g Suppose a 49. L reaction vessel is filled with 1.5 mol of NO. What can you say about the composition of the mixture in the vessel at equilibrium

Suppose a 49.0 L reaction vessel is filled with 1.5 mol of [tex]NO_3[/tex] and 1.5 mol of NO. What can you say about the composition of the mixture in the vessel at equilibrium?

A. There will be very little [tex]NO_3[/tex] and NO.

B. There will be very little [tex]NO_2[/tex]

C. Neither of the above is true.

Answer: The correct option is B. there will be very little [tex]NO_2[/tex]

Explanation:

We are given:

Initial moles of [tex]NO_3[/tex] = 1.5 moles

Initial moles of NO = 1.5 moles

Volume of vessel = 49 L

As, moles of reactants and moles of products are equal, the volume term will not appear in the equilibrium constant expression.

Equilibrium constant for the reaction = 0.74

For the given chemical equation:

[tex]NO_3(g)+NO(g)\rightleftharpoons 2NO_2(g)[/tex]

Initial: 1.5 1.5 -

At eqllm: 1.5-x 1.5-x 2x

The expression of equilibrium constant for the above reaction:

[tex]K_{eq}=\frac{[NO_2]^2}{[NO_3]\times [NO]}[/tex]

Putting values in above equation, we get:

[tex]0.74=\frac{(2x)^2}{(1.5-x)(1.5-x)}\\\\x=-1.13, 0.451[/tex]

Neglecting the negative value of equilibrium constant because concentration cannot be negative

Equilibrium concentration of [tex]NO_3[/tex] = (1.5 - x) = (1.5 - 0.451) = 1.049 moles

Equilibrium concentration of NO = (1.5 - x) = (1.5 - 0.451) = 1.049 moles

Equilibrium concentration of [tex]NO_2[/tex] = 2x = [tex](2\times 0.451)=0.902mol[/tex]

There are 3 possibilities:

If [tex]K_{eq}<1[/tex], the reaction is reactant favored
If [tex]K_{eq}>1[/tex], the reaction is product favored.
If [tex]K_{eq}=1[/tex], the reaction is in equilibrium.

Here, the value of [tex]K_{eq}=0.74[/tex], which is less than 1, therefore the reaction is reactant favored and we can say that there will be very little [tex]NO_2[/tex].

Hence, the correct option is B. there will be very little [tex]NO_2[/tex]