# What makes an increase in temperature change the reaction rate?

Mar 23, 2016

The higher the temperature, the faster the rate of (a non-biological) reaction.

CONCEPTUAL APPROACH

Temperature is directly proportional to the average kinetic energy of the particles in a system.

An increase in temperature increases the average kinetic energy of the particles, which makes them move faster. That makes them collide more frequently and thus react more frequently.

Therefore, the rate of reaction increases at higher temperatures.

MATHEMATICAL APPROACH

A mathematical way of proving this is this. Suppose we have a reaction $i$ with rate ${r}_{i} \left(t\right)$ as a function of time. The rate law for a basic first-order reaction is:

${r}_{2} \left(t\right) = {k}_{2} \left[A\right]$
${r}_{1} \left(t\right) = {k}_{1} \left[A\right]$

Let ${k}_{1}$ and ${k}_{2}$ be rate constants for temperature ${T}_{1}$ and ${T}_{2}$, respectively. Then, using the Arrhenius equation, we have:

$\textcolor{g r e e n}{{k}_{2} = A {e}^{- {E}_{a} \text{/} R {T}_{2}}}$
$\textcolor{g r e e n}{{k}_{1} = A {e}^{- {E}_{a} \text{/} R {T}_{1}}}$

where:

• $A$ is the pre-exponential factor (it won't matter here), because it comes before $e$, the exponential function (kind of obvious where the name came from now, right?)
• ${E}_{a}$ is the activation energy, which does NOT depend on temperature.
• $R$ is the universal gas constant $\text{8.314472 J/mol"cdot"K}$.
• ${T}_{i}$ is the temperature of reaction $i$ in units of $\text{K}$.

Now, if I divide these, I get a ratio between ${k}_{2}$ and ${k}_{1}$, through which we can compare the rates ${r}_{i} \left(t\right)$:

${k}_{2} / {k}_{1} = \left(\cancel{A} {e}^{- {E}_{a} \text{/"RT_2))/(cancel(A)e^(-E_a"/} R {T}_{1}}\right)$

ln\frac(k_2)(k_1) = ln\frac(e^(-E_a"/"RT_2))(e^(-E_a"/"RT_1))

$= \ln {e}^{- {E}_{a} \text{/"RT_2) - lne^(-E_a"/} R {T}_{1}}$

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

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

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

This is a useful equation; if you know the rate constants for a reaction (which you can get experimentally), and you know the starting and final temperature in a temperature ramp, you can determine the reaction's activation energy from the slope.

From this resultant equation, we see that if ${T}_{2} > {T}_{1}$, $\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}$ is larger and negative (since $\frac{1}{{T}_{2}} < \frac{1}{{T}_{1}}$ if ${T}_{2} > {T}_{1}$).

Therefore, since $\ln$ is positive when the argument (${k}_{2} / {k}_{1}$) is greater than $1$, and a negative times a negative is positive, ${k}_{2} > {k}_{1}$.

From this, we see that since ${k}_{2} \leftrightarrow {r}_{2} \left(t\right)$ and ${k}_{1} \leftrightarrow {r}_{1} \left(t\right)$, when ${k}_{2} > {k}_{1}$, it is also true that ${r}_{2} \left(t\right) > {r}_{1} \left(t\right)$.

Thus, the higher the temperature, the faster the rate of (a non-biological) reaction.