hello and welcome to the chemistry
solution this tutorial is on Hess's law
remember that enthalpy is a state
function and that means it depends only
on the current state of the system that
means that the enthalpy change are Delta
H for a reaction depends only on the
initial and final conditions and not the
path taken and so whether a given
reaction occurs in one step or in many
steps the enthalpy change remains the
same and this is the basis for Hess's
law
Hess's law states that the enthalpy
change associated with a reaction
carried out in multiple steps is equal
to the sum of the enthalpy changes for
each individual step so let's look at an
example calculate Delta H for the
following reaction to carbon monoxide
plus o2 goes to two carbon dioxide and
we'll calculate Delta H for this
reaction given the two reactions below
and their Delta H values so the first
step to do this is to manipulate the
given equations so that they most
closely resemble the equation of
interest so let's look at our first
equation in the reaction of interest
carbon monoxide is on the reactant side
and of the two equations given below you
only see carbon monoxide in one of these
equations and it's on the product side
so what we're going to need to do is to
flip this equation around so the carbon
monoxide is on the reactant side so
we'll write the equation below except
we'll flip it around so here we have two
carbon monoxide going to two carbon plus
o2 and because we flipped this reaction
we need to flip the sign on Delta H so
remember the Delta H value for a
reaction written in the opposite
direction is of the same magnitude but
of opposite sign and so if we flip the
equation we flip the sign on Delta H
of to carbon plus o2 gives us two co and
the Delta H value is negative 221
kilojoules per mole then 2 Co going to 2
carbon plus o2 has a delta H value of
positive 221 kilojoules per mole
looking at our reaction of interest I
see that there are two moles of co2 on
the product side and when I look down at
our given equations I see only one of
the equations have co2 the second one it
is on the product side but there's only
one mole of co2 in this reaction and so
for this equation I need to multiply it
by 2 so that I have 2 moles of co2 so it
more closely resembles the reaction of
interest and when I multiply the
reaction by 2 I need to multiply the
whole equation by 2 and I think it helps
to draw a parentheses around the entire
equation so this gives us 2 carbon plus
2 O 2 gives us 2 co2
and again I want to perform the same
operation on Delta H as I did on the
equation so if I multiplied the equation
by 2 I need to multiply the Delta H
value by 2 so if the reaction carbon
plus O 2 gives us co2 had a Delta H
value of negative 300 93.5 kilojoules
per mole then the reaction to carbon
plus 2 O 2 gives us 2 co 2 has a Delta H
value of negative 787 kilojoules per
mole so here I've written my two new
equations that I modified to more
closely resemble my equation of interest
and what you're going to do now is add
these two equations together keeping
everything on the reactant side on the
reactant side and everything on the
product side on the product side so in
my new equation I have to Co plus 2 C
plus 2 O 2 gives me 2 c plus o 2 plus 2
co 2
and because I added the reactions
together I'm going to add the Delta H
values together so the Delta H value for
this equation is negative 566 kilojoules
per mole but you'll notice that I have
some compounds that are the same on both
the reactant side and the product side
and so anything that's the same on both
sides of the reaction arrow can be
canceled out so I can cancel out my two
moles of carbon on both the reactants
and the products side and I can also
cancel out the one mole of ol2 on the
product side with one mole of o2 on the
reactant side leaving me with the final
equation to Co plus o2 goes to two co2
and the last thing you need to do is to
double check that this equation matches
your equation of interest and in this
case it does so the Delta H value for
two CL + o2 goes to 2 CO 2 is negative
566 kilojoules per mole
let's try another example calculate
Delta H for the following reaction - n2
+ 5 o2 goes to 2 and 205 we're given
three reactions listed below and
remember the first step is to modify
these equations to more closely resemble
the equation of interest now I think
it's always helpful to look for a
compound in the reaction of interest
that is found only once in your given
equations and in this case you can see
that n 2 O 5 that's on the product side
and our reaction of interest isn't only
one of the given equations and you'll
see that it's on the reactant side so
the first thing we need to do is to flip
this equation around so that n 205 is on
the product side the other thing we need
to do is to multiply this equation by 2
because our reaction of interest has two
moles of n2o5 on the product side and
remember we perform the same operation
on Delta H as we did on the equate
Asian so if we flip the equation and
multiply it by two we need to flip the
sign on Delta H and multiply it by two
this gives us the new reaction for hno3
goes to two n 2o 5 + 2 h2o
and this reaction has a delta h value of
153 kilojoules per mole
looking at the bottom equation I'll see
that I have n 2 on the reactant side and
in my reaction of interest I also have n
2 on your Acton side but in my reaction
of interest I have 2 moles of n2 on the
reactant side so in order to make this
equation more closely resemble a
reaction of interest I need to multiply
the entire equation by 2 now you'll also
notice in this equation that two moles
of h no.3 are formed on the product side
and we compare that to the equation that
we just modified we have 4 moles of h
no.3 on the reactant side if you look at
our reaction of interest up on the top
you don't see the compound hno3 anywhere
in that equation which means that you're
going to need equal amounts of hno3 on
the reactant side and the product side
in order to cancel out in our final
equation so we already knew that we
needed to multiply this equation by 2 to
give us 2 moles of n2 but you can also
look at it that we need to multiply this
equation by 2 to give us 4 moles of h
no.3 which will eventually cancel out
with the 4 moles of h no.3 on the
reactant side in our first equation that
we modified and when we multiply the
equation by 2 we also need to multiply
the Delta H value by 2 now I save this
top equation for last
and it's because neither h2 nor h2o are
anywhere to be found in our equation of
interest and so we need to compare this
reaction to the reactions that we
already modified in order to figure out
how to modify this equation in order to
cancel out any moles of h2o or h2 gas in
our final equation and what you'll see
is that
we have two moles of h2o on the product
side and we need to modify this third
equation in order to cancel those out so
we'll need to flip this equation and
multiply it by two in order to cancel
out h2o by having it on both the
reactants and the products side and you
see the same thing with h2 gas so when
we flip this equation and multiply it by
two remember that we need to do the same
thing for Delta H flip the sign on Delta
H and multiply it by two now that we
have two moles of water on the reactant
side and the product side those will
cancel out when we add these equations
together and the same thing with our two
moles of h2 on the reactant side and the
product side now hopefully we've
modified all of our equations correctly
so remember we add everything on the
reactant side and everything on the
product side and then we can cancel out
anything that's the same on both sides
now if you don't want to write all of
this out
you can cancel things that are the same
on the reactant side and the product
side right away before adding your
equations together and so you'll see
that we have four moles of H no.3 on the
reactant side and the product side we
have two moles of h2o on the reactant
side and the product side and we have
two moles of h2 on the reactant side in
the product side our one mole of o2 gas
on the product side can cancel with one
mole of o2 gas on the reactant side
leaving us with five moles of o2 on the
reactant side and when we write down
what we're left with we then have an
additional two moles of n2 on the
reactant side and two moles of n2o5 on
the product side and when you check our
new equation now matches the equation of
interest and so because we added up
these three new equations we also need
to add up their Delta H values and so
the Delta H value for this reaction 2 +
2 + 502 goes to 2n 205 is 29 kilojoules
per mole
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tutorial
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