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]
