Be sure to answer all parts. The rate law for 2 NO(g) + O2(g) → 2 NO2(g) is rate = k[NO]2[O2]. The following mechanisms have been proposed: I. 2 NO(g) + O2(g) → 2 NO2(g) II.2 NO(g) ⇌ N2O2(g)[fast] N2O2(g) + O2(g) → 2 NO2(g) [slow] III.2 NO(g) ⇌ N2(g) + O2(g)[fast] N2(g) + 2 O2(g) → 2 NO2(g) [slow] (a) Which of these mechanisms is consistent with the rate law? Check all that apply. I. II. III. None of the above (b) Which of these mechanisms is most reasonable? I. II. III.

a. I

b. I

Explanation:

A rate law is an equation that relates the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers.

In the equation above

2 NO(g) + O2(g) → 2 NO2(g)

rate = k[NO]²[O2]

where k = rate constant

From the mechanism above we see that in;

1: Rate law= k[NO]² [O2]

2: Rate law= k[N2O2][O2] [slow eq determines rate law]

3: Rate law= k[N2][O2]²

We can observe that the resembling equation is 1.

The rate of a chemical reaction is determined by the slowest step.

Rate = k[Concentration of reactants individually raised to their stoichiometric co-efficients]

In mechanism I,

Overall reaction occurs in a single step. Therefore,

rate= k[NO]²[O2]

This is this consistent with the observed rate law.

In mechanism II,

The overall reaction occurs in two steps, through the involvement of an intermediate, N2O2.

Rate of the slowest step should be the overall reaction rate.

Therefore, overall rate= k[N2O2][O2]

Again considering non-accumulation of intermediate, N2O2 in the overall reaction.

Its rate of production will be equal to its rate of decomposition.

Thus, k1[NO]²= k[N2O2][O2]

➡ [N2O2]= (k1/k). [NO]²/[O2]

Overall rate= k(k1/k).([NO]²[O2])/[O2]

=k1[NO]²

So, this is not consistent with the rate law.

Mechanism III,

the overall rate =k[NO]².

Therefore, we see that only mechanism I is is most appropriate , reasonable and consistent with the observed rate law.

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

output?

explanation: