Question

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

Answer 1

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"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 * "exp"(-E_a/(R * T_1))`

and

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

To find `T_2`, divide these two equations

`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 `T_2` on one side of the equation

`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)|)))`