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How does adding oxygen [tex]O_2[/tex] to this reaction change the equilibrium?

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

A. The equilibrium shifts right to produce more [tex]SO_3[/tex] molecules.

B. The equilibrium shifts left to produce more [tex]O_2[/tex] molecules.

C. The equilibrium shifts right because of decreased collisions between [tex]SO_2[/tex] and [tex]O_2[/tex] molecules.

D. The equilibrium shifts left with an increase in [tex]SO_2[/tex] and [tex]O_2[/tex] molecules.

E. The equilibrium shifts left because of increased collisions between [tex]SO_2[/tex] and [tex]O_2[/tex] molecules.



Answer :

GinnyAnswer
To answer the question of how adding oxygen [tex]\((O_2)\)[/tex] to the reaction

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

changes the equilibrium, we need to apply Le Chatelier's Principle. Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

1. Addition of Reactant:
- Adding [tex]\((O_2)\)[/tex] increases the concentration of one of the reactants.

2. Shift in Equilibrium:
- To counteract the increase in [tex]\((O_2)\)[/tex], the system will try to reduce the concentration of [tex]\((O_2)\)[/tex] by shifting the equilibrium to the right, where [tex]\((O_2)\)[/tex] is used up to form more [tex]\((SO_3)\)[/tex].

Correct Answer:
[tex]\[ \boxed{\text{A. The equilibrium shifts right to produce more } SO_3 \text{ molecules.}} \][/tex]

This shift will lead to an increase in the amount of [tex]\(SO_3\)[/tex] produced and a decrease in the amount of [tex]\(SO_2\)[/tex] and [tex]\(O_2\)[/tex] as they are consumed in the reaction.

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