Question

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?

T (K) k (M-1s-1)
203 4.14 x 105
213 7.30 x 105
223 1.22 x 106
233 1.96 x 106
243 3.02 x 106

Answer 1

`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?

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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))`