# A general rule of thumb is that a reaction #"10 K"# higher should be twice as fast. Predict how much faster a reaction is when the temperature is changed from #10^@"C"# to #100^@ "C"#?

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*NOTE: It does depend on the temperature range chosen. This is nontrivial.*

- Truong-Son

*NOTE: It does depend on the temperature range chosen. This is nontrivial.
- Truong-Son*

##### 1 Answer

About

First off, this is an approximation that assumes an activation energy of about

We write the general **rate law** for two trials, each at different temperatures and fixed concentration of

#r_1(t) = k_1[A]^n#

#r_2(t) = k_2[A]^n# where

#r# is the rate in#"M/s"# ,#k# is the rate constant in the appropriate units,#[A]# is the concentration of#A# in#"M"# , and#n# is the order of the reaction with respect to#A# (whatever it may be).

The **Arrhenius equation** relates the rate constants

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

Notice how if we write

#k_2/k_1 = (r_2(t)"/"cancel([A]^n))/(r_1(t)"/"cancel([A]^n))#

#= (r_2(t))/(r_1(t))# ,so what you should note is that

the ratio of the rate constants is the ratio of the rates.That means#k_2/k_1 = r_2/r_1# .

Now, consider a

#ln(r_2/r_1) = -("50 kJ/mol" xx "1000 J"/"1 kJ")/("8.314472 J/mol"cdot"K")[1/("298.15 K") - 1/("288.15 K")]#

#= 0.6999#

Therefore, the **ratio of the rates** is

#e^(ln(r_2"/"r_1)) = r_2/r_1 = color(green)(2.014) ~~ 2# ,

and the reaction **hotter** ran approximately **twice** as fast. Alright, so we chose an appropriate activation energy.

Now, consider a **change** in temperature from

We solve for

#ln(r_2/r_1) = -("50 kJ/mol" xx "1000 J"/"1 kJ")/("8.314472 J/mol"cdot"K")[1/("373.15 K") - 1/("283.15 K")]#

#=> color(blue)(r_2/r_1) = e^(5.1225)#

#=# #color(blue)(bb("167.7 times as fast"))#

Note: here is the same calculation as before, but for a temperature range you are not asked for, to check if the temperature range you are asked for actually matters.

#ln(r_2/r_1) = -("50 kJ/mol" xx "1000 J"/"1 kJ")/("8.314472 J/mol"cdot"K")[1/("343.15 K") - 1/("253.15 K")]#

#=> color(blue)(r_2/r_1) = e^(6.2304)#

#=# #color(blue)(bb("508 times as fast"))#

(Yeah, quite different!)