Cherries Posted April 29, 2018 Report Share Posted April 29, 2018 (edited) Which equation represents the bond enthalpy for the H–Br bond in hydrogen bromide? A. HBr(g) → H(g) + Br(g) B. HBr(g) → H(g) + Br(l) C. HBr(g) → H(g) + 1/2Br2(l) D. HBr(g) → H(g) + 1/2Br2(g) The answer is A. I understand that all substances should be involved should be gaseous, but why cannot the answer be D? Thank you! Edited April 29, 2018 by Cherries 1 Reply Link to post Share on other sites More sharing options...
tim9800 Posted April 29, 2018 Report Share Posted April 29, 2018 The equation in D doesn't balance, even though everything is in gaseous state. Reply Link to post Share on other sites More sharing options...
Cherries Posted April 29, 2018 Author Report Share Posted April 29, 2018 I just realized that the 1/2 did not show up before the Br in C & D, so it is balanced. Reply Link to post Share on other sites More sharing options...
tim9800 Posted April 29, 2018 Report Share Posted April 29, 2018 We know that bond enthalpy is defined as: "The energy required to break one mole of bonds in gaseous molecules in their standard state" D is not the bond enthalpy of HBr because the products contain a species which still has a bond (Br - Br). From my understanding, bond enthalpy requires everything on the products side of the equation to be unbonded i.e. in their isolated states. 1 Reply Link to post Share on other sites More sharing options...
Cherries Posted April 29, 2018 Author Report Share Posted April 29, 2018 Makes sense, thank you! Reply Link to post Share on other sites More sharing options...
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