Calculate the ΔH° for the following reaction.?
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Answer:
Lancenigo di Villorba (TV), Italy You asked :"Using bond energies, calculate the ΔH° for the following reaction. H2(g) + O2(g) => H2O (g) (unbalanced) Bond BE (kJ/mol) H-H 432 O 2 494 H-O 459" The chemical equation has to result 2 H2(g) + O2(g) ---> 2 H2O(g) but since I AM LOOKING FOR COVALENT BONDs, I NEED SHOW IT 2 H-H(g) + O=O(g) ---> 2 H-O-H(g) CHEMISTS STATED THE ENTHALPY REACTION YOU SUBTRACTED ALL Bond Energies Involved in Chemical Products BY MATHEMATICAL SUM OF Bond Energies Involved in Chemical Reactants. I DID IT FOR YOU Delta(H°) = (B.E.)reactants - (B.E.)products Delta(H°) = (2 * 432 + 494) - (2 * 2 * 459) = -456 kJ/mol I hope this helps you.
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Other answers
Lancenigo di Villorba (TV), Italy You asked :"Using bond energies, calculate the ΔH° for the following reaction. H2(g) + O2(g) => H2O (g) (unbalanced) Bond BE (kJ/mol) H-H 432 O 2 494 H-O 459" The chemical equation has to result 2 H2(g) + O2(g) ---> 2 H2O(g) but since I AM LOOKING FOR COVALENT BONDs, I NEED SHOW IT 2 H-H(g) + O=O(g) ---> 2 H-O-H(g) CHEMISTS STATED THE ENTHALPY REACTION YOU SUBTRACTED ALL Bond Energies Involved in Chemical Products BY MATHEMATICAL SUM OF Bond Energies Involved in Chemical Reactants. I DID IT FOR YOU Delta(H°) = (B.E.)reactants - (B.E.)products Delta(H°) = (2 * 432 + 494) - (2 * 2 * 459) = -456 kJ/mol I hope this helps you.
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