Question

reaction adi 2N219) + 50210) 2N2O5() AH = ? den aretle H2O AH = -285.8 kJ H2(g) + X0219) N, 05/9) + H20m 2HNO30) AH = -76.6 kJ N219) +302(0)+ H2(0) 2HNO30) AH = -348.2 kJ Don't add units in the answer! Yanit:

Answer 1

H₂ + 1/2 O₂ ---------> H₂O ----------- (Rxn 1)

N₂O₅ + H₂O -------> 2 HNO₃ ----------- (Rxn 2)

N₂ + 3 O₂ + H₂ ---------> 2 HNO₃ ----------- (Rxn 3)

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Rxn 1 * 2 ====>

2 H₂ + O₂ ---------> 2 H₂O  (ΔH = 2 *-285.8 = -571.6)

Reverse reaction:

2 H₂O  ---------> 2 H₂ + O₂ (ΔH₁ = 571.6)

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Rxn 2 * 2 ====>

2 N₂O₅ + 2 H₂O -------> 4 HNO₃ (ΔH = 2 *-76.6 = -153.2)

Reverse reaction

4 HNO₃ -------> 2 N₂O₅ + 2 H₂O (ΔH₂ = 153.2)

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Rxn 3 * 2 ====>

2 N₂ + 6 O₂ + 2 H₂ ---------> 4 HNO₃ (ΔH₃ = 2 *-348.2 = -696.4)

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Add the final reactions in each step to get required reaction:

2 H₂O  ---------> 2 H₂ + O₂

4 HNO₃ -------> 2 N₂O₅ + 2 H₂O

2 N₂ + 6 O₂ + 2 H₂ ---------> 4 HNO₃

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2 H₂O + 4 HNO₃ + 2 N₂ + 6 O₂ + 2 H₂ ---------> 2 H₂ + O₂ + 2 N₂O₅ + 2 H₂O + 4 HNO₃

_{That is :}

2 N₂ + 5 O₂ ---------> 2 N₂O₅

Which is our required reaction.

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Hence, ΔH for this reaction will be:

ΔH = ΔH₁ + ΔH₂ + ΔH₃

ΔH = (571.6) + (153.2) + (-696.4)

$\small {\color{Red} \boldsymbol{\Delta H=28.4\;kJ}}$ ------------ (**Answer**)

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