Bond Strength and Energy
Name: Sharon K.
Date: July 2006
I have read that spontaneous reactions, such as the
reaction of hydrogen and oxygen to form water is spontaneous
because the bonds in Water (H-O-H) molecules are stronger than
those of Hydrogen (H-H) and Oxygen (O-O) molecules. Because the
(H-O) bonds in water are stronger, they have less chemical energy.
Why do molecules possessing stronger bonds have less chemical
energy? Does this refer to kinetic energy? Potential energy? Also,
what exactly about glucose makes it an energy-rich compound, as
opposed to the carbon dioxide and water from which it is made? I am
trying to understand why chemical reactions happen in terms of the
physical properties of the atoms and molecules as opposed to
general explanations involving free energy changes. Thanks!
Scientists tend to prefer giving explanations involving free energy
changes because, although the concepts are more abstract than
thinking about particular bonds, the free energy explanations are
actually more correct. I will see what I can do talking about
bonding that will not be too misleading.
When we talk about bond strengths or about strong or weak bonds,
what we really mean is the energy required to BREAK the bonds, that
is, to rip the attached atoms apart. Much energy is needed to break
a strong bond, and less energy is needed to break a weak bond. So
the bonds between atoms can be thought of as potential
energy. NEGATIVE potential energy, in fact. So, in other words, a
strong bond has very negative potential energy, while a weak bond
has moderately or slightly negative potential energy. Much more
energy must be dumped into a strong bond to return its atoms to zero
potential energy (that is, separated from each other) than is needed
to dump into a weak bond to bring its atoms to the same zero
potential energy. Make sense?
As for what exactly makes glucose high-energy compared to something
else: make sure you have the correct something else in mind. It is
not quite right to compare glucose alone to water and carbon
dioxide, because glucose does not really decompose into them. The
overall balanced equation for forming glucose from water and carbon dioxide is:
6 CO2 + 6 H2O -> C6H12O6 + 6 O2
So, what is really high-energy compared to (carbon dioxide + water)
is (glucose + OXYGEN).
Looking at it bond-by-bond: in glucose, the hydrogen atoms are
attached either to carbon or to oxygen; in water, they are all
attached to oxygen, which is better (in the sense of lower potential
energy). in glucose, the carbon atoms are attached to oxygen atoms,
carbon atoms, and sometimes hydrogen atoms; in carbon dioxide, they
are all attached to oxygen, which is better. And let us not forget
the oxygen atoms; in glucose, they are all attached to carbon,
hydrogen, or both, and in oxygen gas, they are all attached to other
oxygen atoms. They are better off in water and carbon dioxide.
This analysis is not strictly fit because not all C-O or H-O bonds
are exactly the same, but the general trends are close enough that
you get the right answer for approximately the right reasons.
Explaining spontaneity without referring to Gibbs Free Energy is
hard, because in my mind spontaneity is precisely a measure of
changes in free energy in reference to the 2nd Law of
Thermodynamics. But . . . I will try:
We think of enthalpy change as the difference between the heat that
is put into a system, versus that which is released by the system
(assuming no PV work). In chemical potential energy terms (or
chemical bonds), the enthalpy change of a system is a function of
the difference in the energy required to break bonds (increase the
potential energy) and the energy that is released in forming bonds
(decreases potential energy). That's half the picture.
The other half involves entropy. If we imagine entropy as the number
of possible microstates of a system, then we can imagine that a
small solid sample in a 10L container would have a lower entropy
(fewer microstates, less disorder) then if the solid were turned
into a gas (more microstates, more disorganized). So then, we can
imagine that the transition from solid to gas must be
thermodynamically favored (by second law of thermodynamics) since
entropy increases in this transformation.
The two preceding paragraphs are actually talking about the same
thing - transformations that increase the entropy of the universe -
which is what accompanies spontaneous transformations.
Applying these thoughts to your specific example: assuming that
there is no difference between the microstates of the
hydrogen/oxygen gas system versus that of the water (in gaseous form
system) then we can refer to the enthalpy change that occurs as H-H
and O-O bonds are broken and H-O bonds are formed. Since this
process has a negative enthalpy (energy is released) then it must
mean that the over-all potential energy of the system has gone down
(accounting for the heat released), and the process is spontaneous.
In the case of the combustion of glucose to form CO2 and H2O, you
only need to count molecules on the left and right side of the
equation to realize that the microstates must increase (not to
mention that glucose is a solid). So, entropy increases, the process
is spontaneous. You can also apply the reasoning of the previous
paragraph and approach it from the point of view of enthalpy
So, obviously, I am cheating here. I avoided mention of free energy
by taking the two terms in the Gibbs Free Energy equation: enthalpy
and entropy (which are really just entropy and goes back to the 2nd Law).
Hope this was not needlessly convoluted.
Greg (Roberto Gregorius)
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Update: June 2012