Hannahzg08idn
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What will happen if [tex][tex]$HI$[/tex][/tex] is removed from a reaction vessel containing this reaction at equilibrium?

[tex] H_2 + I_2 \rightleftarrows 2 HI [/tex]

A. There will be no effect.
B. The rate of the backward reaction will increase.
C. The rate of the forward reaction will increase.
D. The reaction will slow down.

Sagot :

GinnyAnsweridn
To understand what will happen if HI is removed from the reaction vessel containing the reaction at equilibrium:

[tex]\[ H_2 + I_2 \rightleftarrows 2 HI \][/tex]

we need to apply Le Chatelier's Principle. This principle states that if a change is made to a system at equilibrium, the system will adjust to counteract that change and restore a new equilibrium.

Here, the removal of HI means that the concentration of HI decreases. According to Le Chatelier's Principle, the equilibrium will shift to the right to oppose the decrease in HI. This shift will favor the formation of more HI from [tex]\(H_2\)[/tex] and [tex]\(I_2\)[/tex], thereby increasing the rate at which [tex]\(HI\)[/tex] is produced.

In detail:
1. The initial equilibrium state has balanced rates for both the forward and backward reactions.
2. By removing HI, we disturb the equilibrium.
3. As the equilibrium shifts to the right to restore balance, more [tex]\(HI\)[/tex] is formed. This occurs by the forward reaction, [tex]\( H_2 + I_2 \to 2 HI \)[/tex], proceeding more quickly than the backward reaction, [tex]\( 2 HI \to H_2 + I_2 \)[/tex].
4. However, the backward reaction needs to counteract the change and replace the removed HI, leading to an increase in the rate of the backward reaction temporarily until a new equilibrium is established.

Thus, the correct answer is:

B. The rate of the backward reaction will increase.
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