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 $NO_2$

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:

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

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 $NO_2$