What does K, the rate constant mean in chemistry?

I'm currently studying chemistry at college and we are doing rates at the moment. We've been doing a lot of calculations with K, rate constant which I understand, and I get about zero, first, second order. Though whilst revising at home, I just wondered what does rate constant mean? When I working out the rate constant, what does the resultant number tell us? I've googled the definition but I don't really get what it means.

I'd be really grateful for any help, thanks!

1 Answer

  • Dr W
    Lv 7
    8 years ago
    Favorite Answer

    the "k" is the constant we insert to solve the proportionality... read on


    the basic idea is this..

    if we start with the reaction

    A --> B + C

    we can define rate as change in amount of A present per time. And if volume is constant, since concentration = amount / volume... rate = change in concentration of A per time

    in chemistry we express concentration with [ ] and in math we express "change in " with the letter "d"

    so that...

    rate = - d[A] / dt


    [A] means concentration of A

    d[A] / dt means change in concentration of A per change in time

    - is because the concentration of A is decreasing. This makes rate +


    in addition, we can say the higher the concentration of A, the faster the rate.

    ie.. rate is proportional to [A]

    ie.. rate α [A]

    now in math we solve proportionalities by inserting a constant to make the equation an equality. In this case, the constant is k


    rate α [A]


    rate = k x [a]

    now because we have different mechanisms in chemical reactions, sometimes 1 A decomposes by itself, sometimes we need 2 or more to collide, we have an exponent on that [A] and..

    rate = k x [A]^n

    where n is the order of the reaction

    n=0 is zero order

    n=1 is 1st order

    n=2 is 2nd order



    combining those 2 equations

    rate = - d[A] / dt = k x [A]^n


    now we have this differential equation that we need to solve

    -d[A] / dt = k x [A]^n


    so how do we do it?

    if n=0

    -d[A] / dt = k x [A]^0

    -d[A] / dt = k.. because X^0 = 1..

    d[A] = -k dt.. rearranging

    and now we need to integrate. And we integrate from the point where t=0 and [A] = [Ao] to the point t = t and [A] = [At]... meaning we start with [Ao] at time = 0 then after some time.. "t".. has elapsed we have concentration = [At]

    ∫d[A] = -k ∫dt.. k is a constant right?


    ... .. ..[A] = [At].. ... .... .... ...t = t

    [A].. .|.. .. ... ... ... .=.-k x t.. |

    ... . .. [A] = [Ao]... ... ... .... ..t = 0

    which becomes

    [At] - [Ao] = -k x (t - 0)

    and finally

    [At] = -kt + [Ao]


    likewise.. if n=1

    -d[A] / dt = k x [A]^1

    1/[A] d[A] = -k dt

    ln[At] = -kt + ln[Ao]

    and if n=2

    -d[A] / dt = k x [A]^2

    1/[A]^2 d[A] = -k dt

    1/[At] = +kt + 1/[Ao]


    so.. what do we do with these 3 equations?

    (1).. [At] = -kt + [Ao]... ... . n=0

    (2).. ln[At] = -kt + ln[Ao]... n=1

    (3)..1/[At] = +kt + 1/[Ao]... n=2


    well if you notice they are all of the form

    y = mx + b

    if we let y = [A], ln[A], and 1/[A] and x = time

    so if...

    n=0... a plot of [A] vs t.. will give a straight line with slope = -k and int = [Ao]

    n=1... a plot of ln[A] vs t.. will give a straight line with slope = -k and int = ln[Ao]

    n=2... a plot of 1/[A] vs t.. will give a straight line with slope = +k and int = 1/[Ao]


    check out my answer here for a detailed problem/solution


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