The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has been determined experimentally that the rate is second order in NO and first order in O2.

Respuesta :

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]

The rate law expression is:

[tex]Rate=k[NO]^2[O_2][/tex]

Now we have to determine the number of molecules of [tex]NO\text{ and }O_2[/tex]

In mixture 1 : There are 5 [tex]NO[/tex] and 4 [tex]O_2[/tex] molecules.

In mixture 2 : There are 7 [tex]NO[/tex] and 2 [tex]O_2[/tex] molecules.

In mixture 3 : There are 3 [tex]NO[/tex] and 5 [tex]O_2[/tex] molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

[tex]Rate=k[NO]^2[O_2][/tex]

[tex]Rate=k(5)^2\times (4)[/tex]

[tex]Rate=k(100)[/tex]

The rate law expression for mixture 2 is:

[tex]Rate=k[NO]^2[O_2][/tex]

[tex]Rate=k(7)^2\times (2)[/tex]

[tex]Rate=k(98)[/tex]

The rate law expression for mixture 3 is:

[tex]Rate=k[NO]^2[O_2][/tex]

[tex]Rate=k(3)^2\times (5)[/tex]

[tex]Rate=k(45)[/tex]

Hence, the mixture 1 has the fastest initial rate.

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