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The reaction below is at equilibrium. What would happen if more [tex]$Na_2O_2$[/tex] were added?

[tex]\[2 Na(s) + O_2(g) \rightleftarrows Na_2O_2(s)\][/tex]

A. The equilibrium position would shift to produce more [tex]$Na_2[/tex] and [tex]$O_2$[/tex].
B. The equilibrium position would stay the same.
C. The equilibrium position would shift to produce more [tex]$O_2[tex]$[/tex].
D. The equilibrium position would shift to produce more [tex]$[/tex]Na_2O_2$[/tex].


Sagot :

To determine the effect of adding more [tex]\( Na_2O_2 \)[/tex] to the system at equilibrium, we can use 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.

The reaction given is:
[tex]\[ 2 Na(s) + O_2(g) \rightleftharpoons Na_2O_2(s) \][/tex]

We are specifically interested in the effect of adding more [tex]\( Na_2O_2 \)[/tex], the product of this reaction. Let's go through the steps:

1. Identify the disturbance:
We are adding more [tex]\( Na_2O_2 \)[/tex], which increases the concentration of the solid product.

2. Apply Le Chatelier's principle:
According to Le Chatelier's principle, the system will respond to this disturbance by shifting the equilibrium position to reduce the effect of the added [tex]\( Na_2O_2 \)[/tex]. This means that the system will try to consume some of the added [tex]\( Na_2O_2 \)[/tex] to restore equilibrium.

3. Determine the direction of the shift:
To consume the added [tex]\( Na_2O_2 \)[/tex], the equilibrium will shift towards the left (the reverse reaction). This shift will produce more [tex]\( Na \)[/tex] and [tex]\( O_2 \)[/tex] from the breakdown of [tex]\( Na_2O_2 \)[/tex].

Therefore, the equilibrium position will shift to produce more [tex]\( Na \)[/tex] and [tex]\( O_2 \)[/tex].

The correct answer is:
A. The equilibrium position would shift to produce more [tex]\( Na \)[/tex] and [tex]\( O_2 \)[/tex].
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