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I noticed that the enthalpy of the products of an exothermic reaction (bonds formed are stronger than bonds broken) is lower than the enthalpy of the reactants. Enthalpy equals internal energy plus $PV$. In a constant pressure environment, change in enthalpy equals to change in internal energy as $PV$ in constant. Internal energy consists of $E_k$ (kinetic energy) of particles and $E_p$ (potential energy) of bonds between particles. When stronger bonds are formed, $E_p$ is bigger. And the energy of the heat released to environment equals to the energy gained in form of potential energy, so why is the internal energy lower? It would be helpful if someone shows me my mistake :)

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  • $\begingroup$ Frequently the convention is $KE$ and $PE$ or if you have a physics bent, $T$ and $U$, instead of $E_{k}$ or $E_{p}$, respectively. $\endgroup$ – Zhe Apr 17 '17 at 18:11
  • $\begingroup$ These are simply conventions. For more grasp, consider it like this: energy goes from high to low, that is energy flow is downwards, which means a negative enthalpy change $\endgroup$ – Pritt says Reinstate Monica Apr 18 '17 at 16:27
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Bonds actually have negative potential energy. Think about it this way: in order to break a bond you have to expend energy so when a bond is created, energy must therefore naturally be released. Hence internal energy is lower.

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When stronger bonds are formed, *Ep* is bigger.

I take issue with this statement. When stronger bonds are formed, the internal energy becomes more negative, not just "bigger". It is this decrease in internal energy that makes the products' bonds so strong, since now the activation energy will be very large for a reaction to occur.

This link might help to explain: http://www.rsc.org/Education/Teachers/Resources/cfb/energy.htm

My favorite passage from this link is, looking at a one-step exergonic reaction: "The height of the peak from the reactants side is a measure of the strength of the bonds in the reactants and the height of the peak from the products is a measure of the strength of the bonds in the product. Thus we can consider the bonds in the product to be stronger than the bonds in the reactants."

If PV constant, this means Ek is the same. Ep gets more negative. So naturally enthalpy decreases (negative sign).

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When energy goes off from a molecule, we take the sign as negative - this implies that the sign of PE which we are considering in formation of stronger bonds is MORE, albeit WITH NEGATIVE SIGN, hence, it is going down in energy coordinate.

The magnitude of change in PE, with formation of stronger bonds is more, this will lower the Internal Energy.

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The change in enthalpy is negative in an exothermic reaction because energy is "lost" through the reaction (because there is more energy on the products side than on the reactants side).

Another way to think about this is by calculating the enthalpy before and after a reaction, for example - and this is a synthesis [and exothermic] reaction:

$$\ce{2H2 + O2 <=> 2H2O}$$ The ΔH [change in enthalpy] is -484 kJ (you can calculate this using standard enthalpy of formation numbers).

You can also think back to the definition of enthalpy - disorder: When two molecules that were 'freely' moving around, come together, there is less disorder [so the disorder in the system has decreased, and consequently is negative].

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