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Rates of reactions

  1. May 2, 2005 #1

    RPN

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    Hello,
    I really want to thank everyone that has helped me in understanding chemistry. I have started a few questions on rates of reactions and would love it if someone could look over my answers and see if I am headed in the right direction.
    1)
    The rate of a simple chemical reaction like,
    NO(g) + ½ O2(g) ---- NO2(g), are most rapid at the beginning of the reaction due to several reasons. First the reaction occurs in a single step and is called a bimolecular reaction. Secondly rate of reactions are affected by the reactants concentrations. In the beginning of a reaction the concentrations of the reactants are at their highest and more collisions between reactant molecules occur. As we know concentrations change most rapidly at the beginning of a reaction. This equation consists of simple ions that combine in a one-to-one ratio. With all these factors combined it can be predicted that this reaction would be rapid in the beginning.

    2)
    In three ways I will dissolve a lump of sugar in water by increasing the rate at which it will dissolve.
    Method one: To increase the rate of dissolution of sugar I will increase the temperature of the water. By increasing the temperature of the water I will in turn be increasing the kinetic energy and the threshold energy (activation energy). When the activation energy is decreased the reaction rate increases. When the kinetic energy is increased this translates to an increased frequency of collisions as the particles are moving more rapidly. This increases their chances for collisions thus increasing the chance of successful collisions. It is said that a temperature increase of 10oC can either double or triple the rate of reaction.
    Method two: I could crush the lump of sugar into granules and thus increasing the surface area. Increasing the surface area of the solid phase in a heterogeneous reaction increases the rate of reaction. Comparing the rate of reaction of the lump of sugar in water to the granules of sugar in water, the lump dissolves more slowly than the granules.
    Method three: I could use a catalyst in order to speed of the rate of reaction. A catalyst is any reagent that increases the rate of reaction but is not consumed during the reaction. By adding a chemical substance we give a reaction a new route to take in order to get the products. Catalysts increase reaction rates by providing a lower activation energy pathway for the reactants. The catalysts are assumed to form an intermediate activated complex that has a lower activation energy.

    3)
    a) The effect of a temperature increase on the rate constant of the forward reaction would be an increase in the reaction. An increase in temperature would increase the energy of the system closer to the activation energy and speed the reaction up.
    b) The effect of a temperature increase on the rate constant of the reverse reaction would also be an increase in the rate of reaction. Increasing the temperature increases the rate of the reverse reaction to a greater extent than it does the rate of the forward reaction.

    Thank you so much
     
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  3. May 3, 2005 #2

    GCT

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    1)most reactions are rapid at the beginning, as the forward reaction progresses, the rate of the reverse reaction increases due to the increase in concentration of the products and this is what causes the slowing down of the overall reaction.

    2)I'm not quite sure if activation energy applies to this question. Some teachers would consider this progress related to a physical change with no chemical changes since no chemical reactions (breakage of bonds, new compounds being formed) are actually occuring. Increasing the temperature will certainly increase solubility. Solubility increases with temperature (not with gaseous substances however) in this case I believe, however this is not true for all cases and the degree of the effect varies with different solvents. I'm not even sure about the case with sugar and water. I would imagine that it does, since sugar is fairly soluble in water, and the energy associated with increasing the temperature would facilitate the dissociation of the crystals.

    3.Where's the question? No question...no answer, I'm not able to deduce the original question in this case...and in all cases you should post the actual question.
     
  4. May 3, 2005 #3

    RPN

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    Thanks for your help with the other answers.
    The original question states:
    What would be the effect of an increase in temperature on the rate constant of
    a) the forward reaction?
    b) the reverse reaction?
     
  5. May 3, 2005 #4

    RPN

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    And another question I have is when I am given the activation energy for a forward reaction and I have found the heat of reaction how do I find the activation energy for the reverse reaction.
     
  6. May 3, 2005 #5

    Gokul43201

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    That the rate is the highest at the beginning follows directly from this last statement; so the following statement is both unnecessary as well as counterproductive in that it assumes the thing that needs to be proved.
    Alternatively, you can look at it from the product side, as GCT has shown.
     
