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Ch 9.4a: Bond breaking and forming | General Chemistry | YouTubeToText
YouTube Transcript: Ch 9.4a: Bond breaking and forming
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Video Summary
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Core Theme
This content explains the energetics of ionic and covalent bond formation and breakage, demonstrating how bond strengths can be used to calculate enthalpies of reaction.
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so the last section of the book that
we're going to cover this semester is
about ionic strength and uh I strengths
of ionic and coal bonds so after this
section you should be able to describe
the energetics of ionic and calent bond
formation uh through form Bond formation
and breakage then we are going to do um
a born habber cycle to compute lattice
energies for IA compounds and we can use
these average bond eny to estimate our
enthalpies of
reactions so just as a reminder bond
breaking is an endothermic reaction so
whenever we break a bond this actually requires
energy so we can start off with hydrogen
we can add energy to it and then we can
actually get two individual hydrogen
atoms so we can break apart that Bond
Bond forming so whenever we form a bond
energy so we can have two hydrogens so
remember when we went way back in terms
of thinking about what happens when we
take two hydrogen atoms bringing them
close together all of a sudden the
energy drops and we form a bond so our
two hydrogens will actually form
hydrogen glass and we will release
energy and just as a rinder since this
is like hess's law and it's a state
function whether you go forward or
backwards the enthalpy is a change so
for our bond breaking this requires
436 kles for every mole of hydrogen gas
or if we look at the bond uh forming
this will release
436 kog for every two hydrogen atoms
that come together
so we can think about these con Bond
strengths in terms of coal bonds and
looking at them so we can think about
lots of these Bond energies and what
happens to these kinds of tables and the
formation so there there's a lot of
Trends in kind of looking at this one of
the trends to think about if we think if
we look at just a single Bond over here
a double bond takes more energy 6111 k a
triple bond is super super strong so in
order to break that Bond it's 300 uh
837 K and remember our bond lengths get
shorter as the bond gets stronger on
there so short bonds are going to be
much stronger and they're going to
require more
energies so in terms of thinking about
our enthalpies and our enthalpies of
reactions we can look at the enthalpy of
reaction and we can estimate it using
these kinds of tables so we can look at
the sum of all the bonds broken minus
the sum of all the bonds that are formed
so in this
example we can think about this hydrogen
we're looking at
hydrogen um and it's going to react with
HCL two of those so in this reaction we
can look at this and think about how
many bonds are actually forming and how
many are breaking so we are going to
break a hydrogen hydrogen bond we're
going to break a chlorine chlorine bond
and we are going to form two HCL bonds
so we can calculate our Delta H uh for
this our Delta H is going to be equal
the number of bonds broken SK so we look
up at the enthalpies of up here and we
have an HH is
436 um and our CL is [Music]
[Music]
243 uh right here so that's sum
436 K plus 243 I'm going to use the
bracket since we have a summation here
and we got to watch out for this
negative sign in the summation we're
going to subtract from that all all the
bonds forms we end up getting two hcls
we look HCL that's
432 so there's two times uh of those
432 and when we sum that up we end up
getting minus
185 kog for every mole of hydrogen for
every mole of chlorine and for every uh
two moles a hydrochloric acid that are
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