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Need help with rate

  1. Apr 15, 2005 #1
    I got this question from my tutor:

    the rate of reaction

    CH3COOC2H5(aq) + H2O -> CH3COOH(aq) + C2H5OH(aq)

    shows 1st order charateristics; that is, rate= k[CH3COOC2H5], even though this is a 2nd order reaction. explain.

    - can i explain it in such a way that its because H2O maybe in low concentration, therefore resulting in the reaction having a 1st order charcteristic?
     
  2. jcsd
  3. Apr 15, 2005 #2

    GCT

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    H2O does not have a concentration, it is the solvent.
     
  4. Apr 15, 2005 #3

    Gokul43201

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    Not entirely true. H2O is more than just a solvent here. It is clearly a reactant, and there will be conditions (which I shall not specify now) where this reaction will be more noticeably second order (or at least nonlinear in [CH3COOC2H5]) .

    The reason it is "pseudo"-first order is not because of the small amounts of H2O, but the opposite. Figure out why this is true.
     
  5. Apr 15, 2005 #4
    Think about how much h20 is actually in there. water is said to be 55.5 Molar. reacting water with CH3COOC2H5 will hardly change the concentration of water at all.
     
  6. Apr 16, 2005 #5

    GCT

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    H20 is the reactant in both hydrolysis and dissociation reactions of acids and bases and yet it is not counted in the equations. Note that concentration is moles of solute/L of water... how would you describe the concentration of water in the equilibrium or the rate equation???
     
  7. Apr 16, 2005 #6

    Gokul43201

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    GCT, how would you answer your own question if the reaction involved two liquids (not aqueous solutions) in roughly equal proportions, miscible or otherwise ?

    A difficulty in the manner of quantifying an effect is not resolved by claiming the non-existence of the effect.
     
    Last edited: Apr 16, 2005
  8. Apr 16, 2005 #7

    GCT

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    Gokul, most often a solvent's role is to participate as a medium for desired chemical reactions, and technically it is present in excess amount...although the latter quality is relatively unimportant since it is better and sufficient to describe it as a solvent.

    Solvents may facilitate some reactions and have many important properties, nevertheless they rarely count in equilibrium/rate equations (I've never heard of a case), even if they react. A common attribute of a solvent is that it always surrounds/solvates its molecules within....you do recall that rate phenomena is dependent on such collision theory, entropical factors apply, but hardly with solvents which constantly surround molecules.

    Although I'm not quite sure, I don't believe that the rate law will apply, same reasoning as above.
     
  9. Apr 16, 2005 #8

    GCT

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    An analogous situation (yet slightly different in nature) is that of concentrated sulfuric acid (mix lots of liquid sulfuric acid with pure water); does the rate laws, equilibrium equation apply in this case? Each would predict a greater acid concentration, yet concentrated sulfuric acid isn't very acidic (why might this be the case?) in the sense of dissociation of its proton (of course, if you pour it over your skin-not good)
     
  10. Apr 16, 2005 #9

    Gokul43201

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    GCT, ethyl acetate and water are mostly immiscible liquids. You will see the activity of water playing a role in the kinetics, if the volume of water in the reaction mixture is of the same order as that of the ester. The only reason you do not see this effect in the given example is because water is in a large excess, not because it is hard to reconcile the measure of the activity of water to your definition of concentration.

    Your explanation of the solvation process helps to show why there is no contribution from water to the reaction rate, when water is in a large excess.
     
  11. Apr 16, 2005 #10

    GCT

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    we are talking about solvents here right? Note that my original post, refers to solvents specifically. I said "water...is the solvent." In low concentrations....it is not a solvent anymore. I think that this is where you've misunderstood me.

    As I said, the definition of solvent is that it surrounds the solvated molecules at all time (taking it to the extreme here a bit), it is the chemical medium (despite what type of liquid it might be).

    You're example of ethyl acetate and water is that of the former's acid properties in water....which takes me back to the sulfuric acid example in my last post. When the concentration of water is low, the rate equation and thus any equilibrium equation related to it thus not apply. You may insist that it does and suggest that one be made, however, we no longer have a chemical medium for the rate dynamics and any other chemical reactions to occur.
     
