To determine what happens to the amount of [tex]\( H_2 \)[/tex] when the reaction shifts to the left, let's consider the balanced chemical equation and the direction of the shift:
[tex]\[ 51.8 \text{ kJ} + H_2(g) + I_2(g) \rightleftharpoons 2 HI(g) \][/tex]
The equation shows that [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] are reactants, and [tex]\( HI \)[/tex] is the product. When the reaction shifts to the left, this means the equilibrium is moving in the direction that favors the formation of the reactants from the products.
In other words, when the equilibrium shifts to the left:
- [tex]\( HI \)[/tex] decomposes back into [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex].
Thus, the decomposition of [tex]\( HI \)[/tex] increases the amounts of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex].
Therefore, as the reaction shifts to the left, the amount of [tex]\( H_2 \)[/tex] increases.
So, the correct answer is:
C. The amount of [tex]\( H_2 \)[/tex] goes up.
