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NO2 (g) reacts with O2 4 NO2 (g) + O2(g) to produce N2 O5 (g) as shown by the equation below: (g) = 2 N2 O5 (g).
Experimentally the rate orders were determined and rate law written as shown below: Rate = k [NO2]2[O2 ].
Calculate the value of k if the initial concentration of NO2 was 0.250 M and initial
concentration of O2 (g) was 0.150 M. The initial rate was 1.25 x 10-2 M.s-1 .

Respuesta :

sebassandin

Answer:

[tex]k=1.33M^{-2}s^{-1}[/tex]

Explanation:

Hello there!

In this case, according to the given information for the definition of the rate law, which is second-order with respect to NO2 and first-order to O2, we can solve for k as both concentrations are given as well as the initial rate of reaction:

[tex]k=\frac{r}{[NO_2]^2[O_2]}[/tex]

In such a way, we can just plug in the given values to obtain the correct rate constant with the appropriate units:

[tex]k=\frac{1.25x10^{-2}Ms^{-1}}{(0.250M)^2(0.150M)}\\\\k=1.33M^{-2}s^{-1}[/tex]

Regards!

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