Solution: Dinitrogen pentoxide (N₂O₅) decomposes to NO₂ and O₂ at relatively low temperatures in the following reaction: 2N₂O₅(aq) → 4NO₂(aq) + O₂(g) This reaction is carried out in a CCl ₄ solution at 45°C. The data are described in the table as well as the plots of the concentration versus t, ln concentration versus t, and 1/concentration versus t and results are below. Determine the rate law and calculate the rate constant. a. Zero order/ 5.3 x10-5 s -1     b. Second order/ 4.8 x 10−4 s −1 M-1      c. First order/5.3 x 10-5 s -1 M-1 d. Second order/ 2.9 x 10−4 s −1 M-1               e. First order/4.8 x 10−4 s −1

Solution: Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction: 2N2O5(aq) → 4NO2(aq) + O2(g) This reaction is carried out in a CCl 4 solution at 45°C.

Problem

Dinitrogen pentoxide (N₂O₅) decomposes to NO₂ and O₂ at relatively low temperatures in the following reaction:

2N₂O₅(aq) → 4NO₂(aq) + O₂(g)

This reaction is carried out in a CCl ₄ solution at 45°C. The data are described in the table as well as the plots of the concentration versus t, ln concentration versus t, and 1/concentration versus t and results are below. Determine the rate law and calculate the rate constant.

a. Zero order/ 5.3 x10^-5 s ^-1     b. Second order/ 4.8 x 10⁻⁴ s ⁻¹ M^-1      c. First order/5.3 x 10^-5 s ^-1 M^-1

d. Second order/ 2.9 x 10⁻⁴ s ⁻¹ M^-1               e. First order/4.8 x 10⁻⁴ s ⁻¹

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