Given the chemical reaction:
[tex]$
51.8 \text{ kJ} + H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)
$[/tex]
We need to determine the direction in which the reaction shifts when [tex]\(I_2\)[/tex] is removed.
To solve this problem, we can use Le Châtelier's principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
Step-by-Step Solution:
1. Understand the Equilibrium:
The given reaction is at equilibrium:
[tex]$
51.8 \text{ kJ} + H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)
$[/tex]
At equilibrium, the rate of the forward reaction (producing [tex]\(HI\)[/tex]) equals the rate of the reverse reaction (producing [tex]\(H_2\)[/tex] and [tex]\(I_2\)[/tex]).
2. Determine the Disturbance:
The problem states that [tex]\(I_2\)[/tex] is removed from the reaction mixture.
3. Apply Le Châtelier's Principle:
According to Le Châtelier's principle, removing [tex]\(I_2\)[/tex] (a reactant) from the system will create a deficiency of [tex]\(I_2\)[/tex]. The system will counteract this disturbance by favoring the reaction that produces more [tex]\(I_2\)[/tex].
4. Shift in Reaction Direction:
To produce more [tex]\(I_2\)[/tex], the equilibrium will shift towards the side that has [tex]\(I_2\)[/tex] as a product. However, in the context of our given reaction, increasing the forward reaction will consume [tex]\(H_2\)[/tex] and [tex]\(I_2\)[/tex] and produce more [tex]\(HI\)[/tex].
Therefore, the reaction will shift to the right or towards the products ([tex]\(HI\)[/tex]) to counteract the removal of [tex]\(I_2\)[/tex] and minimize the change.
5. Conclusion:
When [tex]\(I_2\)[/tex] is removed from the reaction mixture, the reaction shifts to the right towards the products to produce more [tex]\(I_2\)[/tex] and restore equilibrium.
Thus, the correct answer is:
A. The reaction shifts right or to products.
So, the reaction shifts right or to the products when [tex]\(I_2\)[/tex] is removed.
