# Does the rate law change with temperature?

Aug 10, 2015

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:

$r \left(t\right) = {k}_{o b s} {\left[A\right]}^{m} {\left[B\right]}^{n}$

for a reaction such as:

$29 A + 17 B \to 23 C$

which might be represented as, for example:

29A + 53M " "stackrel(k_(1)" ") rightleftharpoons 29A'
${\text{ "" "" "" "" "" }}^{{k}_{- 1}}$

$29 A ' + 17 B \stackrel{{k}_{2} \text{ }}{\implies} 23 C$

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 ${k}_{o b s}$ (observed rate constant).

Note that this observed rate constant is intentionally labeled as such because if a reaction mechanism is complex, ${k}_{o b s}$ can be represented by individual rate constants for each corresponding mechanistic step (for example, maybe ${k}_{o b s} = \frac{{k}_{1} {k}_{2}}{{k}_{- 1}}$, where ${k}_{- 1}$ means rate constant for a reversed initial step).

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

$k = A {e}^{- {E}_{a} \text{/RT}}$

$\implies \textcolor{g r e e n}{\ln k} = \ln A + \ln {e}^{- {E}_{a} \text{/RT}}$

$= \textcolor{g r e e n}{\ln A - {E}_{a} / \left(R T\right)}$

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:

$\textcolor{g r e e n}{\ln {k}_{2} = \ln A - {E}_{a} / \left(R {T}_{2}\right)}$
$\textcolor{g r e e n}{\ln {k}_{1} = \ln A - {E}_{a} / \left(R {T}_{1}\right)}$

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

$\ln {k}_{2} - \ln {k}_{1} = - {E}_{a} / \left(R {T}_{2}\right) - \left(- {E}_{a} / \left(R {T}_{1}\right)\right)$

$= \frac{- {E}_{a}}{R} \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]$

Using log properties and doing some factoring, we get:

$\textcolor{b l u e}{\ln \left({k}_{2} / {k}_{1}\right) = - \frac{{E}_{a}}{R} \left[\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right]}$

which describes the change in the rate constant with the change in temperature. One can graph this by writing it in its $y = m x + b$ form:

$\textcolor{b l u e}{\ln k = - \frac{{E}_{a}}{R} \left[\frac{1}{T}\right] + \ln \beta}$

using $\ln k$ as the y coordinate, $\ln \beta$ as the y-intercept, $\frac{1}{T}$ as the x coordinate, and $- \frac{{E}_{a}}{R}$ as the slope. $\beta$ is an arbitrary constant.