  12. Apr 16, 2005 #11

    GCT

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    ..basic, not acidic, sorry
     
  13. Apr 18, 2005 #12

    Gokul43201

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    This is not my example. It is the exact reaction in the problem posed by the OP.

    This is not the reaction that the OP is asking about.

    Sulfuric acid is, as far as I know an aqueous solution. I have no idea what its natural form is - it may likely be a solid in its dehydrated form, if that even exists. So, I admit there will be no medium that allows the system to be to explore phase space. But - and I repeat - this is not the problem we are discussing here.

    Ethyl acetate is a colorless liquid. It does not need the presence of water to provide "a chemical medium for the rate dynamics and any other chemical reactions to occur". It is fully capable of providing a chemical medium by itself.
     
    Last edited: Apr 18, 2005
  14. Apr 18, 2005 #13

    GCT

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    sulfuric acid exists as a liquid at room temperature



    So you're saying that if we were to add a small amount of water to ethyl acetate, reactions will still occur perfectly (efficiently as the vice versa case), except that...in this case the concentration of ethyl acetate can be neglected and that ethyl acetate is the solvent, right? Hmmm, water dissolving in ethyl acetate....doesn't make sense to me. Water will hydrogen bond with itself, it won't dissolve...no reactions will occur. No solvent, no reaction, it's quite simple really.

    For those who are interested

    hydrolysis of an ester by nucleophilic addition elimination reaction...pertains to the OP's reaction. If you observe the reaction, water participates directly

    http://www.chem.ucalgary.ca/courses/351/Carey/Ch20/ch20-3-3-1.html
     
  15. Apr 18, 2005 #14

    GCT

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    so back to your original post
    not with water as the solvent. With close to excess ethyl acetate the reaction won't occur, water will hydrogen bond with itself ; water is not soluble in ethyl acetate. Second order reaction will never be observed in this case.
     
  16. Apr 18, 2005 #15
    I believe the reason is because that the rate-limiting step is unimolecular. The first step occurs when a hydroxyl anion (OH-) attacks the partial positive carbon atom on the carbonyl group. Now you have a negatively charged activated complex([tex]CH_3CO(OH)OCH_2CH_3^-[/tex] The next step occurs when the best leaving group, the ethoxide anion (OCH2CH3-) departs from the molecule. If this is the rate-limiting step, then the kinetics would be unimolecuar.
     
  17. Apr 18, 2005 #16

    Gokul43201

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    You're right. In this limiting case, there will probably be no hydrolysis taking place.

    But this is not what I'm saying. I was talking about the entire regime of dilutions where not all of the ethyl acetate is solvated, making the amount of water important to the reaction kinetics. So, for instance, the rate will start to lose first order character as the mole ratio (water per ethyl acetate) approaches numbers of order 10 or less.

    Really, the condition for the reaction to be first order is the same as the condition for water to be completely solvating the acetate, which is when you have large mole ratios.
     
  18. Apr 19, 2005 #17

    GCT

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    The first order rate law may become invalid, however, I don't believe that there exists a situation where a first order rate law is modified to an equally valid second order equation in the situation of decreasing solvent; I don't believe that a second order equation would exist. I would say that this aspect of our argument has been concluded.

    as for your other assertion

    A second order reaction, as far as I know is always concentration dependent. Take a huge pool of water...add the two compounds pertaining to the rate equation....except that we make one compound in great excess to other. The reaction will still be second order.
     
    Last edited: Apr 19, 2005
  19. Apr 19, 2005 #18

    Gokul43201

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    The closest I've come to asserting a second order reaction is in saying that it will show more noticeable second-order characteristics (or something of that nature).

    I misuderstood your intent in your first post. I thought it was a response to the stated question rather than (what I now realize it probably was, namely) a rebuttal to the use of the word "concentration" by the OP, when refering to the abundance of water.
     
  20. Apr 20, 2005 #19
    Another possibility is that the reaction occurs in a solvent in which both ethyl acetate and water are miscible. In this case water would indeed have a meaningful activity (concentration).
     
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