Thermochemistry involves reactions that include solutions. These are different form other
reactions, because the heat of formation of the solution includes not only the heat of
formation of the solute, but also the heat of the solution itself, and this heat of solution
is a function of r. r is equal to the moles of solvent, which is usually water, divided
by the moles of solute. And so you look up in an appropriate table to find the heat of
solution that corresponds with your particular r. Let's take a look at a typical example.
We have sulfuric acid, and it's going to be neutralized by sodium hydroxide, and that's
going to leave us with water and a salt, sodium sulfate. One mole of the aqueous H2SO4, and
it's going to be 15% by mol H2SO4. This mol here is aqueous. It's going to be neutralized
by 10 mol% sodium hydroxide, and what we want to find out is what is the Q that's released
by this neutralization. So we're given one heat of solution here, and that's of our sodium
sulfate, and one thing to notice is that heats of solution are given per mole of solute,
not of solution, so let's start by finding the heat's of solution for our aqueous solution.
So this involves solving for r so we can look them up. If we look at our H2SO4 aqueous solution,
in 1 mole there's 0.85 moles of water, and .15 moles of sulfuric acid, so our r is going
to equal 5.7, and when we look this up we find our heat of solution is equal to -60
kJ per mole of H2SO4. Again the solute, not the solution. Now we're going to do the same
for the aqueous sodium hydroxide, and here our ratio is 0.9 to 0.1, since it's 10% sodium
hydroxide, and therefore our r is going to be equal to 9, and again we look up this heat
of solution for sodium hydroxide with an r of 9, and we find that it is -41 kJ per mole
of sodium hydroxide. Now we need to determine the Q, and our Q is going to equal this extent
of reaction times the heat of this particular reaction, so how do we find the heat of reaction?
That's equal to the sum of the heats of formation of our products minus the sum of the heats
of formation of our reactants. What makes this different from other reactive problem
is that this heat of formation is equal to not only the heat of formation of the solute,
but also the heat of solution, or the heat caused by dissolving the solute in water.
One way of looking at it is first we form the solute, which is that heat of formation,
then we dissolve it in its solvent, which is the heat of solution, and the combination
is the heat of formation of the aqueous solution. So now let's solve for the heats of formation
for all of our reactants and products, and we'll start again with our H2SO4 aqueous,
and that's going to equal its heat of formation plus it's heat of solution, and so our heat
of formation for aqueous H2SO4 is -871 kJ per mol of H2SO4. Now let's do the same
with the rest of our aqueous solutions. However, we have to be careful to not leave out the
water, even though it's not dissolved it still has a heat of formation so we need to include
that in our reaction. So now that we have all of our heats of formation we need to solve
for the extent of reaction, and I did it using our H2SO4, and so coming out we have no H2SO4,
coming in we have 0.15 moles of H2SO4 minus the extent of reaction, so the extent of reaction
is just equal to 0.15 moles. Finally we put this whole thing together, so our Q is going
to equal our extent of reactions, and now we're going to multiply it by our sum of our
enthalpys of our products, minus the sum of our enthalpys of our reactants. What's important
though is we have to multiply each one of these heats of formation by it's stoichiometric
coefficient, and when we add and subtract these all together we end up with -22.7 kJ.