In part one we covered the common conversion factors in
enthalpy stoichiometry and how to use them. So if
you missed that video you can watch that here. Now let’s
go over this practice problem. We are told that ethanol combusts
to produce 1235 kJ of heat energy. We’ll find the balanced equation first.
balanced equation first. The keyword to figure
out what type of reaction this is, is the word combust. That means this is a combustion
reaction which always has an O2 as our reactant and our products are always
carbon dioxide and water. Our first reactant is ethanol since that was provided in the question.
Now we’ll balance this out by separating each element and counting how much of each element there is
on either side. On the reactants side we have 2 carbons, 6 hydrogens since we have
5 here and 1 more at the end of ethanol, and we have a total of 3 oxygens since we have
1 here and 2 here. On the products side we have 1 carbon, 2 hydrogens
and 2 plus 1 so 3 oxygens.
Always start with balancing everything other than hydrogen and oxygen and always leave
oxygen for the very end. So we’ll place a 2 in front of CO2 and recount
our carbons and oxygens, so we have 2
carbons and a total 5 oxygens since 2 x 2 is 4 plus this oxygen so 5.
Now we’ll balance the hydrogen, this side has 6 the other side has 2 so 3 times 2 will give us 6,
that’s why we’ll place a 3 in front of water.
Recount the hydrogens and oxygens. We now have 6 hydrogens and
4 plus 3 oxygens so 7. Now for the oxygens
since there is one oxygen here and 2 here we will make this oxygen a 6
by placing a 3 in front of the O2 and now we have 6 plus 1 so 7 oxygens. This is now balanced.
Next let’s see what sign our enthalpy change is. Our keyword is produce, if heat is produced then our enthalpy change is negative.
Now we’ll use this for part b.
Part b asks, how much heat is released when 50 grams of ethanol combusts?
So since we are going from grams of ethanol to heat. Our plan will be to go from grams of ethanol to moles of
ethanol using the molar mass of ethanol and then use the enthalpy change given to get to kJ.
We’ll start with our given of 50 grams of ethanol. Now let’s find the molar mass of ethanol.
There’s 2 carbons, multiply 2 by the mass of carbon so we get 24.02,
now for the hydrogens we have 6
multiply that by the mass of hydrogen and we get 6.06 and there’s only
one oxygen so our mass is 16. Add all these up to get the molar mass of ethanol
which is 46.08 grams per mole.
We’ll place the grams of ethanol we just found on the bottom since we want grams and grams to cancel,
and 1 mole of ethanol is left on top. Now that we are at moles of ethanol we can use
the enthalpy change as our conversion factor. Where -1235 kJ is on top and it’s divided by the
moles found in the balanced equation for ethanol. Align the units so they can cancel and we get
-1340 kJ. There are two different ways you can write your answer.
The first answer is with a negative sign because heat is released. We could have also written
our answer as positive 1340 kJ is released and the fact that we said released
that accounts for the negative so we don’t have to put it.
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