Question #d10dd

#E_a = "143.48 kJ/mol"#

This is just asking you to use the Arrhenius equation. In fact, it takes you through every major step.

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

• #k# is the rate constant at some temperature #T# in #"K"#.
• #A# is the frequency factor in #"s"^(-1)#.
• #E_a# is the activation energy.
• #R# is the universal gas constant.

Given two temperatures, one can obtain the #ln# form of the equation, assuming the frequency factor is the same for two different temperatures.

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

And one can readily see that taking the ratio is useful.

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

And so,

#ln(k_1/k_2) = -E_a/(RT_1) + E_a/(RT_2)#

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

is the #ln# form. You should have this written down in your notes somewhere already...

#A)#

This is simple enough; convert temperature from #""^@ "C"# to #"K"# and plug them in.

#25^@ "C" + 273.15 = "298.15 K"#

#43^@ "C" + 273.15 = "316.15 K"#

Therefore:

#1/T_2 - 1/T_1 = [1/316.15 - 1/298.15] "K"^(-1)#

#= -"0.0001910 K"^(-1)#

#B)#

The left-hand side is then:

#ln(k_1/k_2) = ln(0.100/2.70)#

#= -3.296#

#C)#

And so the activation energy is:

#color(blue)(E_a) = Rln(k_1/k_2) cdot [1/T_2 - 1/T_1]^(-1)#

#= "0.008314472 kJ/mol"cdotcancel"K" cdot (-3.296) cdot 1/(-0.0001910) cancel"K"#

#=# #color(blue)("143.48 kJ/mol")#

And this corresponds to an ordinary reaction rate.