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Practice Problem - Free Energy and Nonstandard Conditions

WEBVTT
Kind: captions
Language: en

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okay let's look at a problem regarding
free energy and non-standard conditions
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consider the following reaction and the
non-standard conditions below so we have
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two NH3 yielding three H2 plus N2 and we
have the standard free energy change as
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being equal to 33.0 kilojoules per
mole now here are the non-standard
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conditions we were referring to we have
a temperature of 25 degrees Celsius and
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then we have some pressures for these
three gases for ammonia that's twelve
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point nine atmospheres for hydrogen
that's zero point two five zero
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atmospheres and for nitrogen that's zero
point eight seven zero atmospheres now
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we call these non-standard conditions
because we are not at equilibrium we are
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also not at standard temperature so the
free energy change is not going to be
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the standard free energy change it's
going to have a different value because
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the standard free energy change is
calculated at standard conditions if we
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are not under standard conditions we
have a different Delta G value so we are
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going to want to calculate that here
what is the free energy change for this
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process so that's all the data we need
go ahead and give this a try.
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so let's put our reaction up top and let's put
all of our information right there so we
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are going to need to use this equation
we know that Delta G meaning the free
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energy change of some process under some
conditions that are not standard
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conditions is going to be equal to the
standard free energy change plus RT ln Q
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where Q is the reaction quotient so
let's plug in what we know we know that
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the standard free energy change is 33.0
kilojoules per mole that was given to us
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then for R that's the gas constant we
are going to use the version of the gas
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constant that is 8.314 joules per mole
kay and in particular we are going to
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want to convert this into kilojoules per
mole K because we need it to agree with
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the free energy change value so that's
in kilojoules per mole so let's take
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this and divide by a thousand that gives
us zero point zero zero eight three one
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four kilojoules per mole K then
temperature as we said is 25 degrees
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Celsius which as always we convert into
Kelvin so that's 298 K now for Q that's
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the reaction quotient so that is going
to be equal to the concentrations of the
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products raised to the powers of their
stoichiometric coefficients divided by
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the concentration of the reactants
raised to the powers of their
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stoichiometric coefficients so here
rather than concentrations we can just
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use pressures and so let's take the
pressure of H2 which is zero point two
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five zero and let's raise it to the
third power
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because we have three H2 and let's
multiply by the value for N2 which is
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zero point eight seven zero and then we
will divide by the value for NH3 which
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is twelve point nine and we will square
that because we have two NH3 in the
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equation and so if we do the arithmetic
Q is going to be equal to eight point
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one seven times ten to the negative five
now to get Delta G let's first take our
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standard free energy change value that's
thirty three point zero and we'll
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subtract from that RT ln Q so we can put
that in the calculator all at once that
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will come out to twenty three point
three and that gives us a delta g value
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of nine point seven kilojoules per mole.

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