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    Why does temperature affect the rate constant K
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    (Original post by SIMZZ)
    Why does temperature affect the rate constant K

    take the reaction: A + B ----> C

    with rate equation Rate = K[A][B]

    If you increase the temperature, the rate would increase due to higher kinetic energy of molecules etc. The concs of A and B remain constant, so the only thing that can alter is K, which increases to fit the equation. Similarly if temperature drops, K drops. Hope this helps.
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    (Original post by noggin)
    take the reaction: A + B ----> C

    with rate equation Rate = K[A][B]

    If you increase the temperature, the rate would increase due to higher kinetic energy of molecules etc. The concs of A and B remain constant, so the only thing that can alter is K, which increases to fit the equation. Similarly if temperature drops, K drops. Hope this helps.
    thanks for the help, i understand now.
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    (Original post by SIMZZ)
    thanks for the help, i understand now.
    goodo - i also understand it better now having explained it!!
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    On a similar theme, could someone explain about second order reactions. I understand they increase the rate of reaction x4 when they are doubled, but what about when they are tripled etc?
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    (Original post by noggin)
    goodo - i also understand it better now having explained it!!
    lol!
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    (Original post by kingpong)
    On a similar theme, could someone explain about second order reactions. I understand they increase the rate of reaction x4 when they are doubled, but what about when they are tripled etc?
    it increases by a factor of 9. when doubles it increases by a factor of because 2 squared is 4 and when tripled the rate increases by a factor of 9 because 3 squared is 9. Basically the rate is proportional to concentration squared.
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    (Original post by kingpong)
    On a similar theme, could someone explain about second order reactions. I understand they increase the rate of reaction x4 when they are doubled, but what about when they are tripled etc?
    The orders are the powers to which concentrations are raised in the rate equation. Hence for a first order reaction, doubling or trebling the concentration of a reactant increases rate by a factor of 2 or 3 respectively, while in a second order reaction factors of 4 or 9, and so on.
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    (Original post by kingpong)
    On a similar theme, could someone explain about second order reactions. I understand they increase the rate of reaction x4 when they are doubled, but what about when they are tripled etc?
    i got stuck with this just yesterday - i found the easiest way to make it make sense was to write out the rate equation:

    rate = k[A][B]^2

    now if you bung in random numbers, say to start with k=1, A=1, B=1
    so the rate at this point = 1

    now if you double B keeping A constant, see what happens to the rate:
    rate = 1x1x2^2
    = 4 (rate has quadrupled)

    now if you triple B, see what happens....
    rate = 1x1x3^2
    = 9 (rate has ninedrupled!! is that a word?!)
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    Ah, makes perfect sense now, don't know why I didn't notice that.


    Cheers all.
 
 
 
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