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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Answer from: 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.

Answer from: Quest

c most likely

hop this let me know if it is wrong

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.