# If the temperature coefficient #alpha# for a certain reaction is #2.5#, then if the reaction was run at #283^@ "C"# and #293^@ "C"#, what is the activation energy in #"kJ/mol"#?

##### 1 Answer

I got about:

#E_a = 6.5xx10^3# #"kJ/mol"#

From Wikipedia, the definition of the **temperature coefficient**

#(dk)/(k) = alphadT#

If we integrate this from state 1 to state 2 on the left side, and

#int_((1))^((2)) 1/k dk = int_(T_1)^(T_2) alphadT#

When we assume that the temperature coefficient stays constant across this small temperature range, we can pull

#ln(k_2) - ln(k_1)#

#= color(green)(ln(k_2/k_1)) = alpha(T_2 - T_1)#

#= (2.5)(293^@ "C" - 283^@ "C")#

#= color(green)(25)# where we used the fact that intervals in the celsius temperature scale are equal to intervals on the Kelvin temperature scale.

Thus, we have indirectly calculated **Arrhenius equation** that demonstrates the *temperature dependence* of the rate constant:

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

Therefore, we can now algebraically solve for and calculate the **activation energy**

#color(blue)(E_a) = -(Rln(k_2/k_1))/(1/T_2 - 1/T_1)#

#= -(Rln(k_2/k_1))/((T_1 - T_2)/(T_1T_2))#

#= -Rln(k_2/k_1)[(T_1T_2)/(T_1 - T_2)]#

#= -("0.008314472 kJ/mol"cdot"K")(25)[(("556.15 K")("566.15 K"))/(556.15 - 5"66.15 K")]#

#= color(blue)(6.5_(45)xx10^3)# #color(blue)("kJ/mol")# where the subscripts indicate digits past the last significant figure.

This means that the reactants have about a