The enthalpy change for the reaction
2 H2 + O2 > 2 H20
is -571.6 kJ. Determine the enthalpy change for the decomposition of 24.0g H2O.
My Process
-571.6 is the enthalpy of 2 mols of H2O.
So the enthalpy of 1 mol of H2O will be -285.8.
Since it's decomposition, -285.8 will become +285.8.
Then I would have to convert the 24g into mols. 24/18.02= 1.3 mols
Now I am lost fro here on out
5 answers
See your CH4 post. It's done the same way but make sure that's for 2 mols H2O.
would I multiply my answer from 1.3(285.8) by 2 then?
To follow the same pattern I followed above it would be
+571.6 kJ for 36 g H2O so how much for 24 g?
(571.6/36) x 24 = ?
You've already divided the 571.6 by 2 so your step is ok as is EXCEPT you rounded and I don't know why? 24/18 = 1.33 mol.
+571.6 kJ for 36 g H2O so how much for 24 g?
(571.6/36) x 24 = ?
You've already divided the 571.6 by 2 so your step is ok as is EXCEPT you rounded and I don't know why? 24/18 = 1.33 mol.
I was trying to keep sig figs. If I were trying to do it my way, I would just multiply 1.33 by 285.8 and that would get me the same answer(almost) at 380.11. Yours is 381.07
If that 24 is correct (and not 24.0) then you're right and I'm wrong and the number of significant figures is 2 so 1.3 is appropriate.
But no matter which way you do the problem you should end up with the same answer if you use the same numbers for the molar masses.
But no matter which way you do the problem you should end up with the same answer if you use the same numbers for the molar masses.
5 answers