Chemistry, 06.05.2020 00:29, tiiiiffany

The rate law for 2NO(g) + O2(g) >2NO2(g) is rate=k[NO]^2[O2]. In addition to the mechanism, the following have been proposed: I. 2NO(g) +O2(g) ===> 2NO2(g) II. 2NO(g) ===> N2O2(g) [fast] N2O2(g) +O2(g) ===> 2NO2(g) [slow] III. 2NO(g) ===>N2(g) + O2(g) [fast] N2(g) +2O2(g) ===> 2NO2(g) [slow] Which of these mechanisms is consistent with the rate law?

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answered: mdaniella522

Mechanism (I).

Explanation:

The given chemical reaction:

2NO(g) + O₂(g) → 2NO₂(g)

The given rate equation for the reaction:

rate = k [NO]² [O₂]

The rate equation of a given chemical reaction is equal to the product of the concentration of reactants raised to the power the partial orders of reaction.

The overall rate of a given multi-step chemical reaction depends on the slowest step in the reaction mechanism.

Therefore, the slowest step of a given chemical reaction is called the rate-determining step (RDS).

The proposed reaction mechanisms:

(I). 2NO(g) + O₂(g) → 2NO₂(g)

In an elementary (single step) reaction, the rate equation can be written as:

rate = k [NO]² [O₂]

(II). 2NO (g) ===> N₂O₂ (g) [fast]

N₂O₂(g) +O₂(g) ===> 2NO₂ (g) [slow] ⇒ RDS

In an complex (multi-step) reaction, the slow step is the rate determining step (RDS). So, the rate equation can be written as:

rate = k [N₂O₂] [O₂]

(III). 2NO(g) ===>N₂(g) + O₂(g) [fast]

N₂(g) + 2O₂(g) ===> 2NO₂(g) [slow] ⇒ RDS

In an complex (multi-step) reaction, the slow step is the rate determining step (RDS). So, the rate equation can be written as:

rate = k [N₂] [O₂]²

Thus, the rate equation of the reaction mechanism: (I). is the same as the given rate law of the chemical reaction.

Therefore, reaction mechanism (I). is consistent with the rate law of the given chemical reaction.

answered: Guest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.