# How do you solve the Arrhenius equation for #T_2#?

##### 1 Answer

Here's how you could do that.

#### Explanation:

The **Arrhenius equation** does not include a

However, you can use the Arrhenius equation to determine a **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

*rate constant* for a given reaction

*pre-exponential factor*, specific to a given reaction

*activation energy* of the reaction

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

You can use Arrhenius equation to write

#k_1 = A * "exp"(-E_a/(R * T_1))#

and

#k_2 = A * "exp"(-E_a/(R * T_2))#

To find

#k_1/k_2 = color(red)(cancel(color(black)(A)))/color(red)(cancel(color(black)(A))) * ("exp"(-E_a/(R * T_1)))/("exp"(-E_a/(R * T_2)))#

Use the property of exponents

#color(purple)(|bar(ul(color(white)(a/a)color(black)(x^a/x^b = x^((a-b)), AA x !=0)color(white)(a/a)|)))#

to rewrite the resulting equation as

#k_1/k_2 = "exp"[E_a/R * (1/T_2 - 1/T_1)]#

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

#ln(k_1/k_2) = ln("exp"[E_a/R * (1/T_2 - 1/T_1)])#

This will be equivalent to

#ln(k_1/k_2) = E_a/R * (1/T_2 - 1/T_1)#

Finally, do some algebraic manipulation to isolate

#ln(k_1/k_2) = E_a/R * 1/T_2 - E_a/R * 1/T_1#

#1/T_2 = ln(k_1/k_2) + E_a/R * 1/T_1#

Therefore,

#color(green)(|bar(ul(color(white)(a/a)color(black)(T_2 = (R * T_1)/(R * T_1 * ln(k_1/k_2) + E_a))color(white)(a/a)|)))#