Question

011 1.0 points What energy change is associated with the reaction to obtain one mole of H2 from one wang (ew22622) – 19D Thermodynamics: First Law – ozuna – (92127) 3 mole of water vapor? The balanced equation is 2 H2O(g) → 2 H2(g) + O2(g) and the relevant bond energies are H H : 436 kJ/mol; H O : 467 kJ/mol; O O : 146 kJ/mol; O O : 498 kJ/mol. 1. −436 kJ 2. +249 kJ 3. −425 kJ 4. +425 kJ

Answer 1

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Answer:

249 kJ

Step-by-step explanation:

To obtain the energy change of the reaction:

H₂O → H₂ + ¹/₂ O₂

It is necessary to obtain the difference between bond energy of the products and bond energy of the reactant, thus:

Energy of products:

1 mol of H-H bond × 436 kJ/mol = 436 kJ

¹/₂ mol of O=O bond × 498 kJ/mol = 249 kJ

Energy of reactant:

2 mol of H-O bond × 467 kJ/mol = 934 kJ

Energy change of the reaction is:

934 kJ - (436 kJ + 249 kJ) = 249 kJ

Step-by-step explanation:

Answer 2

249 kJ

Step-by-step explanation:

To obtain the energy change of the reaction:

H₂O → H₂ + ¹/₂ O₂

It is necessary to obtain the difference between bond energy of the products and bond energy of the reactant, thus:

Energy of products:

1 mol of H-H bond × 436 kJ/mol = 436 kJ

¹/₂ mol of O=O bond × 498 kJ/mol = 249 kJ

Energy of reactant:

2 mol of H-O bond × 467 kJ/mol = 934 kJ

Energy change of the reaction is:

934 kJ - (436 kJ + 249 kJ) = 249 kJ

I hope it helps!