To determine how the equilibrium will shift for the given reaction [tex]\( H_2O_2(l) \rightleftharpoons H_2(g) + O_2(g) \)[/tex], we need to apply Le Chatelier's Principle, which states that the equilibrium will shift in the direction that counteracts any imposed change.
1. Increase [tex]\( H_2 \)[/tex]:
- Increasing [tex]\( H_2 \)[/tex] will cause the equilibrium to shift towards the reactants to counteract the increase in [tex]\( H_2 \)[/tex] concentration by producing less [tex]\( H_2 \)[/tex] and [tex]\( O_2 \)[/tex], and more [tex]\( H_2O_2 \)[/tex].
2. Decrease [tex]\( O_2 \)[/tex]:
- Decreasing [tex]\( O_2 \)[/tex] should cause the equilibrium to shift towards the products to counteract the decrease by forming more [tex]\( H_2 \)[/tex] and [tex]\( O_2 \)[/tex], and using up more [tex]\( H_2O_2 \)[/tex].
3. Add a catalyst:
- Adding a catalyst does not shift the equilibrium position; it only speeds up the rate at which equilibrium is achieved for both forward and reverse reactions.
4. Decrease the temperature:
- Since the forward reaction is endothermic (absorbs heat), decreasing the temperature will shift the equilibrium towards the exothermic direction, which is the reverse reaction, forming more [tex]\( H_2O_2 \)[/tex].
5. Increase [tex]\( H_2O_2 \)[/tex]:
- Increasing [tex]\( H_2O_2 \)[/tex] will shift the equilibrium towards the products to counteract the increase in [tex]\( H_2O_2 \)[/tex] concentration by producing more [tex]\( H_2 \)[/tex] and [tex]\( O_2 \)[/tex].
Given these considerations, the changes resulting in the equilibrium shifting towards the reactants are:
I. Increase [tex]\( H_2 \)[/tex]
IV. Decrease the temperature
Thus, the correct answer is A. I and IV.
