1
$\begingroup$

Consider the paraphrase from Ebbing:

It takes energy to break bonds & energy is released when bonds are formed

Let's look at the first statement It takes energy to break bonds.

  1. Is this true of spontaneous processes?
  2. If yes, is this true only in a closed-environment sense, where the energy required for a reaction to occur between two molecules is the energy contained within those molecules; i.e. no outside energy is necessary?
  3. If yes to 2), is this energy, outside the kinetic energy of the molecules themselves, potential energy (PE) stored in the bonds?

Then for the second statement, energy is released when bonds are formed. When the PE of the reactants is greater than the PE of the products, heat is released and we have an exothermic process. Consider the conversion of ethane to ethene. This is an endothermic reaction requiring approximately 220 kJ/mol of energy (bonds broken (2X C-H) - bonds formed (C=C).

  1. As the PE of ethene is greater than that of ethane, why is this not reflected when considering the sum of bond energies in each molecule? For example, for ethane we have 2,812 kJ/mol and for ethene we have 2,246 kJ/mol.
  2. Given the bond energy of C=C (602 kJ/mol) is higher than C-C (346 kJ/mol), why would it be incorrect to assume the energy necessary to drive the reaction is stored in the C=C bond, given the PE of ethene is greater than that of ethane?

Lastly, when teaching cellular respiration, we speak of energy being stored in the phosphate bonds when ADP is converted to ATP and in photosynthesis, of the energy from sunlight being stored in the bonds of glucose.

  1. Is this line of thinking correct, if not why?

  2. If that line of thinking is correct, how do we reconcile this fact with "Energy is released when bonds are formed?

  3. Would it be appropriate to say energy is both stored and released as bonds are formed?

I am a tutor hoping to master this content to best help my students. Please lend a hand to whichever questions you feel most confident in answering.

$\endgroup$
  • $\begingroup$ The planetary model is of course incorrect, but think of the electrons "dropping" into lower orbitals. Being closer to the nucleus the electron "picks up" energy. However the electron can't have that energy in the closer orbital, so the electron has to shed the energy somehow. $\endgroup$ – MaxW Oct 14 '19 at 20:01
  • $\begingroup$ In the ethane/ethene case don't forget hydrogen. $\endgroup$ – Buck Thorn Oct 14 '19 at 20:30
2
$\begingroup$

Let's start with the specific examples. For the conversion of ethene to ethane, you need a reaction partner to provide the extra hydrogen, e.g. dihydrogen:

$$\ce{H2C=CH2 + H2 -> H3C-CH3}$$

In the reaction, a single bond and the pi-bond of the double bond are broken, and two single bonds are formed. If you look up the four bond dissociation energies (two broken and two formed), you get a fairly good estimate of the reaction enthalpy.

$$\ce{ADP + phosphate -> ATP + H2O}$$

I'm leaving out the charges and the protonation state of phosphate to make things simpler. If we look a bit closer, we can write:

$$\ce{RO3P-OH + H-OPO3R' -> RO3P-OPO3R' + H-OH}$$

So when ATP is formed, we are not only making the phospho-anhydride bond (and a bond within water), but also breaking two bonds.

Lastly, when teaching cellular respiration, we speak of energy being stored in the phosphate bonds when ADP is converted to ATP and in photosynthesis, of the energy from sunlight being stored in the bonds of glucose.

  1. Is this line of thinking correct, if not why?

You could say the energy is stored in the reaction, and when the reaction runs in the other direction (hydrolysis), it gets released again. Similarly, glucose is a source of free energy only if oxygen is available as reaction partner. Breaking glucose into atoms is not a source of free energy.

Let's look at the first statement It takes energy to break bonds.

  1. Is this true of spontaneous processes?

Here is the reaction this refers to:

$$\ce{molecule -> free atoms}$$

The entropy of reaction is expected to be positive (more particles on product side). If the reaction were also exothermic, these molecules would not be stable under any condition (Gibbs energy of reaction would be negative at any temperature). So bonds of molecules that actually exist are expected to have a positive bond dissociation energy.

The other questions

The answers to the rest of the questions follow from the information above.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I like your use of logic in the last paragraph, even though I've yet to encounter a bond with a negative bond dissociation energy :-) $\endgroup$ – Buck Thorn Oct 15 '19 at 16:49
1
$\begingroup$

On breaking a bond: dissociation will occur if the energy to overcome the attraction that is holding the atoms together is somehow made available to the bond, such that it can be excited into a higher energy state leading to dissociation. "Spontaneity" would imply access to the excess energy that will drive the process forward. No energy, no dissociation. The energy can come from laser photons with just the right energy to excite the appropriate vibrational mode and trigger dissociation, or, just as well, the energy might be provided by a heat bath that results in energy being transferred through collisions into the molecule and eventually into the vibrational mode that results in dissociation. If you had only two molecules, dissociation would require inter- or intramolecular transfer of energy into the cleaved bond. The molecules would evidently have to be excited, loaded with excess energy that when transferred would raise the vibrational state of the cleaved bond well above the ground state. Ultimately the energy would be stored as potential energy of the now dissociated bond plus kinetic energy from the excess above the required dissociation energy. The heat bath would provide kinetic energy to the products.

Molecular process are reversible, which means bond formation could proceed through precisely the same path, but in reverse. Atoms would come together to form a highly vibrationally excited product that could shed that energy by distributing it to all available degrees of freedom, including eventually into surrounding molecules.

Regarding whether energy can be stored in bonds: yes, it can, in the same way a ball may come to rest in a higher energy dip. However some of these questions might be better dealt with using a thermodynamic treatment that addresses the free energy of the substances in specific states. These questions are answered in other posts.

| improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.