Answer :
To determine which changes will shift the equilibrium to the right for the given reaction:
[tex]\[ \text{HC}_2\text{H}_3\text{O}_2(\text{aq}) \leftrightarrow \text{H}^+(\text{aq}) + \text{C}_2\text{H}_3\text{O}_2^-(\text{aq}) \][/tex]
we need to recall Le Chatelier's Principle. This principle states that if an external change is applied to a system at equilibrium, the system adjusts itself to partially counteract that change and re-establish equilibrium. Let’s analyze the given options:
1. Increase [tex]\(\left[\text{HC}_2\text{H}_3\text{O}_2\right]\)[/tex]:
- Increasing the concentration of the reactant ([tex]\(\text{HC}_2\text{H}_3\text{O}_2\)[/tex]) will shift the equilibrium to the right to produce more products ([tex]\(\text{H}^+\)[/tex] and [tex]\(\text{C}_2\text{H}_3\text{O}_2^-\)[/tex]) in order to reduce the added reactant.
2. Add solid NaOH:
- Adding solid NaOH to the solution will dissociate into [tex]\(\text{Na}^+\)[/tex] and [tex]\(\text{OH}^-\)[/tex] ions. The [tex]\(\text{OH}^-\)[/tex] ions will react with [tex]\(\text{H}^+\)[/tex] ions to form water [tex]\((\text{H}_2\text{O})\)[/tex]. This reaction reduces the concentration of [tex]\(\text{H}^+\)[/tex] ions in the solution. As a result, the equilibrium will shift to the right to produce more [tex]\(\text{H}^+\)[/tex] ions.
3. Decrease [tex]\(\left[\text{H}^+\right]\)[/tex]:
- Directly decreasing the concentration of the product ([tex]\(\text{H}^+\)[/tex]) will also cause the equilibrium to shift to the right to produce more [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{C}_2\text{H}_3\text{O}_2^-\)[/tex]).
4. Decrease [tex]\(\left[\text{C}_2\text{H}_3\text{O}_2^-\right]\)[/tex]:
- Similarly, decreasing the concentration of the other product ([tex]\(\text{C}_2\text{H}_3\text{O}_2^-\)[/tex]) will shift the equilibrium to the right to produce more [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{C}_2\text{H}_3\text{O}_2^-\)[/tex]).
From this detailed analysis, we can see that all the given changes will shift the equilibrium to the right. Therefore, the answer is:
- All of the above
- Increase [tex]\(\left[\text{HC}_2\text{H}_3\text{O}_2\right]\)[/tex]
- Add solid NaOH
- Decrease [tex]\(\left[\text{H}^+\right]\)[/tex]
- Decrease [tex]\(\left[\text{C}_2\text{H}_3\text{O}_2^-\right]\)[/tex]
These modifications all result in:
[tex]\[ ( \text{True}, \text{True}, \text{True}, \text{True} ) \][/tex]
Thus, each option listed will indeed shift the equilibrium to the right.
[tex]\[ \text{HC}_2\text{H}_3\text{O}_2(\text{aq}) \leftrightarrow \text{H}^+(\text{aq}) + \text{C}_2\text{H}_3\text{O}_2^-(\text{aq}) \][/tex]
we need to recall Le Chatelier's Principle. This principle states that if an external change is applied to a system at equilibrium, the system adjusts itself to partially counteract that change and re-establish equilibrium. Let’s analyze the given options:
1. Increase [tex]\(\left[\text{HC}_2\text{H}_3\text{O}_2\right]\)[/tex]:
- Increasing the concentration of the reactant ([tex]\(\text{HC}_2\text{H}_3\text{O}_2\)[/tex]) will shift the equilibrium to the right to produce more products ([tex]\(\text{H}^+\)[/tex] and [tex]\(\text{C}_2\text{H}_3\text{O}_2^-\)[/tex]) in order to reduce the added reactant.
2. Add solid NaOH:
- Adding solid NaOH to the solution will dissociate into [tex]\(\text{Na}^+\)[/tex] and [tex]\(\text{OH}^-\)[/tex] ions. The [tex]\(\text{OH}^-\)[/tex] ions will react with [tex]\(\text{H}^+\)[/tex] ions to form water [tex]\((\text{H}_2\text{O})\)[/tex]. This reaction reduces the concentration of [tex]\(\text{H}^+\)[/tex] ions in the solution. As a result, the equilibrium will shift to the right to produce more [tex]\(\text{H}^+\)[/tex] ions.
3. Decrease [tex]\(\left[\text{H}^+\right]\)[/tex]:
- Directly decreasing the concentration of the product ([tex]\(\text{H}^+\)[/tex]) will also cause the equilibrium to shift to the right to produce more [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{C}_2\text{H}_3\text{O}_2^-\)[/tex]).
4. Decrease [tex]\(\left[\text{C}_2\text{H}_3\text{O}_2^-\right]\)[/tex]:
- Similarly, decreasing the concentration of the other product ([tex]\(\text{C}_2\text{H}_3\text{O}_2^-\)[/tex]) will shift the equilibrium to the right to produce more [tex]\(\text{H}^+\)[/tex] and [tex]\(\text{C}_2\text{H}_3\text{O}_2^-\)[/tex]).
From this detailed analysis, we can see that all the given changes will shift the equilibrium to the right. Therefore, the answer is:
- All of the above
- Increase [tex]\(\left[\text{HC}_2\text{H}_3\text{O}_2\right]\)[/tex]
- Add solid NaOH
- Decrease [tex]\(\left[\text{H}^+\right]\)[/tex]
- Decrease [tex]\(\left[\text{C}_2\text{H}_3\text{O}_2^-\right]\)[/tex]
These modifications all result in:
[tex]\[ ( \text{True}, \text{True}, \text{True}, \text{True} ) \][/tex]
Thus, each option listed will indeed shift the equilibrium to the right.
