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sera2018

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May 20, 2016
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  1. Pre-Medical
    I'm having a hard time conceptually making sense of high E bonds and energy release.

    I've learned that bonds store energy and so was under the impression that breaking a bond would release Energy but it's the opposite.

    In ATP hydrolysis, it's favorable and exergonic. I've read another thread explaining that E released by ATP hydrolysis is actually due to a small amt of E being put into breaking the weak bonds of ATP followed by the release of a large amt of E due to formation of stable, low E bonds, thus resulting in a net energy release making the rxn favorable.

    Why are the ATP phosphanhydridic bonds considered weak but high E?
    Also, if this is true, are all metabolic rxns that use ATP to push a rxn, getting the E from not the breaking of the bonds but the formation of a bond with part of the ATP?
     

    pine138

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    Dec 11, 2014
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    1. Pre-Medical
      The way I understand bond formation and breaking is through the gibbs free energy equation
      G = H - TS

      In bond formation there is a decrease in entropy, which is unfavorable, but can be overcome with a favorable (negative) enthalpy change.
      In the case of ATP, formation of the phosphoanhidride bond between the beta and gamma phosphate is unfavorable from an enthalpy point of view because putting a highly negative phosphate beside a highly negative phosphate is unfavorable. this is why formation of ATP is endergonic and thus, the hydrolysis of this bond is exergonic.

      In this same regard, metabolic reaction couple the free energy released from the hydrolysis of ATP -> ADP (exergonic) in order to drive endergonic reactions
       
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      aldol16

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        I've learned that bonds store energy and so was under the impression that breaking a bond would release Energy but it's the opposite.

        Never take a process on its own. When you're breaking a bond, you're usually making another bond as well. The overall favorability of that process depends on the relative energies of the two bonds. If you're breaking a weak bond to make a stronger bond, then that is favorable.
         
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        sera2018

        Full Member
        5+ Year Member
        May 20, 2016
        30
        64
        New York
        1. Pre-Medical
          The way I understand bond formation and breaking is through the gibbs free energy equation
          G = H - TS

          In bond formation there is a decrease in entropy, which is unfavorable, but can be overcome with a favorable (negative) enthalpy change.
          In the case of ATP, formation of the phosphoanhidride bond between the beta and gamma phosphate is unfavorable from an enthalpy point of view because putting a highly negative phosphate beside a highly negative phosphate is unfavorable. this is why formation of ATP is endergonic and thus, the hydrolysis of this bond is exergonic.

          In this same regard, metabolic reaction couple the free energy released from the hydrolysis of ATP -> ADP (exergonic) in order to drive endergonic reactions

          Thanks! That makes sense. I guess I forgot to make the connection that those bonds are actually unfavorable
           
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