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# 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 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"

so that...

rate = - d[A] / dt

where...

[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

and..

rate α [A]

becomes...

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

etc.

*******

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

right?

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?

becomes..

... .. ..[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