# If the activation energy of a certain reaction is #"125 kJ/mol"#, and the reaction is at #"312.00 K"#, at what new temperature would its rate constant be twice as large?

##### 2 Answers

The rate constant will double at 317 K.

#### Explanation:

The **Arrhenius equation** gives the relation between temperature and reaction rates:

#color(blue)(bar(ul(|color(white)(a/a)k = Ae^(-E_"a"/(RT))color(white)(a/a)|)))" "#

where

If we take the logarithms of both sides, we get

#lnk = lnA - E_"a"/(RT)#

Finally, if we have the rates at two different temperatures, we can derive the expression

#color(blue)(bar(ul(|color(white)(a/a)ln(k_2/k_1) = E_"a"/R(1/T_1 -1/T_2)color(white)(a/a)|)))" "#

In your problem,

Now, let's insert the numbers.

I got

This is a fairly straightforward problem; the challenge is to interpret what the variables we are given mean, which equation to use, and how to use it.

#E_a = "125 kJ/mol"#

#k_1 = "0.200 s"^(-1)#

#k_2 = 2k_1#

#T_1 = "312.00 K"#

#T_2 = ?#

Each of these variables can be found in the **Arrhenius equation**

#k = Ae^(-E_a"/"RT)# ,

which can be written in the form of an initial

#k_1 = Ae^(-E_a"/"RT_1)# ,

#k_2 = Ae^(-E_a"/"RT_2)# ,

noting that for the **same** reaction at two **different** temperatures **identical**.

Therefore, we can solve for

#k_1/k_2 = e^(-E_a"/"RT_1)/e^(-E_a"/"RT_2)#

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

#= ln(e^(-E_a"/"RT_1)) - ln(e^(-E_a"/"RT_2))#

#= -(E_a)/(RT_1) + (E_a)/(RT_2)#

#= -(E_a)/(R)[1/(T_1) - 1/(T_2)]#

which we can recognize as the Arrhenius equation for *two different temperatures*

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

Now, setting

#ln(cancel(k_1)/(2cancel(k_1))) = -(E_a)/(R)[1/(T_1) - 1/(T_2)]#

#-ln(2) = -(E_a)/(R)[1/(T_1) - 1/(T_2)]#

Moving

#(Rln(2))/(E_a) = 1/(T_1) - 1/(T_2)#

Subtracting

#(Rln(2)T_1)/(E_aT_1) - (E_a)/(E_aT_1) = -1/(T_2)#

Reciprocating both sides of the equation:

#=> T_2 = -1/[(Rln(2)T_1)/(E_aT_1) - (E_a)/(E_aT_1)]#

#= -1/[(Rln(2)T_1 - E_a)/(E_aT_1)] = 1/[(E_a - Rln(2)T_1)/(E_aT_1)]#

Therefore:

#color(blue)(T_2) = [E_aT_1]/(E_a - Rln(2)T_1)#

#= [("125 kJ/mol")("312.00 K")]/(("125 kJ/mol") - ("0.008314472 kJ/mol"cdot"K")ln(2)("312.00 K"))#

#=# #color(blue)("316.55 K")#

(Note that we needed five sig figs to get the necessary precision. If we had written