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Ch 9.3c: Standard state | General Chemistry | YouTubeToText
YouTube Transcript: Ch 9.3c: Standard state
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Video Summary
Summary
Core Theme
This content explains the concept of standard state enthalpy changes, specifically focusing on standard enthalpy of combustion and formation, and demonstrates how to use these values in calculations.
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so when we think about enthalpy we have
some definitions that we use and one of
them is what we call the standard state
so the standard State we are going to
indicate our Delta H with a not sign
above it so this indicates standard
State what does that actually mean
according to
IUPAC standard state is one bar and one molar
molar
concentrations however if you look at
most Labs at the end of your chap at
your end of your book there's actually
an appendix G they actually list um all
of their information um in a slightly
different way so most tables with
thermodynamic uh information they list
everything as one atmosphere and this is
really really close to a
bar and since we're doing things under
constant pressure typically it usually
doesn't matter so this is approximately
concentrations and our temperature is it
always listed but we usually do all of
our temperature all of our reactions at room
room
or
298.15 kelvin when we do our conversion
so we will see these Nots in a lot of
our thermodynamic kinds of calculations
so that just means standard state so you
can assume these values but if you don't
see a not sign if you don't see the zero
that means these are non standard
conditions so you'll have to read the
problem very
carefully so we also have some of the
other values that we will see in our
tables that are very common so our
standard enthalpy of combustion is going
to be indicated with a Delta H knot so
we're at standard conditions and you'll
see a lowercase C to it so that just
indicates that we are thinking about a
combustion so remember when we are
talking about reactions a combustion
reaction reacts oxygen with some sort of
organic compound that contains
CH uh kind of elements in it and we
always produce carbon dioxide and water
so we can use this to actually calculate
some of our enthalpies and if we look up
in the uh appendix G uh we find out the
enthalpy of combustion is uh 13668
13668
K so what if we had 1.0 lers of
ethanol so e is ethanol um and then how
many kles of energy is going to be
created by burning ethanol so our kogs
are going to be trying to calculate this
we have to go ahead and convert this to
liter uh
milliliters for one liter and then if we
look up the density of
ethanol uh it's
789 grams per
milliliter and then we multiply that or
divide that by the molecular weight so
it has um 46.0
7 G for every one mole and then we can
now convert this uh to the to the
ethanol because this value over here
indicates a one for the ethanol so
that's that many kles for one mole of of
ethanol so that's minus
13668 k for every one mole of
e and we can again a value and remember
this is a combustion so we should be
releasing heat or heat is produced so
exothermic reaction we should get a
negative Q for this or negative enthalpy
uh for this reaction so it's uh minus 2
23,000 408 K if we go ahead and do it to
the right sig figs minus
2.34 * 10 4 K of energy
and that's to compare to something like
octane um has an energy per liter is
about minus
3.31 * 10 4
K in ISO octane so you might uh
recognize the word octane because that
is gasoline so this is one of the reason
why our gasoline is made of a
hydrocarbon with eight different uh
carbon atoms in there because it has
more energy density per liter than
something like
ethanol we can also look into our
standard kind of uh values our standard
enthalpy of formation and in your book
this is in
appendix G so you can look this up all
these values are in there so you'll see
this as a Delta H knot and with a
lowercase f and our main concern for
enthalpy of formation is that we are
looking for one
mole of compound
formed um and it is formed from free elements
and what we'll do is something we'll
think about something like carbon
dioxide gas and looking at the enthalpy
of formation you look this book up in
the back of the book you'll see your Del
Delta H knot uh formation equals
-31 3.5 K but how do you actually write
a reaction for that so we are going to
form one mole of that material so we
have to form it from the element so we
have have carbon and this is in the
solid state at standard conditions plus
oxygen so oxygen is diatomic and at its
standard State oxygen is a gas and that
will give us our equation for the
enthalpy of formation and one more thing
is typically we will look at
elements your Delta
HF is is going to equal zero so for free
State your enthalpy of formation is zero
so keep that in mind for what we need to
do uh when you're writing these equations
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