Question
4. Calculate the enthalpy change of the following reaction between nitrogen gas and oxygen gas, given the thermochemical equations below: 2N2(g) + 5O2(g) → 2N2O5(g) 1) 2H2(g) + O2(g) → 2H2O(l) ΔH° = -572 kJ 2) N2O5(g) + H2O(l) → 2HNO3(l) ΔH° = -77 kJ 3) 1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l) ΔH° = -174 kJ Show the calculations
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Answer to a math question 4. Calculate the enthalpy change of the following reaction between nitrogen gas and oxygen gas, given the thermochemical equations below: 2N2(g) + 5O2(g) → 2N2O5(g) 1) 2H2(g) + O2(g) → 2H2O(l) ΔH° = -572 kJ 2) N2O5(g) + H2O(l) → 2HNO3(l) ΔH° = -77 kJ 3) 1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l) ΔH° = -174 kJ Show the calculations
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To find the enthalpy change ( \Delta H^\circ
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 forming2HNO_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}
We have three given reactions:
1.
2.
3.
By utilizing reactions 2 and 3, we can determine the enthalpy change for forming
Reversing reaction 2 to get it in line with the target reaction:
Total enthalpy change is:
Therefore, the enthalpy change for the target reaction is
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