starpaste Posted October 16, 2008 Share Posted October 16, 2008 For the equation change in enthalpy = bonds broken - bonds made. What if the compound makes a double bond. Meaning in the reactants there is a C--O bond, in the product it becomes C=O, do i add for bonds made the value for a single C-O bond (since 1 C--O bond already exists) or the value for a C=O (the final product)? Link to comment Share on other sites More sharing options...
hermanntrude Posted October 16, 2008 Share Posted October 16, 2008 assuming you've been given the data for the C-O and C=O bonds, I'd say you're safest using the values for C-O in "bonds broken" and C=O in "bonds formed". You'll probably also notice that the value for C=O is NOT twice the value for C-O Link to comment Share on other sites More sharing options...
starpaste Posted October 17, 2008 Author Share Posted October 17, 2008 the thing is that the C--O is not being broken to make to C==O bond, because I am comparing an intermediate to the product it makes, so only electrons are being shifted in the mechanism. wouldn't it be the value of c==o - c--o? Link to comment Share on other sites More sharing options...
hermanntrude Posted October 18, 2008 Share Posted October 18, 2008 There's a law which you may or may not have heard of, called hess's law. It states that the heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps. In other words, it doesnt matter if you measure the enthalpy change by breaking the C-O bond and then forming a C=O bond or if you just jump from C-O directly to C=O. the advantage of the former method is that you can do it using the numbers you've been given. essentially it works out the same. Link to comment Share on other sites More sharing options...
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