  7. May 4, 2005 #6

    GCT

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    in most cases, the rate constants of both forward and reverse reactions will increase with an increase in temperature, more energy is given for both reactions to occur. However, the relative increase in rate in most cases is not the same. Rates always increase with temperature, as far as I know.

    Draw yourself a energy diagram. In general (activation energy-heat of reaction), remember to use correct signs. Do you understand now?
     
  8. May 12, 2005 #7
    Confused..

    Hi, I've been working on the same material and I just can't understand how to determine the heat of reaction. Unfortunately I have not been supplied with a textbook to refer to and as most of the pages in my question booklet tell me to refer to page so and so, i'm utterly lost. I've tried to read things on the internet about it, but it doesn't help me. I've seen many people refer to using the Arrhenius equation to determine it. I dont know where to link the equation to the answer, so any help would be appreciated. Thank you
     
  9. May 13, 2005 #8

    Gokul43201

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    When asking about a specific question, please reproduce the question here exactly, lest there be any ambiguity about what is required. You have not told us what you are given. I imagine it is the forward and reverse reaction rates of activation energies, but unless you specify this, it would be wasteful to "describe" how to draw/interpret diagrams of reaction energetics.
     
  10. May 13, 2005 #9

    GCT

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    If you're referring to determining defference enthalpies of reactions, heat of reaction, activation energy, reverse activation energy etc...the answers are in the previous posts.
     
  11. May 13, 2005 #10

    GCT

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    [tex] Activation~energy_{forward} - \Delta H_{foward} = Activation~ energy_{reverse} [/tex] It is quite essential for you to comprehend this from the energy diagram perspective.
     
  12. May 13, 2005 #11
    The question.

    The question is:

    For the reaction

    CO + NO2 --> CO2 + NO

    the activation energy for the forward reaction is 135 kJ/mol of CO reacted.

    Determine the heat of reactoin

    Determine the activation energy for the reverse reaction

    Draw and label a potential energy diagram for the reaction


    ... if anyone has any good sources on drawing potential energy diagrams, I'm sure I can take it from there, but I'd also like to know how you get the heat of reaction..is it through the diagram? or through an equation? I'm lost in that part of the question and I have no text to support me, thanks again.
     
  13. May 13, 2005 #12

    GCT

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    You'll need find each compound's standard enthalpy of formation, which should be listed explicitly in your text. From this you'll need to find the net standard enthalpy of reaction. The rest should be simple.

    I don't understand why you're having such a difficult time finding a energy diagram, I'm quite positive that they should at least have an example in your text.

    Your text should also describe the general process on how to finish up this whole problem. You don't even need directions, one peek at the energy diagram and you should be able to figure everything out.
     
  14. May 18, 2005 #13
    That's exactly the problem, I do not have a text, I am doing an independent learning course to get my Chem 12 credit. When I went to apply they gave me the booklet with the questions, which constantly tells me to read certain pages in a textbook, unfortunately they also told me that I would not be able to get this textbook from them because they are out of stock. Quite a stupid system if you ask me.... to learn from a book that might as well not exist. Anyways, I guess I'll just try to look up the process on the internet as I have been with all the other questions. I just find it tedious because alot of the websites present only general information without having in depth explanations to the processes of actually getting the answers. Thank you for your help though, I'll try to use the enthalpy stuff to figure this out.
     
  15. May 18, 2005 #14

    Gokul43201

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    [tex]E_{act} (forward) = E(AC) - E(Reac)[/tex]
    [tex]E_{act} (reverse) = E(AC) - E(Prod)[/tex]

    [tex]\Delta H(rxn) \equiv E(Prod) - E(Reac) = E(Prod) - [E(AC) - E(AC)] - E(Reac) = E_{act} (forward) - E_{act} (reverse) [/tex]
     

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  16. Jun 14, 2005 #15
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