okay let's try a question about Hess's law. so given the thermochemical data
below those are the three reactions at the bottom we want to find the change in
enthalpy for the following reaction so we have ClF + F2 yields ClF3
and that Delta H value is unknown but we have three somewhat related equations at
the bottom with known thermo chemical data see how we have those Delta H's
there so we can use a technique called Hess's law to solve this problem and if
you're unfamiliar go ahead and check out my tutorial on Hess's law that will
explain everything you need to know to solve this problem and when you're ready give it a try.
ok so up top we have what we need to figure out and then on the
bottom we have all of our given data that's the stuff that we can use that's
what we can work with and manipulate in order to get the Delta H that we want. so
where are we going to start? let's start out by looking at what we need in our
given equation and then finding where those substances are in the given
equations so we know we need ClF we need ClF we need to find where that is in
these equations down here well the second equation has 2ClF so there's
that whole equation but it has ClF in it that's the only place we have it so
we know we need to use this equation the problem is we have to ClF we only
want one ClF so what should we do well what we can do is we can cut it in half
we're gonna cut it in half meaning we can cut the stoichiometric coefficients
in half and we will therefore also cut the Delta H value in half so we've done
this here 2ClF becomes ClF, O2 becomes 1/2 O2, Cl2O becomes
one half Cl2O, and OF2 becomes 1/2 OF2. now we mustn't forget to cut that
Delta H in half as well because we're cutting all the stoichiometric
coefficients in half that means that Delta H will be half as much as well so
two O five point six becomes one oh two point eight. now what's next we need F2
we need F2 as a reactant so where's F two well over here we have 2F2 and we
have it as a product and that's the only place F2 is so we know we need to use
this equation so what are we going to do we need F2 as a reactant but here we
have 2F2 as a product so what we need to do is we need to take this first
reaction we have to flip it so that the products become reactants and reactants
become the products and we also have to cut it in half so we could do two
manipulations here so let's do that products become the reactants. so O two
as a product becomes one half O2 as a reactant and then 2F2 as a product
becomes F2 as a reactant and then 2OF2 as a reactant becomes one OF2 as
a product so we have flipped it and we have cut it in half. and then don't
forget we have to also manipulate the Delta H value in the corresponding way
so when we flip it whenever we make the products the reactants in the reactants
to the products all we have to do is flip the sign so we took our negative
forty nine point four and we made it positive forty nine point four but we
also cut everything in half so from positive forty nine point four we went
down to positive 24 point seven so that is the Delta H value for this reaction
and lastly we need ClF3 and we need it as a product so where can
we find it here well this third reaction has ClF3 but it has it as a reactant so
what can we do all we have to do is flip the third reaction so let's do that now
so one-half Cl2O as a product becomes a reactant
three-halves OF2 becomes a reactant and then ClF3 which was reactant is
now a product and O2 is also now a product and once
again we can't forget we have to take our Delta H value it was positive to
sixty six point seven because we flipped it it is now negative to sixty six point
seven kilojoules so we've got everything we need so let's just check and make
sure what cancels out let's make sure everything that we don't want in there
is going to cancel out so what do we have well on the Left we
have a half O2 in the first reaction we have a half O2 in the second reaction
those are both reactants and then we have O2 as a product and the third
reaction so half and half makes one and then we have one on the right so O2's
all cancel out. then we have one-half Cl2O
as a product in the first reaction and we have one-half Cl2O
as a reactant in the third reaction so those will also cancel out nicely then
on the right we have one half OF2 in the first reaction we have one OF2 in
the second reaction and then in the third reaction we have three halves OF2
as a reactant so three halves OF2 total as product and three halfs
as reactants and so those will also cancel nicely and what are we left with
we are left with just what we wanted we've got ClF we've got F2 and we've
got ClF3 so that is what we wanted we do end up with the reaction as listed at
the top so all we have to do is now add up the corresponding enthalpy data and
so for the first one we've got one or two point eight for the second one we've
got 24 point seven and for the third one we've got negative two sixty six point
seven so we put those in the calculator add them up and we end up with negative
one hundred thirty nine point two kilojoules that will be the Delta H for the reaction in question.