Let's analyze the scenario using Le Chatelier's Principle, which states that if a dynamic equilibrium system experiences a disturbance, the system will adjust itself to counteract the disturbance and establish a new equilibrium.
The given reaction is:
[tex]\[ 51.8 \text{ kJ} + H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \][/tex]
When HI gas is removed from the system, the equilibrium will be disturbed. According to Le Chatelier's Principle, the system will shift in the direction that counteracts the disturbance—in this case, the removal of HI.
To counteract the removal of HI, the reaction will shift to the right, towards the products side. This means more [tex]$HI$[/tex] will be produced to replace the removed [tex]$HI$[/tex]. As the system shifts to the right, it uses up some of the [tex]$H_2$[/tex] and [tex]$I_2$[/tex] to form more [tex]$HI$[/tex].
Therefore, as the reaction shifts to produce more [tex]$HI$[/tex], the concentrations of [tex]$H_2$[/tex] and [tex]$I_2$[/tex] will decrease, because they are the reactants being consumed in the formation of more [tex]$HI$[/tex].
The correct answer is:
The reaction shifts to the right (products) and the concentrations of [tex]$I_2$[/tex] and [tex]$H_2$[/tex] decrease.
