To find the enthalpy change ( \Delta H^\circ ) for the given reaction:
2N_2(g) + 5O_2(g) \rightarrow 2N_2O_5(g)
We have three given reactions:
1. 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \quad \Delta H° = -572 \, \text{kJ}
2. N_2O_5(g) + H_2O(l) \rightarrow 2HNO_3(l) \quad \Delta H° = -77 \, \text{kJ}
3. \frac{1}{2}N_2(g) + \frac{3}{2}O_2(g) + \frac{1}{2}H_2(g) \rightarrow HNO_3(l) \quad \Delta H° = -174 \, \text{kJ}
By utilizing reactions 2 and 3, we can determine the enthalpy change for forming 2HNO_3 :
4 \times (-174 \, \text{kJ}) = -696 \, \text{kJ}
Reversing reaction 2 to get it in line with the target reaction:
2HNO_3 \rightarrow N_2O_5 + H_2O
Total enthalpy change is:
-696 \, \text{kJ} (4 \times \text{Reaction 3}) + (-77 \, \text{kJ}) (\text{Reaction 2}) = \boxed{-773 \, \text{kJ}}
Therefore, the enthalpy change for the target reaction is -773 \, \text{kJ} .