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!
- Dr WLv 78 years agoFavorite 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"
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?
-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
[At] - [Ao] = -k x (t - 0)
[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
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