[narrator] Now that we’ve talked about a calorimetry experiment measuring the heat
either gained or released by an object placed in a calorimeter let’s look at what happens
when we mix two reactant solutions together inside of a calorimeter and what the measurement
looks like.
What are we looking for with heat of reaction?
I’ve got ΔH of reaction of this process that looks a lot like the heat of formation
of water vapor.
But, it’s actually not the heat of formation.
Heat of formation only applies to one mole of a substance.
We’ve got steam here, H2O gas.
The heat of formation--the heat of reaction that corresponds to the heat of formation
is actually one half of this particular reaction.
The one-half would go out in front of the oxygen.
From its constituent elements one mole of a substance would end up being half of the
enthalpy of reaction of the process that I’ve got written above.
It’s not the heat of formation but it is a true balanced equation that is related to
the heat of formation expression.
How do we find this ΔH of reaction?
What is it, first of all?
ΔH of reaction is the amount of heat that is released by--released or absorbed by the
reaction but we don’t measure the reaction.
We measure the water based system inside the calorimeter.
It’s the negative heat of the water divided by one mole of reaction.
One mole of reaction could correspond to two moles of H2, one mole of O2, or two moles
of H2O.
All the different parts of the balanced equation we have to pay attention to what these are
when we determine what a mole of reaction is for a particular process.
The easiest thing to do is to look at an example.
This ‘-q’ of water, we’re going to be needing to find the mcΔT value.
We’re looking for a measurement of each one of these factors plus how to determine
mole of reaction.
Let’s go ahead and look at this fictional reaction, below.
I’ve kind of set it up so that we can solve.
What do we need?
We need, first, to find the ‘q’ of water.
We’re going to multiply that by -1, because it’s not the system.
We’ve defined the system as the reaction and the surroundings are what we have to measure,
as far as the ‘q’.
Again, no calorimeter constant but we would add the calorimeter constant inside parentheses
if such a constant existed for this problem.
But, here, we just have the ‘q’.
What is the ‘-q’, the mcΔT of the water?
We look at our problem and so we see some information about the reactant in the first
sentence.
Then we say, “Oh, wait, here’s a coffee cup calorimeter having 100 mL of water at
20 °C and what’s the enthalpy change if the solution is raised in temperature by five
degrees?”
We’ve got what we need, now, for this portion.
We’ve got 100 mL of water and 10 grams of substance so that resulting solution is going
to have a mass of 100 grams of water plus the 10 grams of our reactant that we have
added.
That total solution, we’re ignoring the mass of y -- it’s not given to us, but a
total mass of that solution at the end, to our best estimate, is 100 grams of water from
the milliliters, 1 gram per milliliter, plus the 10 grams of reactant that we’re given
in the initial problem.
We add those things together to get the mass of solution that we’re measuring.
We’re stirring and we’re measuring the mass of the solution inside the coffee cup
calorimeter, at the end.
Then the specific heat.
In a question like this it usually tells us, “Assume the specific heat of the resulting
solution is the same as pure water,” and so we will assume that here.
Even if it’s not written that’s our best estimate.
Joules per gram °C. It is intended for us to use that value.
That’s why we have to memorize it, like we said talking about the calorimetry and
specific heat measurement of an object.
Then ΔT, the solution, the enthalpy of reaction if the solution is raised in temperature by
five degrees.
They’ve gone ahead and given us the ΔT value so we have this value for ‘q’, of
the water, that we can solve for now.
When we punch everything in we get -2299.
Grams cancels, degrees Celsius cancels, and joules are left over.
We’ll round at the end once we get our full complement of significant figure values, here.
We have three in each of the terms here so we just keep in the back of our mind that
this is three sig figs and we’ll come back to it.
What about mole of reaction?
Or the ‘-q’ of water?
Water receives the energy and so we’ve actually lost energy from the reaction.
What about moles of reaction?
Well, in our reaction we have 10 grams of substance that’s 100 grams per mole.
We can get into terms of moles, we’ve got 10.0 grams of ‘x’ and it is a mole, molar
mass, per 100 grams.
To make it a little easier.
When we cancel here we’re going to get grams of ‘x’ out and then moles of ‘x’ left
over.
What do we have?
1/10th.
0.100 moles of ‘x’.
But, this is not moles of ‘x’ we’re after.
We’re after moles of reaction.
How many--we want to get rid of moles of ‘x’--how many moles of ‘x’ per reaction?
We look at the coefficient.
There are two moles of ‘x’ per one mole of reaction.
That cancels and we get 0.050 moles of reaction.
Now we’re ready to calculate our ΔH.
It’s ‘-q’ of the water, divided by moles of reaction, and so that’s -2299 joules.
We’ve already determined this as that value so we don’t have to do another negative
here, we’ve already taken that into account in the calculation above, divided by 0.050,
to three significant figures, here.
Moles--and so then we can divide by 0.05 and we get -45,980 joules per mole, or we can
convert that to kilojoules.
-45.98, so rounds up to -46.0 kJ per mole.
That’s our final value for ΔH of reaction.