# A chemist's rule of thumb is that #10^@ "C"# of difference results in a two-fold change in the reaction rate. If the temperature is varied from #35^@ "C"# to #45^@ "C"# for the rxn of crystal violet with hydroxide, by what factor does the rate multiply?

##### 1 Answer

You can use the Arrhenius equation.

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

#A# is the frequency factor,#E_a# is the activation energy,#R# is the universal gas constant, and#T# is the temperature in#"K"# .

For the same reaction at most temperatures and pressures, the activation energy is assumed to not change. It is certainly true that

Thus, all that can vary are

#r_i(t) = k_i[A]_i^m[B]_i^n#

When we assume the concentrations are held constant as well to observe the effects of temperature change, we thus have:

#(r_i(t))/(r_j(t)) = (k_icancel([A]_i^m[B]_i^n))/(k_jcancel([A]_i^m[B]_i^n))#

for trial *Thus, the initial rates of reaction are directly proportional to the rate constants.*

As a result, we can therefore find the ratio of the rate constants, which is also the ratio of the initial rates.

#k_i/k_j = (cancel(A)e^(-E_a"/"RT_i))/(cancel(A)e^(-E_a"/"RT_j))#

#= e^(-E_a"/"RT_i)*e^(E_a"/"RT_j)#

#= e^(-E_a"/"RT_i + E_a"/"RT_j)#

#= e^(-E_a/R(1/(T_i) - 1/(T_j))#

#= "exp"[-E_a/R(1/(T_i) - 1/(T_j))]#

If we take

A typical experimental result for the activation energy for this reaction of

#k_i/k_j = color(blue)((r_i(t))/(r_j(t)))#

#= "exp"[-"50000 J/mol"/("8.314472 J/mol"cdot"K")*(1/"318.15 K" - 1/"308.15 K")]#

#= color(blue)(1.847)#

So it's not a bad estimate. It's somewhat close to