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Factors Affecting the Rate of a Reaction

  1. Jun 6, 2009 #1
    I've just completed a simulated experiment concerning the reaction between iodine ions and persulphate ions:

    2I-(aq)+S2O82-(aq)-->I2(aq)+2SO42-(aq) (1)

    The iodine produced from this reaction, in turn reacts with thiosulphate ions:

    I2(aq)+2S2O32-(aq)-->2I-(aq)+S4O62-(aq) (2)

    Reaction (2) is the more rapid of the two, which means that reaction (1) is the rate-determing step.

    My question is, how do I determine the rate law of this reaction with the follwing experimental data:

    Mixture----[I-] (mol/L)----[S2O82-] (mol/L)----Time (s)

    --1---------0.10--------------0.050------------20
    --2---------0.075-------------0.050------------28
    --3---------0.050-------------0.050------------41
    --4---------0.025-------------0.050------------84
    --5---------0.10--------------0.038------------25
    --6---------0.10--------------0.025------------39
    --7---------0.10--------------0.013------------82

    My attempt at a solution:

    From the data, I've concluded that as the concentration of the persulphate ions is kept constant, doubling the concentration of the iodine ions resuls in halving the time taken for a reaction to occur. Also, as the concentration of the iodine ions is kept constant, the time is again halved each time the concentration of the persulphate ions is doubled.

    Therefore, the rate law equation will take the following form:

    r=k[I-]m[S2O82-]n

    From this point I am somewhat lost. I'm assuming that the equation can be re-written to solve for m and n as:

    log r = log k + m log[I-] + n log[S2O82-]

    However, I'm not entirely sure how to apply the experimental data to this equation. Some guidance would be appreciated.

    Thanks in advance.
     
  2. jcsd
  3. Jun 7, 2009 #2

    Redbelly98

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    If the reaction time is halved, then the reaction rate is ____? That should help with finding m and n.
     
  4. Jun 7, 2009 #3
    If the reaction time is halved then the reaction rate is doubled right?

    Does this mean that the rate is increased by a factor of 2 meaning that the rate of reaction is proportional to the square of the concentration of both the iodide ions and the persulphate ions?
     
  5. Jun 7, 2009 #4

    Redbelly98

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    Yes, exactly.

    No. If it were proportional to the square, then doubling the concentration would cause the rate to quadruple:

    r1 = k' c12
    r2 = k' c22

    c2 = 2c1r2 = k' (2c1)2 = 4 k' c12 = 4r1


    What must the exponent be in order to get:
    r2 = 2r1
    ?
     
  6. Jun 7, 2009 #5
    Oh, so does that just mean that the rate of reaction is proportional to the concentration of both iodide and persulphate ions? meaning that the exponents m and n, must be 1:

    r = k[I-]1[S2O82-]1

    r = k[I-][S2O82-]

    Therefore, as the concentration of either substance is doubled, the rate of reaction is also doubled?
     
  7. Jun 7, 2009 #6

    Redbelly98

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    Yes, you got it! :smile:
     
  8. Jun 7, 2009 #7
    Awesome! Thanks a lot.

    P.S. Breaking up the rate law equation into two separate equations really helped.
     
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