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Reaction: Heat will slow down?

  1. Feb 23, 2006 #1
    I had a dispute with a professor where he argued that adding heat to a reaction will shorten the time it takes to reach equilibrium. I feel as though adding heat will increase collisions and and therefore increase the rate which A+B goes to C. However, if you are looking at the reverse, where C goes to A+B, I think that adding heat will cause the A+B reaction to speed up, causing the process to take longer. Can any one see a flaw in my analysis or if it is logical, point out any reactions where adding heat will actually slow the it down? Thanks.
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  3. Feb 23, 2006 #2


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    What do you mean causing the A+B reaction to speed up? It sounds like you're talking about a decomposition....
  4. Feb 23, 2006 #3
    Sorry, my question was unclear. What I mean is the reaction between A+B --> C will eventually reach an equilibrium. Adding heat to the forward reaction will cause this to happen sooner. Are there any reactions which will reach equilibrium slower with the addition of heat?
  5. Feb 23, 2006 #4


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    Please make a distinction between "heat" and "temperature," and restate your question.
  6. Feb 23, 2006 #5
    What kind of equlibrium?


  7. Feb 23, 2006 #6
    http://www.wpbschoolhouse.btinterne...we speed up or slow down chemical reactions?"

  8. Feb 23, 2006 #7
    I realize I am still being unclear, let me explain the question asked which caused this dilemma:

    3I- (aq) + S2O82- (aq) ---> I3- (aq) + 2 SO42- (aq)

    (sorry for not being familiar with the proper way to write this on this forum, but with the S2O82-, the 2 and 8 and subscripts and the 2- is the charge on the oxygen, the I3- is I with a -3 charge, the 2SO42-, 4 is the subscript and the 2- is the charge on it).

    During this redox reaction, we are assuming that the reaction goes to completion.

    "In an experimentm equal volumes of 0.0120 M I- and 0.0040 M S2O82- are mixed at 25ºC. The conentration of I3- over the following 80 minutes is shown in the graph below."
    - a graph showing the molar concentration of the I3- is on the y-axis andtime in minutes on the x-axis. As the time progresses, the concentration starts to increase fast then levels off at about 35 minutes, where it has reached equilibrium. The graph looks like a graph of y=square root of x, increasing in that fashion.

    Now it is asked, if the temperature is raised to 35ºC, how would the graph change. The answer is it would be steeper at the beginning, reaching the same equilibrium level at an earlier time.

    Now my question is: Is there ever a time when increasing temperature would cause the reaction (increasing [I3-] until equilibrium) to slow down. I am not speaking particular of this question, but in general, are there any instances when increasing temperature would cause the equliibrium point to be reached slower (later) ?
  9. Feb 24, 2006 #8
  10. Feb 24, 2006 #9


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    adding "heat" energy to a reaction medium will help in surpassing the activation energy barrier for both forward and reverse reactions. The reaction will speed towards equilibrium, however, the equilibrium constant may change. I may add to this later, but this should make things a lot clearer; you don't want to conclude on such absolutes though, in fact, under the right context perhaps the time it takes to reach equilibrium may slow down in a sense.
  11. Feb 26, 2006 #10


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    Adding heat to an exothermic reaction would cause a shift towards the reactants.

    Here is a restatement of what GCT wrote, which is right on:

    For a reaction to reach equillibrium you have to have enough energy around to get past the activation barriers. Adding heat gives more energy to the system, and therefore there is more energy to overcome the reaction barriers. How much heat is added can change where the equillibrium lies, but it will always speed up how fast equillibrium is reached because the barriers are more accessible.
  12. Feb 26, 2006 #11
    Thank you, I do understand that completely. What I was wondering though, is whether or not there will ever be a time when adding heat will cause equilibrium to be reached slower? I was thinking that if A+B gives you C and adding heat will cause more collisions, then of course the equilibrium will speed up, but if you started with all of C and wanted to reach equ. breaking C into A and B, will adding heat slow down how long it takes because as C breaks into A and B, more heat will cause more collisions of A and B going back to C? Was that clear?
  13. Feb 26, 2006 #12


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    That doesn't make any sense. You're focusing too much on one aspect of the reaction,

    the reaction between A and B is dependent upon the decomposition of C, that is every C that decomposes, produces A and B, and only from there A and B can combine to produce C Thus the decomposition of C is somewhat of a "limiting" rate.

    That is you've got to establish an equilibrium rate, that is when both forward and reverse rates are the same. You start with a higher rate of decomposition of C, the A+B reaction doesn't have an initial rate. As more A+B is produced, its rate of reaction can increase until it reaches the rate of decomposition of C (which is constant, assuming a simple reaction).
    Last edited: Feb 26, 2006
  14. Feb 26, 2006 #13
    But as the formation of A+B is increasing, adding heat would increase collisions and force A+B to go back to C before equilibrium hits.. but eventually the rate of C ->A+B will equal A+B --> C (equilibrium). So is it possible that adding heat during the decomposition of C would make the time it takes to reach equilibrium slow down?
  15. Feb 27, 2006 #14


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    no, not "before equilibrium hits" but rather a new equilibrium will be established. Remember equilibrium constants change with temperature. The equilibrium is fixed for a particular temperature, if you add heat, it will change.
  16. Feb 28, 2006 #15


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    I think the part that you are neglecting is that in your example adding more heat would speed up the decomposition of C into A + B. This decomposition could be unimolecular and therefore could be initiated by adding energy to the system and increasing bond rotations, etc. that might lead to decomposition to A + B. Also, the decomposition could be initiated by a collision of two molecules of C to give 2A + 2B or C + A + B; in either case increasing the temperature would increase the frequency of the collisions between two molecules of C.

    Also, the rate of the C -> A + B reaction will decrease as the concentration of C decreases, so it's not entirely that the rate of A + B -> C increases toward equilibrium, but also that C -> A + B decreases toward equillibrium.
  17. Mar 1, 2006 #16
    Google for "LE CHATELIER'S PRINCIPLE" ...

    ... and discover that adding heat (and therefore increasing temperature) displaces an equilibrium in the heat-consuming direction of the (equilibrium) reaction.

    This maybe linked to your question since displacing an equilibrium also mean "activating" or "accelerating" a reaction. Therefore the word "activity" used sometimes in analysing Gibbs energy changes under a reaction.

    This is only valid around an equilibrium state. For example, "activating" an explosive triggers a heat-producing reaction. This is because the initial state is not a thermodynamic equilibrium. Obviously, this is also what justifies the "LE CHATELIER'S PRINCIPLE", starting from the definition of an equilibrium. Tautology, isn't it?

    Hope it heats you up !
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