For B, we have to calculate the enthalpy, delta H, in joules per mole for the formation
of water.
Here we have to figure out the enthalpy of water.
We're going to say delta H of reaction equals Q of solution divided by the moles of whatever
they're asking us to find.
Here, they’re asking for the formation water so it’s moles of water.
We already have half of this because we just figured out what the Q of solution was, so
we’ll just plug it in.
Now what we have to do is we have to figure out how many moles of water we have.
What we should realize here is I give you information on NaOH and HCl.
Remember, the word of when it’s in between two numbers means multiply.
Remember, when I say big M that means molarity.
Molarity equals moles over liters.
When I say 0.100 molar NaOH, that really means 0.100 moles of NaOH over 1 liter.
We always assume it's always moles over liters.
If you guys don’t remember this too well, go back to a few videos past where we talked
about molarity and molarity in stoichiometry because to do this question, it's essential
that you remember those steps from earlier.
We’re going to work it out down here.
Since I gave you information on two reactants and I'm asking you how much product do you
have in moles, this is really a limiting reactant type of question.
We're going to have to first do it with the molarity of NaOH, see how many moles of water
that gives us then do the same thing with HCl.
Since this is a limiting reactant type of question, the one that gives us a smaller
amount of moles of water, those are those moles we’re going to use.
We have 50 ml of 0.100 molar NaOH.
I need to change my ml into liters. 1 ml on the bottom, 10 to the negative 3 liters on
the top.
Mls cancel out and I finally isolate liters.
Now that I have liters, I can multiply it times the molarity to find the moles of NaOH.
We need the moles of water though.
We need to go from these moles I've given that we just found to our moles of unknown.
Remember when we do that jump, we do a mole to mole comparison.
We look at the coefficients in the balanced equation.
According to our balanced equation, for every one mole of NaOH, we have one mole of water.
NaOH on the bottom and mole of H2O what we’re looking for in the top.
Since this is in 10 to the negative 3, put it in brackets.
When we work it out, we get 0.005 moles of water.
Now we have to do the same thing with HCl.
We have 75 ml of 0.100 molar HCl.
Same thing, we change the mls into liters first then we multiply it times the molarity.
For every one mole of HCl, we have one mole of H2O.
Liters cancel out, moles cancel out.
We have 0.0075 moles of water.
Remember, since this is the limiting reactant type of question, we're going to go with the
smaller moles of water.
Those are the moles we're going to plug down here to find our enthalpy.
When we plug that in, we get back negative 2.73 times 10 to the 6 joules per mole.
This question had a lot of parts to it.
It became even harder because we had to incorporate concepts that we learned a few videos back.
Just because we passed molarity and stoichiometry involved with them doesn't mean we don't have
to see it again this semester.
Those types of ideas always resurface at some point.
We had to use them in order to solve this question.
Remember, if we’re at constant pressure, that's when we use a coffee cup calorimeter.
Its whole purpose is to help us find the enthalpy of a reaction.
Remember, if your enthalpy, I may not have said that but if your enthalpy is positive,
that means you’re absorbing energy or absorbing heat so you're endothermic.
If you're releasing heat or releasing energy by the system, then it's exo.
It’s negative delta H. Just remember the concepts that we covered here and the approaches
we take.
When you see a question like this, you'd be better prepared to do it.