The rate constant for the reaction: NO2(g)+O2(g)---->NO3(g)+O2(g) was determined over a range of 40.0 K, with the following results. Determine the activation energy for the reaction (kJ/mol) and calculate the rate constant of the reaction at 300K?

#E_a = "20.37 kJ/mol"#
#k = 2.051 xx 10^7 "M"^(-1)"s"^(-1)#

Does it make physical sense that #k# is higher at higher temperature? Why? If #E_a# is typically #50 - 150# #"kJ/mol"#, is this reaction fast or slow?

You should plot the data in Excel. Reactions following Arrhenius kinetics obey the Arrhenius equation.

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

Show that

#ln k = ln A - E_a/(RT)#

so that:

#barul|stackrel(" ")(" "ln k = -E_a/R 1/T + ln A" ")|#

Then notice how #y = ln k# and #x = 1/T#... from here, the slope is #-E_a/R# and y-intercept is #ln A#. Now make the graph:

Add the slope and y-intercept.

And obtain them. Did you finish the graph?

So then... since the slope is

#-E_a/R = -"2450.6 K"#,

#color(blue)(E_a) = -Rcdot"slope" = -"8.314 J/mol"cdot"K" cdot -"2450.6 K"#

#=# #"20374.3 J/mol"#

#= ulcolor(blue)"20.37 kJ/mol"#

The y-intercept is given as #ln A#, so...

#A = e^(25.005) = ul(7.237 xx 10^(10) "M"^(-1)"s"^(-1))#

And therefore, the rate constant at #"300 K"# is:

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

#= e^(25.005) "M"^(-1)"s"^(-1) cdot e^(-"2450.6 K"//"300 K")#

#= color(blue)ul(2.051 xx 10^7 "M"^(-1)"s"^(-1))#