Answer :
To determine how Reaction 1 should be manipulated to achieve the goal reaction, we can analyze the provided reactions and their changes in enthalpy (ΔH).
### Provided Reactions:
1. [tex]\( \text{Rxn 1: } \mathrm{N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g)} \)[/tex]
2. [tex]\( \text{Rxn 2: } \mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)}, \Delta H^{\circ} = -543.0 \, \frac{\text{kJ}}{\text{mol}} \)[/tex]
3. [tex]\( \text{Rxn 3: } \mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}, \Delta H^{\circ} = -484.0 \, \frac{\text{kJ}}{\text{mol}} \)[/tex]
### Goal Reaction:
[tex]\[ \mathrm{N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g)} \][/tex]
### Steps for Manipulation:
1. Keep Reaction 1 the same:
[tex]\[ \mathrm{N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g)} \][/tex]
2. Reverse Reaction 2:
[tex]\[ \mathrm{2H_2O(g) \rightarrow 2H_2(g) + O_2(g)}, \Delta H = +543.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
3. Double Reaction 3:
[tex]\[ 2 \times \left( \mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)} \right) \][/tex]
Hence:
[tex]\[ \mathrm{2N_2(g) + 6H_2(g) \rightarrow 4NH_3(g)}, \Delta H = 2 \times -484.0 \, \frac{\text{kJ}}{\text{mol}} = -968.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
### Combining the Reactions:
- Reaction 1 + Reversed Reaction 2 + Doubled Reaction 3:
[tex]\[ \begin{aligned} &\mathrm{N_2H_4(l) + O_2(g)} &\mathrm{\rightarrow N_2(g) + 2H_2O(g)} & \quad\text{Reaction 1} \\ &\mathrm{+ 2H_2O(g)} &\mathrm{\rightarrow 2H_2(g) + O_2(g)} & \quad\text{Reversed Reaction 2} \\ &\mathrm{+ 2N_2(g) + 6H_2(g)} &\mathrm{\rightarrow 4NH_3(g)} & \quad\text{Doubled Reaction 3} \\ \end{aligned} \][/tex]
### Cancel Out Common Terms:
- [tex]\( \mathrm{O_2(g)} \)[/tex] and [tex]\( \mathrm{2H_2O(g)} \)[/tex] cancel out:
[tex]\[ \mathrm{N_2H_4(l) + 6H_2(g) \rightarrow 4NH_3(g)} \][/tex]
But our goal reaction is only [tex]\( \mathrm{N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g)} \)[/tex], so we observe that all the given manipulations lead back to verifying that reversing Reaction 2 achieves a necessary step:
### Thus, the correct choice:
[tex]\[ 2) \text{ reverse the reaction} \][/tex]
Answer: 2
### Provided Reactions:
1. [tex]\( \text{Rxn 1: } \mathrm{N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g)} \)[/tex]
2. [tex]\( \text{Rxn 2: } \mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)}, \Delta H^{\circ} = -543.0 \, \frac{\text{kJ}}{\text{mol}} \)[/tex]
3. [tex]\( \text{Rxn 3: } \mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}, \Delta H^{\circ} = -484.0 \, \frac{\text{kJ}}{\text{mol}} \)[/tex]
### Goal Reaction:
[tex]\[ \mathrm{N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g)} \][/tex]
### Steps for Manipulation:
1. Keep Reaction 1 the same:
[tex]\[ \mathrm{N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g)} \][/tex]
2. Reverse Reaction 2:
[tex]\[ \mathrm{2H_2O(g) \rightarrow 2H_2(g) + O_2(g)}, \Delta H = +543.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
3. Double Reaction 3:
[tex]\[ 2 \times \left( \mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)} \right) \][/tex]
Hence:
[tex]\[ \mathrm{2N_2(g) + 6H_2(g) \rightarrow 4NH_3(g)}, \Delta H = 2 \times -484.0 \, \frac{\text{kJ}}{\text{mol}} = -968.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
### Combining the Reactions:
- Reaction 1 + Reversed Reaction 2 + Doubled Reaction 3:
[tex]\[ \begin{aligned} &\mathrm{N_2H_4(l) + O_2(g)} &\mathrm{\rightarrow N_2(g) + 2H_2O(g)} & \quad\text{Reaction 1} \\ &\mathrm{+ 2H_2O(g)} &\mathrm{\rightarrow 2H_2(g) + O_2(g)} & \quad\text{Reversed Reaction 2} \\ &\mathrm{+ 2N_2(g) + 6H_2(g)} &\mathrm{\rightarrow 4NH_3(g)} & \quad\text{Doubled Reaction 3} \\ \end{aligned} \][/tex]
### Cancel Out Common Terms:
- [tex]\( \mathrm{O_2(g)} \)[/tex] and [tex]\( \mathrm{2H_2O(g)} \)[/tex] cancel out:
[tex]\[ \mathrm{N_2H_4(l) + 6H_2(g) \rightarrow 4NH_3(g)} \][/tex]
But our goal reaction is only [tex]\( \mathrm{N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g)} \)[/tex], so we observe that all the given manipulations lead back to verifying that reversing Reaction 2 achieves a necessary step:
### Thus, the correct choice:
[tex]\[ 2) \text{ reverse the reaction} \][/tex]
Answer: 2
