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Can we determine whether a reaction is endothermic or exothermic? For example if we are given the following reactions:

$$\ce{2SO2 + O2 -> 2SO3}$$

$$\ce{H2 + I2 ->2HI}$$

$$\ce{N2 + 3H2 -> 2NH3}$$

Can we determine weather these reactions are endothermic or exothermic or do we have to just memorize this? Are there any general trends that could make memorizing this information more manageable?

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I edited your question to format the chemistry look like chemistry using the MathJax plugin. You can learn how to do this yourself from the help center: chemistry.stackexchange.com/help/notation –  Ben Norris Jul 22 '13 at 1:54
    
thanks. i will try to learn it. –  Rafique Jul 22 '13 at 2:17
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2 Answers 2

As anglinb pointed out, the only completely reliable method of determining whether the reactions are endo- or exothermic is by calculation. That said, certain generalizations may allow reasonable predictions about changes in thermodynamic parameters to be made:

  • Reactions that produce a net increase in $\sigma$-bonds between particular pairs of atoms that were $\pi$-bonded in the reactants tend to be exothermic, since $\sigma$-bonds tend to be more stable than $\pi$-bonds. Conversely, reactions in which the generation of $\pi$-bonds and reduction in number of equivalent $\sigma$-bonds occurs tend to be endothermic.
  • Reactions that involve a net reduction in the total number of molecules of products as compared to reactants, and/or involve the generation of products in more highly ordered states (for example, the generation of solids from liquids or gases), tend to occur with a reduction in entropy. Reactions that involve the generation of heteroatomic compounds from homoatomic substances tend to increase entropy. The spontaneity of a reaction is governed by the equation $\Delta G = \Delta H - T\Delta S$, where $\Delta G < 0$ for a spontaneous process. Therefore, if a reaction occurs spontaneously while the entropy change ($\Delta S$) is negative, the reaction must be exothermic by default; an endothermic process with negative entropy change cannot be spontaneous.

Remember that these are very rough guidelines, and ultimately only performing calculations with concrete values for the relevant thermodynamic parameters can be totally definitive. The first and third reactions from your question can be predicted to have negative $\Delta S$ by the above guidelines, and therefore must be exothermic (if they're supposed to be spontaneous at all). Crunching the numbers using thermodynamic data does in fact confirm that.

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In the first guideline, you have mentioned that if there is a net increase in the sigma bonds between particular pairs of atoms that were pi bonded in the reactants then reactions tend to be exothermic. But to break pi bond, we will need to provide energy so, shouldn't the reaction be endothermic then? –  Rafique Jul 23 '13 at 23:08
    
Yes, some energy is required to initially break the bonds (which is factored into the energy of activation), but the formation of bonds always releases energy; the only question is whether that energy released upon completion is greater than the initial input of energy required to set the reaction in motion. When a sigma bond between some generic pair of atoms forms, it's typical that more energy is released than would be required to break a pi bond between the same two atoms, so the reaction is exothermic overall. –  Greg E. Jul 24 '13 at 1:56
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In general just by looking at these equations there is no surefire way to know which is endo or exothermic. However you can look at the bond strength or the ∆H of formation for each molecule. In short you have to know something.

enter image description here

The energy stored in the molecules(specifically in the bonds) is lower before the reaction in an endothermic reaction and higher in an exothermic reaction. This means if you know relatively how strong the bonds or how much energy it takes to form each of the molecules then you can know whether the reaction is endo or exothermic.

If you have a few ∆H of formation values you can use Hess's Law

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