To connect Reaction 1 with the goal reaction, let’s analyze and determine the appropriate manipulation steps:
Given Reactions:
1. [tex]\[ \text{Rxn 1: } N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2 H_2O(g) \quad \Delta H^{\circ} = -543.0 \frac{\text{kJ}}{\text{mol}} \][/tex]
2. [tex]\[ \text{Rxn 2: } 2 H_2(g) + O_2(g) \rightarrow 2 H_2O(g) \quad \Delta H^{\circ} = -484.0 \frac{\text{kJ}}{\text{mol}} \][/tex]
3. [tex]\[ \text{Rxn 3: } N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \quad \Delta H^{\circ} = -92.2 \frac{\text{kJ}}{\text{mol}} \][/tex]
Goal Reaction:
[tex]\[ N_2H_4(l) + H_2(g) \rightarrow 2 NH_3(g) \][/tex]
Steps to Manipulate Reactions:
1. Rxn 1: No change is required. This reaction is already in the desired direction.
[tex]\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2 H_2O(g) \quad \Delta H^{\circ} = -543.0 \frac{\text{kJ}}{\text{mol}} \][/tex]
2. Rxn 2: Reverse the reaction. Reversing this reaction will change the sign of [tex]\(\Delta H^{\circ}\)[/tex].
[tex]\[ 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) \quad \Delta H^{\circ} = 484.0 \frac{\text{kJ}}{\text{mol}} \][/tex]
3. Rxn 3: Double this reaction to match the required product quantity in the goal reaction.
[tex]\[ 2 N_2(g) + 6 H_2(g) \rightarrow 4 NH_3(g) \quad \Delta H^{\circ} = 2 \times -92.2 \frac{\text{kJ}}{\text{mol}} = -184.4 \frac{\text{kJ}}{\text{mol}} \][/tex]
Combining these steps, we get:
- From Rxn 1, we have: [tex]\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2 H_2O(g) \][/tex]
- From reversed Rxn 2, we have: [tex]\[ 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) \][/tex]
- From doubled Rxn 3, we have: [tex]\[ 2 N_2(g) + 6 H_2(g) \rightarrow 4 NH_3(g) \][/tex]
When adding these reactions together, the terms involving [tex]\(O_2\)[/tex], [tex]\(H_2O\)[/tex], and [tex]\(N_2\)[/tex] will cancel out appropriately, yielding:
[tex]\[ N_2H_4(l) + H_2(g) \rightarrow 2 NH_3(g) \][/tex]
This matches our goal reaction.
Therefore, the answer choice number is:
[tex]\[ \boxed{3} \][/tex]
