A well-known aphorism in chemistry is that for every 10° increase in temperature, the rate of a reaction will increase twofold. If the reaction were performed at 45° C predict whether it would occur at twice the rate of the reaction performed at 35° C ?

Yeah, it would, more or less. But it depends on its activation energy, and whether it is close to #"50 kJ/mol"# or not... You'll be within #10%# error if #E_a = 45 - "60 kJ/mol"#.

Well, let's first determine what average activation energy is required to fit to this "aphorism". The well-known Arrhenius equation states:

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

and we know that #k prop r(t)#, the rate of reaction. Hence,

#(r_2(t))/(r_1(t)) = k_2/k_1 = e^(-E_a/R [1/T_2 - 1/T_1])#

since #E_a# is hardly dependent on temperature. We assume that #k_2/k_1 = 2#, and solve for the #E_a# such that this is true.

#color(green)(E_a) = -(Rln(2))/(1/T_2 - 1/T_1) = -(Rln2)/(1/(T+10) - 1/T)#

#= -(Rln2)/(T/(T(T+10)) - (T+10)/(T(T+10)))#

#= -(Rln2)/(-10/(T(T+10))#

#= ((T+10)/10)RTln2#

#= color(green)((T/10 + 1)RTln2)#

where #T# is a lower temperature and #T+10# is #"10 K"# higher. #R = "8.314472 J/mol"cdot"K"#.

Do NOT be confused into thinking that #E_a# depends on the temperature.

All this says is that in order to manage a rate at #T+10# to be twice the rate of #T#, the reaction has to be chosen so that the activation energy is a certain value (i.e. via a catalyst), and depending on the temperature, only certain reactions will follow this "rule".

If we pick #T = "298.15 K"#, then #E_a = "52.95 kJ/mol"#.

We simplify this to find that at standard temperatures, this rule typically holds true if #E_a ~~ "50 kJ/mol"#, a relatively fast reaction.

So let's see if #k_2/k_1 = 2# if we choose a different temperature range. We begin from mid-derivation...

#E_a = (T/10 + 1)RTln(k_2/k_1)#

#ln(k_2/k_1) = E_a/((T/10 + 1)RT)#

#=> color(blue)(k_2/k_1) ~~ "exp"[("50000 J/mol")/(("308.15 K"/10 + 1)("8.314472 J/mol"cdot "K" cdot "308.15 K"))]#

#= color(blue)(1.92)#

Not bad... yeah, it's pretty close to #2#. Verify that if #E_a = "52.95 kJ/mol"#, it would be exactly #2#.