# How do you solve the Arrhenius equation for T_2?

Mar 19, 2016

Here's how you could do that.

#### Explanation:

The Arrhenius equation does not include a ${T}_{2}$, it only includes a $T$, the absolute temperature at which the reaction is taking place.

However, you can use the Arrhenius equation to determine a ${T}_{2}$, provided that you know a ${T}_{1}$ and the rate constants that correspond to these temperatures.

The Arrhenius equation looks like this

color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" ", where

$k$ - the rate constant for a given reaction
$A$ - the pre-exponential factor, specific to a given reaction
${E}_{a}$ - the activation energy of the reaction
$R$ - the universal gas constant, useful here as $8.314 {\text{J mol"^(-1)"K}}^{- 1}$
$T$ - the absolute temperature at which the reaction takes place

So, let's say that you know the activation energy of a chemical reaction you're studying.

You perform the reaction at an initial temperature ${T}_{1}$ and measure a rate constant ${k}_{1}$. Now let's say that you're interested in determining the temperature at which the rate constant changes to ${k}_{2}$.

You can use Arrhenius equation to write

${k}_{1} = A \cdot \text{exp} \left(- {E}_{a} / \left(R \cdot {T}_{1}\right)\right)$

and

${k}_{2} = A \cdot \text{exp} \left(- {E}_{a} / \left(R \cdot {T}_{2}\right)\right)$

To find ${T}_{2}$, divide these two equations

${k}_{1} / {k}_{2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{A}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{A}}}} \cdot \left(\text{exp"(-E_a/(R * T_1)))/("exp} \left(- {E}_{a} / \left(R \cdot {T}_{2}\right)\right)\right)$

Use the property of exponents

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{x}^{a} / {x}^{b} = {x}^{\left(a - b\right)} , \forall x \ne 0} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to rewrite the resulting equation as

${k}_{1} / {k}_{2} = \text{exp} \left[{E}_{a} / R \cdot \left(\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right)\right]$

Next, take the natural log of both sides of the equation

$\ln \left({k}_{1} / {k}_{2}\right) = \ln \left(\text{exp} \left[{E}_{a} / R \cdot \left(\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right)\right]\right)$

This will be equivalent to

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

Finally, do some algebraic manipulation to isolate ${T}_{2}$ on one side of the equation

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

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

Therefore,

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{T}_{2} = \frac{R \cdot {T}_{1}}{R \cdot {T}_{1} \cdot \ln \left({k}_{1} / {k}_{2}\right) + {E}_{a}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$