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    This question is quite easy save for the last part, but I need to show the earlier parts otherwise you won't be able to help.

    The Following reaction
    2N2O5 ---> 4NO2 + O2

    is described as first order with respect to N2O5

    a) Write down the rate expression for the reaction

    answer: rate = k [N2O5]

    b) Explain what is meant by the term half life for this reaction
    answer: Time taken for the concentration of N2O5 to fall to half its initial value

    c) State what is characteristic about the half life of a first order reaction
    answer: the half life is constant

    d) The rate constant at a certain temperature for the reaction is 0.0052 (standard units) calculate the time taken for the concentration of N2O5 to decrease from 0.1 mol dm-3 to 0.01 mol dm-3

    answer: ?? (my intelligent guess is 5.2 seconds..)
    I would be very grateful for any help!
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    (Original post by andymt)
    This question is quite easy save for the last part, but I need to show the earlier parts otherwise you won't be able to help.

    The Following reaction
    2N2O5 ---> 4NO2 + O2

    is described as first order with respect to N2O5

    a) Write down the rate expression for the reaction

    answer: rate = k [N2O5]

    b) Explain what is meant by the term half life for this reaction
    answer: Time taken for the concentration of N2O5 to fall to half its initial value

    c) State what is characteristic about the half life of a first order reaction
    answer: the half life is constant

    d) The rate constant at a certain temperature for the reaction is 0.0052 (standard units) calculate the time taken for the concentration of N2O5 to decrease from 0.1 mol dm-3 to 0.01 mol dm-3

    answer: ?? (my intelligent guess is 5.2 seconds..)
    I would be very grateful for any help!
    You have already ascertained that the reaction is first order and that the half life of a first order reaction is constant.

    Using the expression t1/2 = 0.693/k

    you get t1/2 = 133.3 s


    now take it from there...
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    answer: ?? (my intelligent guess is 5.2 seconds..)
    I would be very grateful for any help![/QUOTE]

    I think as mentioned by charco,

    1/100 = (1/2)^n x 1/10

    Final conc = no.of half lifes x initial concentration

    you would get like n = ln10/ln2

    half life = ln2/k(or ln2 =0.693, but better be exact)

    so total time taken = no.of half lifes x n
    = ln10/k

    I think i might have done that correctly, but if someone could tell me what i have done wrong, that'd be appreciated too.
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    so you need to work out how many half lives the decrease in concentration represents? how do you do that?
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    OK the form is ln(N0/Nt) = -kt

    Where N0 is the initial amount
    and Nt is amount remaining at time t

    therefore the time taken is ln(N0/Nt)/k

    = ln(0.1/0.01)/0.0052 = 442.8 s
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    (Original post by charco)
    OK the form is ln(N0/Nt) = -kt

    Where N0 is the initial amount
    and Nt is amount remaining at time t

    therefore the time taken is ln(N0/Nt)/k

    = ln(0.1/0.01)/0.0052 = 442.8 s
    Thanks a lot
    but why is the minus sign relevant?
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    (Original post by andymt)
    Thanks a lot
    but why is the minus sign relevant?
    You got me thinking and I noticed that I put the relationship upside down!

    It should be ln(Nt/N0) = -kt

    The way I wrote it before gives you a negative answer (that's why the negative sign is relevant)!

    ln(0.01/0.1)/0.0052 = -t

    time taken = 442.8 s
 
 
 
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