Question

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

Answer 1

The question is incomplete, the complete question is:

At a certain temperature, the equilibrium constant
`K_(eq)` for the following reaction is 0.74:

`NO_3(g)+NO(g)\rightleftharpoons 2NO_2(g)`

Suppose a 49.0 L reaction vessel is filled with 1.5 mol of
`NO_3` 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
`NO_3` and NO.

B. There will be very little
`NO_2`

C. Neither of the above is true.

Answer: The correct option is B. there will be very little
`NO_2`

Step-by-step explanation:

We are given:

Initial moles of
`NO_3` = 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:

`NO_3(g)+NO(g)\rightleftharpoons 2NO_2(g)`

Initial: 1.5 1.5 -

At eqllm: 1.5-x 1.5-x 2x

The expression of equilibrium constant for the above reaction:

`K_(eq)=([NO_2]^2)/([NO_3]* [NO])`

Putting values in above equation, we get:

`0.74=((2x)^2)/((1.5-x)(1.5-x))\\\\x=-1.13, 0.451`

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

Equilibrium concentration of
`NO_3` = (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
`NO_2` = 2x =
`(2* 0.451)=0.902mol`

There are 3 possibilities:

• If
  `K_(eq)<1`, the reaction is reactant favored
• If
  `K_(eq)>1`, the reaction is product favored.
• If
  `K_(eq)=1`, the reaction is in equilibrium.

Here, the value of
`K_(eq)=0.74`, which is less than 1, therefore the reaction is reactant favored and we can say that there will be very little
`NO_2`.

Hence, the correct option is B. there will be very little
`NO_2`