JANET: In this screencast we're gonna use
the extent of reaction method for two reactions.
So, here we have 100 moles of ethane.
They react to form ethylene and hydrogen gas.
So, C2H6 goes to C2H4 plus H2.
However, some of the ethane, C2H6,
reacts with the ethylene, C2H4,
which forms propylene, C3H6, and methane.
So, if the fractional conversion of ethane is 0.7,
and we'll get back to that,
as well as the selectivity of ethylene to propylene is five,
we want to, and make sure you read the question,
find the molar composition of the product gas.
So, the first thing that let's do is draw a picture.
Here we have two equations, or reactions,
so we're gonna need two squiggles, one for each reaction.
So, for this first reaction we'll use squiggle one,
for the second reaction we'll use squiggle two.
So, we have five components,
we need five different equations.
And the equation we're going to use is
that the number of moles of species i coming out is equal
to the number of moles of species i coming in,
plus or minus squiggle one, plus or minus squiggle two.
So, the plus is if you have a product,
the minus is if you have a reactant.
Squiggle one is if the component
is involved in the first reaction,
and squiggle two is if it's involved
in the second reaction.
So, using that equation
let's write it out for all five of our components.
We'll start with ethane.
So, the number of moles of ethane
coming out is gonna be equal
to the number of moles coming in,
which we know is 100.
Now, it's involved in the first reaction,
it is a reactant, and the stoichiometric coefficient is one,
so it's gonna be minus squiggle one,
minus squiggle two.
So, now let's do that for the rest of our components.
Note, that the rest of our components
do not have any moles coming in,
so this term for all of them will be zero.
As you can see, we have seven unknowns.
We have the five products and the two squiggles,
therefore we need two more equations.
So, let's go back to the problem statement,
and I've mentioned this in the beginning
that we're gonna use other information.
So, we have right here
that the fractional conversion of ethane is 0.7,
and the selectivity of ethylene to propylene is five.
So, let's write these out as equations.
So, 0.7, here's our fractional conversion,
is equal to the number of moles of ethane
coming in minus the number of moles of ethane coming out,
divided by the number of moles of ethane coming in.
And if you remember from our problem statement,
or you look at this picture,
you know that this is equal to 100.
Therefore, we can solve for the number
of moles of ethane that come in,
and that's gonna be equal to 30 moles.
The other piece of information
that we have is the selectivity,
and what that means is that five is equal
to the number of moles of ethylene
divided by the number of moles of propylene.
So, let's see now how we can use
these extra pieces of information.
So, we'll call this one, two, three, four, and five.
So, we're gonna start with equation one,
and we can rewrite this as that 70 moles
is equal here to our squiggle one plus our squiggle two.
Now, look at our selectivity
and substitute in for equations two and three.
So, now our five is equal to squiggle one
minus squiggle two divided by squiggle two.
And if we rearrange this we can find that squiggle one
is actually equal to six times squiggle two.
Now, we're gonna go back to equation one,
and when we put this piece of information
into equation one what we find is that squiggle two
is equal to 10 and squiggle one is equal to 60.
Now that we have both of our extents of reaction
we can solve for each one of these components.
So, we find that our C2H4
is equal to squiggle one minus squiggle two,
so that's equal to 50 moles.
We find again that our nH2,
which is equal to squiggle one,
and so that's equal to 60 moles.
And now the other two components are equal to squiggle two,
and our squiggle two we solve for as 10.
So, this is equal to 10 moles as is our methane.
But note what we were asked for in the product statement
is the molar composition of the product gas.
And so, what that means is we need to find
the mole fractions of each one of these components.
So, y, which is the mole fraction,
so y sub i is equal to the number of moles of species i
divided by the total moles.
So, let's find the total moles.
So, this is just summing up all of the different components,
so it's equal to 30 moles, plus 50 moles, plus 60 moles,
plus 10, plus 10, and that's gonna equal 160 moles.
So, I'll show you how to do the first one,
which is the ethane.
So, our y of the ethane
is going to be equal to 30 number of moles of ethane
divided by the total number of moles,
and this is gonna be equal to 0.19.
So, let's see the values of the other components.
And yes, they should and do add up to one.
So, hopefully you can see through this video
how to use the extent of reaction to calculate for unknowns
in a material balance involving multiple reactions.
And you can see the more reactions we have
the more complicated the problem gets
as you introduce a new variable,
another squiggle, for each reaction.