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

1 Answer
Mar 19, 2016

Answer:

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)|)))#