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Stability of Products: Nuclear binding energy vs Enthalpy

  1. Jul 20, 2011 #1
    I want to get my head around this....

    Why is that in nuclear fusion, the formed nucleus is more stable because its nuclear binding energy/nucleon is HIGHER than the sum of its reactants


    in general chemical reactions products are more stable if their enthalpy is LOWER than the sum of their reactants

    I know that in both cases: bonds are formed, the products are more stable, and energy is released...so it has something to do with the definitions of what is +ve and -ve

    But I'm trying to connect the dots here, and would appreciate any help :)

  2. jcsd
  3. Jul 20, 2011 #2


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    Binding energy refers to the energy needed to pull the nucleus apart. It is the inverse of Enthalpy which is a measure of the total energy of something. The higher the binding energy of a nucleus, the less energy it contains compared to its parts before fusion, as it released more energy during binding.
  4. Jul 20, 2011 #3
    I think i get it know...the bolded part is the key.

    Binding energy is actually equivalent to what we would call 'bond strength' in general chemical species.

    What is the equivalent of enthalpy for a nucleus though? Can we say more stable (higher binding energy) nucleus has lower 'nuclear enthalpy' too?
  5. Jul 20, 2011 #4


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    I would say yes. The mass of an atom of Deuterium is LESS than a proton, neutron, and electron combined when they are free.
  6. Jul 21, 2011 #5
    Can you say less mass would imply less enthalpy?
  7. Jul 21, 2011 #6


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    Only in the right context. Deuterium has much less mass than Iron, but it is able to give off energy when it is fused, while Iron does not. So it just depends on what you are comparing and how.
  8. Jul 21, 2011 #7
    cool....thanks for the replies
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