<p>I'm confused on one of the examples given on sparknotes.</p>
<p><a href="http://www.sparknotes.com/testprep/books/sat2/chemistry/chapter9section5.rhtml%5B/url%5D">http://www.sparknotes.com/testprep/books/sat2/chemistry/chapter9section5.rhtml</a></p>
<p>On the first example, they say DHrxn is 28.8/2 = 14.4. But what happened to the -.08 from the first equation? DH's from the second two equations add up to 28.8...shouldn't you subtract .08 from this to get 28.72/2 = 14.36? </p>
<p>I know that in this particular problem it's very negligible and you can mostly round on all of the math. But is this the correct way to solve a prob like this? You can't just ignore all of the negative enthalpies...</p>
<p>Thanks!</p>
<p>You only have to know Hess and its descriptive definition...you do not need to know how to use it THIS thoroughly...just know how to calculate if reaction mechs and enthalpy values are given..</p>
<p>I know it's extremely picky on this prob, but in general, we do subtract the negative enthalpies, right?</p>
<p>Yes, you sum all of the enthalpy values together. I think for this problem it was discounted because the number was so small...</p>
<p>Thanks for the prompt replies, guys!</p>
<p>Not a formula or anything. Just states that the enthalpy of a reaction such as </p>
<p>H2O(s) --> H2O(g) </p>
<p>is equal to the sum of the enthalpies of all the steps you must go thru to get from H2O solid to H2O gas. </p>
<p>enthalpy of...
H2O at starting temp --> H2O at 0 degrees C, plus
H2O(s) -- > H2O(l), plus
H2O at 0 degrees --> H2O at 100 degrees, plus
H2O(l) --> H2O(g)</p>
<p>...equals total enthalpy.</p>