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Sagot :
To determine how to manipulate reaction [tex]\(2\)[/tex] to connect with reaction [tex]\(1\)[/tex] and achieve the goal reaction, let's carefully analyze and break down the steps.
We have the following reactions:
1) Reaction 1:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -543.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
2) Reaction 2:
[tex]\[ 2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -484.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
3) Reaction 3:
[tex]\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -92.2 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
The goal reaction is:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow 2\text{NH}_3(g) \][/tex]
### Step-by-Step Solution:
1. Identify the need to manipulate Reaction 2:
- From the goal reaction, we need [tex]\(\text{H}_2(g)\)[/tex] to be produced somehow, as it is not available in Reaction 1, but required for the goal reaction.
2. Reversing Reaction 2:
- Reversing Reaction 2 will produce [tex]\(\text{H}_2(g)\)[/tex] and [tex]\(\text{O}_2(g)\)[/tex], which we need to match the goal reaction setup.
[tex]\[ 2\text{H}_2\text{O}(g) \rightarrow 2\text{H}_2(g) + \text{O}_2(g) \][/tex]
[tex]\[ \Delta H^{\circ} = +484.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
3. Adjusting the reversed Reaction 2:
- To match the exact amount of [tex]\(\text{H}_2(g)\)[/tex] needed for the goal reaction, we need to half the reversed Reaction 2.
[tex]\[ \text{H}_2\text{O}(g) \rightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \][/tex]
[tex]\[ \Delta H^{\circ} = +242.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
Therefore, to connect Reaction 2 with Reaction 1 and achieve the goal reaction, we should:
[tex]\[ \boxed{4} \][/tex]
Answer: 4) half the reaction
We have the following reactions:
1) Reaction 1:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -543.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
2) Reaction 2:
[tex]\[ 2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -484.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
3) Reaction 3:
[tex]\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -92.2 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
The goal reaction is:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow 2\text{NH}_3(g) \][/tex]
### Step-by-Step Solution:
1. Identify the need to manipulate Reaction 2:
- From the goal reaction, we need [tex]\(\text{H}_2(g)\)[/tex] to be produced somehow, as it is not available in Reaction 1, but required for the goal reaction.
2. Reversing Reaction 2:
- Reversing Reaction 2 will produce [tex]\(\text{H}_2(g)\)[/tex] and [tex]\(\text{O}_2(g)\)[/tex], which we need to match the goal reaction setup.
[tex]\[ 2\text{H}_2\text{O}(g) \rightarrow 2\text{H}_2(g) + \text{O}_2(g) \][/tex]
[tex]\[ \Delta H^{\circ} = +484.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
3. Adjusting the reversed Reaction 2:
- To match the exact amount of [tex]\(\text{H}_2(g)\)[/tex] needed for the goal reaction, we need to half the reversed Reaction 2.
[tex]\[ \text{H}_2\text{O}(g) \rightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \][/tex]
[tex]\[ \Delta H^{\circ} = +242.0 \, \frac{\text{kJ}}{\text{mol}} \][/tex]
Therefore, to connect Reaction 2 with Reaction 1 and achieve the goal reaction, we should:
[tex]\[ \boxed{4} \][/tex]
Answer: 4) half the reaction
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