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

[tex]\[ \text{Rxn 1:} \quad N_2H_4 (l) + O_2 (g) \rightarrow N_2 (g) + 2 H_2O (g) \][/tex]
[tex]\[ \Delta H^{\circ} = -543.0 \frac{kJ}{mol} \][/tex]

[tex]\[ \text{Rxn 2:} \quad 2 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (g) \][/tex]
[tex]\[ \Delta H^{\circ} = -484.0 \frac{kJ}{mol} \][/tex]

[tex]\[ \text{Rxn 3:} \quad N_2 (g) + 3 H_2 (g) \rightarrow 2 NH_3 (g) \][/tex]
[tex]\[ \Delta H^{\circ} = -92.2 \frac{kJ}{mol} \][/tex]

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

A. Reaction 1 stays the same
B. Reaction 2 reverse the reaction
C. Reaction 3 double the reaction
D. Reaction 4 half the reaction

Sagot :

GinnyAnswer
To solve for the goal reaction [tex]\( \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow 2\text{NH}_3(g) \)[/tex], we need to manipulate the given reactions and combine them appropriately. Let's go through the steps in detail:

Step 1: Use Reaction 1 as is
[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]

Step 2: Reverse Reaction 2
[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]

Step 3: Double Reaction 3
[tex]\[ 2(\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)) \][/tex]
[tex]\[ 2\text{N}_2(g) + 6\text{H}_2(g) \rightarrow 4\text{NH}_3(g) \][/tex]
[tex]\[ \Delta H^\circ = 2 \times (-92.2) = -184.4 \frac{\text{kJ}}{\text{mol}} \][/tex]

Combining the reactions:

Now, let's add up these reactions:

1. [tex]\[ \text{N}_2\text{H}_4(l) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(g), \Delta H^\circ = -543.0 \frac{\text{kJ}}{\text{mol}} \][/tex]

2. [tex]\[ 2\text{H}_2\text{O}(g) \rightarrow 2\text{H}_2(g) + \text{O}_2(g), \Delta H^\circ = +484.0 \frac{\text{kJ}}{\text{mol}} \][/tex]

3. [tex]\[ 2\text{N}_2(g) + 6\text{H}_2(g) \rightarrow 4\text{NH}_3(g), \Delta H^\circ = -184.4 \frac{\text{kJ}}{\text{mol}} \][/tex]

We can see that:
- The [tex]\(\text{O}_2(g)\)[/tex] cancels out because 1 mole from Reaction 1 and 1 mole from the reversed Reaction 2.
- The [tex]\(2\text{H}_2\text{O}(g)\)[/tex] from Reaction 1 cancels with [tex]\(2\text{H}_2\text{O}(g)\)[/tex] from the reversed Reaction 2.
- The [tex]\(\text{N}_2(g)\)[/tex] cancels out because 1 mole from Reaction 1 and 1 mole from doubled Reaction 3.

This leaves us with the desired reaction:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow 2\text{NH}_3(g) \][/tex]

By adding up the enthalpy changes:
[tex]\[ \Delta H^\circ = -543.0 + 484.0 - 184.4 = -243.4 \frac{\text{kJ}}{\text{mol}} \][/tex]

So the combined reaction and the enthalpy change are:
[tex]\[ \text{N}_2\text{H}_4(l) + \text{H}_2(g) \rightarrow 2\text{NH}_3(g) \][/tex]
[tex]\[ \Delta H^\circ = -243.4 \frac{\text{kJ}}{\text{mol}} \][/tex]
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