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maxonik [38]
2 years ago
11

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 ves

sel at equilibrium
Chemistry
1 answer:
irakobra [83]2 years ago
7 0

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.

<u>Answer:</u> The correct option is B. there will be very little NO_2

<u>Explanation:</u>

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}=\frac{[NO_2]^2}{[NO_3]\times [NO]}

Putting values in above equation, we get:

0.74=\frac{(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\times 0.451)=0.902mol

There are 3 possibilities:

  • If K_{eq}, 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

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