Gibbs free energy change - Thermodynamics Video Lecture - Class 11

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before watching the video please make
sure that you will have watched part 1 -
part 25 in Gibbs energy some new
THERMOTRONIC function we have to revise
just now so it is denoted by G and G is
nothing but H minus TS so it is an
extensive property and a state function
similar to most of the other thermalize
property and Delta G system is nothing
bad ed system - cheater dust this is the
whole Delta G the whole surrounding so
we use this term Delta G is equal to
Delta H minus T Delta s right and by
definition it is nothing but the amount
of energy available for a system and a
given set of condition that can be put
into useful cough and there isn't my
life on this Delta G was birth was
difficult to find s total in prophethood
and we saw that Delta G was nothing but
minus T Delta s I think total so the
sign changes if Delta G is negative the
process is spontaneous the Delta G is
positive the process contains mouse
button please note for the s total it
was if s total is greater than zero it
was spontaneous right so since this
Delta G is nothing but Delta G is
nothing but minus T Delta s T that was
we saw in the last slide so the sign
changes here in this case so if Delta G
is negative the process is spontaneous
if Delta G is greater than zero or
positive the process is not spontaneous
if Delta G is zero the process isn't
right and with Delta Z begin tella
consider into your reaction also it is
noted time the reaction feasibility is
depend on temperature that there's a
process which is spontaneous at
temperature t1
may not be spontaneous at temperature t2
it can happen okay so you know that
exothermic is nothing but temperature
which in exothermic reaction we say that
higher temperature favors spontaneity
and an endothermic exothermic we say
that lower temperature
and spontaneity I'll explain this just
hold on for a second what I am trying to
say is the spontaneity of the reaction
is depend on division because Delta G
you see has a conversion factor alright
so if you see if I make use of Delta H
and Delta s let's suppose Delta H is
positive Delta H is negative and Delta s
is positive and Delta s is negative four
options we can have see Delta H can have
positive or negative and Delta s can
have positive negative so we'll make
this kind of structure so if Delta s is
positive and Delta H is negative that
means I have this case the formula I
have is Delta H minus T so if Delta H is
negative that means negative minus Delta
s is positive that is into some positive
number so the whole thing is negative
right see if you see Delta H is negative
minus X right and Delta s is policy sub
plus y the whole thing becomes negative
it is always spontaneous ok let's take
this analogous or let's take this only
if Delta H is negative and Delta s is
negative so if my Delta H is negative
that is minus X and Delta s is also
negative it is T into minus y so this
becomes Y minus X right so Y minus sorry
Y into T Y into t minus X so if
temperature is very high Delta G will
become positive you see this is nothing
but T into y minus X Y because this
minus minus cancel this becomes C into
violence a tenth that if t temperature
is very high the whole thing will become
positive when the
we'll be non spontaneous so only at low
temperature reaction is spontaneous so
it is no temperature is right it's
Cygnus and Delta H is positive so I have
Delta H minus T Delta s right is the
formula I have so Delta H is positive so
I have X minus T into and this guy also
positive Y let's suppose entropy to be
some number Y so this guy for this guy
to be negative temperature has to be
very high right because if the
temperature is low suppose it is 1 or 0
this thing is positive the whole thing
is positive but for reaction to be
spontaneous this G has to be negative so
for this X minus T 1 has to be negative
G has T has to be high so I can say that
high temperature favors spontaneity
right this and we will take Delta H is
positive minus T into Delta s is
negative so let's take minus 4 here so
this become X plus T by you put any
value of T it is always greater than 0
right so it is always non spontaneous
right so I have for option in this case
it's always spontaneous it case lower
temperature spontaneous high temperature
spontaneous always wanted and that's
what I have says is exothermic that is
blue one it favors sorry exothermic is
this one Delta H will do is inform
against in to commit right so exothermic
this guys at low temperatures excuse me
and this is anything right so this is
going is equally possible this
so by endothermic my net light is
positive so high temperature will favour
spontaneity exothermic my Delta H is
negative low temperature favors that is
why I wrote this for Gibbs energy change
as a formula and Delta G is nothing but
Delta G standard plus RT Ln K where
Delta G is nothing but Gibbs free energy
change and Delta G with our standard
Gibbs free energy change is a standard
and RT log K something you know this is
a formula they do not have derivation
for this please assume in this formula
Delta G standard Delta G is the tip
speed machine of the gut
dylaney standard plus RT Ln C Gibbs free
energy change for equilibrium reaction
is zero there is a reaction a B plus the
Librium or you can write this as a plus
the equilibrium the cypher the same in
addition you can use equilibrium little
mystery more of this in the next chapter
when we talk about the equilibria we
have one full Chuck body fluid so there
the Gibbs free energy is so if you
put this in the form earlier formula we
had the reform like we have as well
Delta G gives phases of the word Delta G
naught plus RT Ln K this becomes zero so
what we can see that is the standard
Gibbs free energy change is nothing but
minus 2.303 RT log this is how you find
the standard Gibbs free energy change
and then we can use to find other Gibbs
free energy we take one example for the
Gibbs free energy change calculate the
standard Gibbs free energy the standard
gives the zero standard Gibbs finishing
the conversion of oxygen to ozone this
is the reaction we have oxygen to ozone
and the KP is given for this conversion
is this well study more about this key
factor in the next chapter equilibrium
sample so the formula is pretty simple
we have the standard Gibbs free energy
change is nothing but minus 2.303 RT log
of K right Q so you have Ln we are using
log so I'm multiplying at 3.30 3 this
becomes 2 point 3 0 3 R will take 8
point 3 1 4 T is 298 and then log off
this guy's 2 point 4 7 into 10 to power
minus 20 so if you solve this you get 1
6 3 tipple 0 do almost other way this is
nothing but 1 63 kilo Joule per and that
is nice
the question says at 60 degree Celsius
my Trojan tetroxide is 50% be saluted so
that means I are n2f it is in
equilibrium with you know calculate the
standard free energy sheen at this
temperature and it one atmospheric
pressure 180 m and 60 degree Celsius 60
Hertz is nothing but 333 Kelvin right so
let's see this you start with one mole
initially you start with one mole of
this there is ZERO mole of no.2 when
equilibrium we have one mole of 0.5 more
of this this is 50% associated right
support with 0.5 it is 50% is 0.5 we
have now and 4.5 how much how many moles
will get here one mole is correct
because for one mole of n2 for you get 2
mole of energy so we get 1 love this is
2 into 0.5 so total more here is 1 total
mole here is 1.5 so total atmospheric
pressure is 1 1 atm so pressure by n2 a
whole will be what 0.5 by 1.5 ATM and
pressure by no.2 will be what 1 by 1
point 5 so here the KB was not given so
we will use this formula for mine KP
which we'll be learning more in next
chapter but I just use this no.2 so
pressure where do the two moles here
square by pressure of n2o4 so product by
the attend and you put to the bar the
number of rows involved
this becomes no.2 is this guy 1 by 1
point 5 to the power 2 by Phi n Phi to
the power 1.5 you solve this you get 1
point 3 3 that is the shaping because
they are 280 in here 180 in here divide
to get ki p is this so we can find Delta
G naught minus 2 point
3:03 rd log of k put the values here
minus 2 point 3 0 3 r is 8.314 Joule per
Kelvin or more in the T is 333 log of 1
point 3 point you saw this you get minus
seven six sixty three point eight kilo
joule per that is ask here the KB was
not given so we have found up key to
this method and we'll will learn more
about these are finding K the next
chapter Cleary of chapter there when we
solve more cushion of this thank you
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