Standard enthaply of reaction

  • Thread starter zeshkani
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  • #1
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this is my question , i just dont know how to get the standard enthaply of reaction
of this problem iam sure its easy, but i just dont get it, any help is welcome

14. Given the reactions (1) and (2) below, determine (a) ΔrH and ΔrU for reaction (3), (b) ΔrH for both HBr(g) and H2O(g) all at 298 K.
(1) H2(g) + Br2(l) → 2 HBr(g) ΔrH = - 72.80 kJ/mol
(2) 2 H2(g) + O2(g) → 2 H2O(g) ΔrH = –483.64 kJ/mol
(3) 4 HBr(g) + O2(g) → 2 Br2(l) + 2 H2O(g)
 

Answers and Replies

  • #2
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Look at reaction (3) as two separate phases, first reaction (1) happens in reverse and then reaction (2) happens. This isn't what happens but all the energies involved here are state functions which means that it doesn't matter how you got to that state, the energy is always the same. This way complex problems can be simplified.

PS. This also works for example when you want to calculate the difference in volume of a gas when you both heat it and change the pressure. First you calculate the change in volume due to the pressure change and than with the new volume you do the same with the new temperature (or vise versa, it doesnt matter). It's neat, give it a go ;)
 

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