madey21
Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

How should Reaction 1 be manipulated to connect with the goal reaction?

Given reactions:

1) [tex]\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(g) \][/tex]
2) [tex]\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -543.0 \, \frac{kJ}{mol} \][/tex]
3) [tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \][/tex]
[tex]\[ \Delta H^{\circ} = -484.0 \, \frac{kJ}{mol} \][/tex]

Goal:
[tex]\[ N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g) \][/tex]

Options:
1. stays the same
2. reverse the reaction
3. double the reaction
4. half the reaction

Enter the answer choice number.


Sagot :

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