# Does the rate law change with temperature?

##### 1 Answer

No, only the **rate constant** does. The form of the rate law is always the same.

Normally, at higher temperatures, the rate constant increases. (In some cases though, like thermal denaturation of a protein, the rate constant decreases.)

Some people confuse the rate *law* with the rate *constant*. The rate *law* is something like the following:

#r(t) = k_(obs)[A]^m[B]^n#

for a reaction such as:

#29A + 17B -> 23C#

which might be represented as, for example:

#29A + 53M " "stackrel(k_(1)" ") rightleftharpoons 29A'#

#" "" "" "" "" "" "^(k_(-1))#

#29A' + 17B stackrel(k_2" ")(=>) 23C# where

#M# is a catalyst and#A'# is an intermediate. The absurd stoichiometric coefficients only emphasize that#m# and#n# don't correspond to these.

The rate *constant* is the

Note that this *observed* rate constant is intentionally labeled as such because if a reaction mechanism is complex,

That aside, the rate *constant* changes with temperature. We can derive the equation starting from the initial form of the Arrhenius equation:

#k = Ae^(-E_a"/RT")#

#=> color(green)(ln k) = lnA + lne^(-E_a"/RT")#

#= color(green)(ln A - E_a/(RT))# where

#A# is the "pre-exponential factor",#E_a# is the activation energy (energy barrier),#R# is the universal gas constant, and#T# is temperature.

Thus, you get:

#color(green)(ln k_2 = ln A - E_a/(RT_2))#

#color(green)(ln k_1 = ln A - E_a/(RT_1))#

...for the same reaction at two different temperatures. If you subtract these, you get:

#ln k_2 - lnk_1 = - E_a/(RT_2) - (- E_a/(RT_1))#

#= (-E_a)/(R)[1/(T_2) - 1/(T_1)]#

Using log properties and doing some factoring, we get:

#color(blue)(ln (k_2/k_1) = -(E_a)/(R)[1/(T_2) - 1/(T_1)])#

which describes the change in the rate *constant* with the change in temperature. One can graph this by writing it in its

#color(blue)(ln k = -(E_a)/(R)[1/T] + lnbeta)#

using