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Sagot :
The question is incomplete, the complete question is:
At a certain temperature, the equilibrium constant [tex]K_{eq}[/tex] for the following reaction is 0.74:
[tex]NO_3(g)+NO(g)\rightleftharpoons 2NO_2(g)[/tex]
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]
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