Hess's Law, Multiplying Coefficients

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Mara Lockhart 3J
Posts: 53
Joined: Fri Sep 26, 2014 2:02 pm

Hess's Law, Multiplying Coefficients

Postby Mara Lockhart 3J » Wed Jan 14, 2015 4:57 pm

On problem 7.63, we have to find the balanced equations of three reactions and for one of the reactions, we find it to be H2(g) + 1/2O2(g) ---> H20. We then have to multiply the reaction by 2, leading us to multiply the delta H by two as well.

However, why can't we just have the reaction as 2H2 + O2----> 2H2o in the beginning, which would mean we would not end up multiplying the delta H by 2? How do you know to have the 1/2O2 rather than the 2H2 and 2H20 from the start?

Sanmeet Atwal 1D
Posts: 17
Joined: Fri Sep 26, 2014 2:02 pm

Re: Hess's Law, Multiplying Coefficients

Postby Sanmeet Atwal 1D » Wed Jan 14, 2015 8:16 pm

We have to write the reaction like this in the beginning:
H2(g) + 1/2 O2(g) ----> H20 (l) ΔHformation: -286 kJ
because the standard enthalpy of formation of water for one mol is: -286 kJ so in the reaction you want one mol of water which means 02 has to have 1/2 as the coefficient for it to be balanced correctly.
Thats why you have to multiply the ΔHformation of water by 2: 2(-286 kJ/mol) when the above reaction is written:
2H2 (g) + O2 (g) ----> 2H20 (l) ΔHformation: (-286 kJ/mol)*(2mol) = -572 kJ
Hope this helps!

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

Re: Hess's Law, Multiplying Coefficients

Postby Neil DSilva 1L » Wed Jan 14, 2015 11:39 pm

Problem 7.63 deals with the which is the enthalpy of combustion for one mole of . (This is the same equation and enthalpy as the reaction for the formation of one mole of water, but since the problem specifically deals with the enthalpy of combustion, I just wanted to leave this note here, so you're not confused.)


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