy of the surroundings. What do you think? What do you know about entropy of the surroundings?
Footnotes for the sadomasochist:
* In chemistry there are always exceptions to an explanation that makes sense. We encounter here again the ambiguity surrounding the term "polar solvent". I'm using dielectric constant as a metric for polarity and trying to correlate the difference in dielectric constants of two solvents to the decrease in solution entropy change. I think when molecular structures are fairly similar, this is a fine approach, but the isoproponal data, which I asked you to ignore, doesn't fit this correlation. You'll notice that for most alcohol mole fractions, the enthalpy of mixing for water-isoproponal mixtures is lower than than of water-propanol mixtures, suggesting that water-isopropanol has more favorable intermolecular interactions than water-proponal. Which would, based on my argument, lead you to conclude that isopropanol is more polar than propanol. However, the dielectric constant of propanol is higher than that of isopropanol, which indicates the opposite. Yet, the dipole moment of propanol is higher than isoproponal, suggesting that propanol may be higher polarity. The take home message here is that it's important to choose a metric, but also to understand its limitations. Here the limitations show up because isopropanol is not a primary alcohol, and so straight comparisons may not be appropriate. Either there's a weakness in the metric we're choosing to represent polarity that becomes more prominant here, or there are other subtle factors that determine the enthalpy of mixing besides just polarity matching. We could speculate on what those may be, but I think that would be taking us beyond what would be helpful to you. So again: let's just ignore isopropanol for the time being.
**I mean, if you want an extreme case: if you mix water with water, the enthalpy of mixing is zero, despite the fact that the polarity of water is exactly the same as the polarity of water, the best polarity match possible!. That may seem like a silly trivial example, but it shows that relative polarity of the solvents does not directly determine the enthalpy of mixing. What you are really after here is a way to guess at how well the solvent molecules interact with each other in the mixture compared to how they interact with each other as separated, pure substances. This requires a complex analysis of intermolecular forces in three separate cases (two solvents by themselves and one mixture). In a water-alcohol system, while addition of alcohol to water may be favorable from water's point of view, from the alcohol's point of view it may be unfavorable (or vice-versa). The balance is somewhere in the middle, with the final favorability or unfavorability determined by which solvent wins or loses the most. This is why the enthalpy of mixing is dependent on mole fraction, by the way. When you're adding a little alcohol to water, that's a different situation than adding a little water to alcohol, because who wins out the most is dependent on how many players there are on each competing team. Nevertheless, looking at relative polarties of the solvents in a mixture is at least a good general guide to what you should expect if you're trying to predict how one solvent mixture will perform compared to another.
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Topic: Non-ideal solution (Read 12371 times)