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Consider the reaction represented by the equation?2SO2(g) + O2(g) 2SO3(g).?For the system at chemical equilibrium, which of the following explains what happens after the volume of the reaction mixture is increased (assume constant temperature)?


a. The amount of SO3(g) increases and the value for K increases.
b. The amount of SO3(g) decreases and the value for K increases.
c. The amount of SO3(g) stays the same and the value for K decreases.
d. The amount of SO3(g) decreases and the value for K stays the same.
e. The amount of SO3(g) increases and the value for K stays the same

Respuesta :

sebassandin

Answer:

b. The amount of SO3(g) decreases and the value for K increases.

Explanation:

Hello,

In this case, for the given reaction:

[tex]2SO_2(g) + O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

The change in the stoichiometric coefficient is:

[tex]\Delta \nu=2-2-1=-1[/tex]

In such a way, since the reagents have more moles than the products, based on Le Chatelier's principle, if the volume is increased, the side with more moles is favored. In addition, since the formation of reagent is favored, K is diminished based on the law of mass action shown below:

[tex]K=\frac{[SO_3]_{eq}^2}{[SO_2]_{eq}^2[O_2]_{eq}}[/tex]

Therefore the answer is:

b. The amount of SO3(g) decreases and the value for K increases.

Best regards.

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