Use the bond enthalpies tabulated below to estimate thenthalpy of formation of H

Question

Use the bond enthalpies tabulated below to estimate thenthalpy of formation of H2O(g). H2(g) + 1/2 O2(g) -> H2O(g)     Bond                         Bond Enthalpy (kj/mol*rxn)     H-H                           436     O-H                            463     O-O                          146     O=O                          498 Use the bond enthalpies tabulated below to estimate thenthalpy of formation of H2O(g). H2(g) + 1/2 O2(g) -> H2O(g)     Bond                         Bond Enthalpy (kj/mol*rxn)     H-H                           436     O-H                            463     O-O                          146     O=O                          498     H-H                           436     O-H                            463     O-O                          146     O=O                          498

Explanation / Answer

In this equation we are breaking 1 H-H bond, 1/2 O-O bond andforming 2 H-O bonds. Based on this: Since the numebrs given are for forming we need to reversethem for those bonds that we are breaking H-H is being broken this Enthalpy is -436 .5 O-O is being broken thus .5 X -146 = -73 2 H-O bonds are forming thus 2 X 463. Add all the values andyou will get the answer: Answer is 417 KJ/Mol Please Rate Please Rate

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