grayfalcon89 Posted October 20, 2009 Posted October 20, 2009 Hi, I'm having hard time convincing myself to two answers from my inorganic chemistry course's last year's exam. 1. Which of the following statements are valid concerning ideal gas mixtures: (there are three choices, two of which make sense to me, but this is the one I disagree) * Each gas present in the mixture behaves independently of the others. The solution says that it is valid, but I do not agree. Like, in the textbook, it mentioned how oxygen gas would not react with water, but NH3 would readily dissolve in the water - which clearly show that gases like NH3's behaviors are dependent on other molecules. Or, am I reading something into the question? 2. Ethanol (C2H5OH) and dimethyl ether (CH3OCH3) have the same empirical formula but different molecular structures. Based on the IMF, which of the following statements would be predicted to be true? (there are five choices, but I narrowed down to two) b) Ether has a higher surface tension than ethanol. e) Ethanol has a higher enthalpy of vaporization than ether. Main confusion with this problem is that I'm not sure if dimethyl ether can exhibit hydrogen bonding or not. Like, is hydrogen bonding defined as a heteroatom having a lone pair OR is it having a heteroatom with hydrogen attached to it? If it's latter, then I assume that dimethyl ether does not have hydrogen bonding, so its IMF are lower. If not, I don't know how to distinguish them based on IMF. As far as the choices are concerned, I thought that the higher IMF, the higher surface tension, along with higher MP and higher BP. Isn't this true? I'm not sure how to understand why E is correct. I really attempted them, so hopefully you guys can provide hints to guide me to the right direction (or point out flaws in my reasoning!). Thank you!
UC Posted October 20, 2009 Posted October 20, 2009 Water vapor and ammonia are not ideal gases. Ideal gases make the assumption that there is no interaction at all between the gas molecules. This is only a model (a flawed one for most cases) but is a good approximation of the noble gases and things like nitrogen or hydrogen, especially at low pressures and high temperatures. Dimethyl ether does not have a hydroxy group. The hydrogens that it possesses are bound to carbons and are essentially nonpolar. Ethanol, however has an OH group, which is polar and can participate appreciably in hydrogen bonding. Think about it. Benzene has hydrogens, but it won't dissolve in water, which has oxygen lone pairs. If the benzene hydrogens could participate in hydrogen bonding, dissolving in water would be quite favorable, but that is clearly not the case. The enthalpy of vaporization of a liquid represents the energy needed to counteract all the IMF in the liquid state added to the energy needed to expand the resulting gas to it's final volume. For liquids that H-bond, the intermolecular forces represent the vast majority of the energy needed.
Horza2002 Posted October 21, 2009 Posted October 21, 2009 The ideal gas modle is an aproximation that assumes the gases don't interact with each other. Obviously this is not true...but it only meant to simplify the system so that it can be solved.
D H Posted October 21, 2009 Posted October 21, 2009 Regarding question 1: The answer is you are "reading something into the question". That individual constituents in a mixture do behave independently of the others is what validates the use of concepts like partial pressure.
grayfalcon89 Posted October 21, 2009 Author Posted October 21, 2009 Thank you all for answers! Everything makes so much more sense now. Just curious, UC mentioned that ideal gas considers cases like low pressure and high temperature. I agree with the low pressure, but how does high temperature fit in this case as well? Or is it just a direct application of Charles's Law?
hermanntrude Posted October 21, 2009 Posted October 21, 2009 Thank you all for answers! Everything makes so much more sense now. Just curious, UC mentioned that ideal gas considers cases like low pressure and high temperature. I agree with the low pressure, but how does high temperature fit in this case as well? Or is it just a direct application of Charles's Law? at a high temperature, the molecules have more kinetic energy, which makes any attractions between the molecules less significant when compared to that kinetic energy. Think of two people running past each other, and one tries to grab the other. The faster the people are running, the less effect the grab will have.
grayfalcon89 Posted October 28, 2009 Author Posted October 28, 2009 Ah okay. Thank you all! And exam tonight.
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