To understand what happens to the amount of [tex]\( H_2 \)[/tex] as the reaction shifts to the left, let's first analyze the given chemical equilibrium:
[tex]\[
CO_2(g) + H_2(g) + \text{Energy} \rightleftharpoons CO(g) + H_2O(g)
\][/tex]
In this equilibrium, the reaction can proceed in either direction:
- Forward: From [tex]\( CO_2(g) + H_2(g) + \text{Energy} \)[/tex] to [tex]\( CO(g) + H_2O(g) \)[/tex]
- Reverse: From [tex]\( CO(g) + H_2O(g) \)[/tex] to [tex]\( CO_2(g) + H_2(g) + \text{Energy} \)[/tex]
When we say that the reaction shifts to the left, we mean that it favors the reverse reaction. Since the left side of the reaction is being favored, more reactants ([tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex]) are being formed, and the products ([tex]\( CO(g) \)[/tex] and [tex]\( H_2O(g) \)[/tex]) are being consumed.
Now, let's explore what happens to [tex]\( H_2 \)[/tex]:
1. Shift to the Left Explained: The shift to the left means that the reverse reaction is taking place more frequently, converting [tex]\( CO(g) \)[/tex] and [tex]\( H_2O(g) \)[/tex] back into [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex].
2. Formation of [tex]\( H_2 \)[/tex]: In the reverse reaction, [tex]\( CO \)[/tex] and [tex]\( H_2O \)[/tex] combine to form [tex]\( CO_2 \)[/tex] and [tex]\( H_2 \)[/tex]. Therefore, the concentration of [tex]\( H_2 \)[/tex] will increase as the reaction shifts left.
Therefore, we conclude that:
The amount of [tex]\( H_2 \)[/tex] goes up.
Hence, the correct answer is:
A. The amount of [tex]\( H_2 \)[/tex] goes up.
