Welcome back to part 2. In this video we're going to focus mostly on this science understanding. The energy released in combustion of fuels can be determined experimentally. You'll need to be able to use experimental data to determine the enthalpy of combustion of a fuel and undertake thermochemical calculations involving enthalpy changes and temperature changes of a specified mass of water given the necessary data.
As a bit of a revision from stage 2 chemistry, in subtopic 2.1, Rates of Reactions. We learn that enthalpy is the heat or the energy content of a system at constant pressure and volume. The word system essentially refers to the reactants and products.
Therefore, the enthalpy change is the heat energy that's released or absorbed in a chemical reaction under constant pressure. And another way of thinking of it is it's the difference in the enthalpy or the energy contained within the chemical... bonds of our products versus our reactants.
So we can express it as an equation where delta H equals the enthalpy of our products minus the enthalpy of our reactants. Calorimetry is a study and measurement of heat changes in chemical reactions. We were introduced to this back in stage one chemistry in topic four, looking at energy in reactions.
Note that combustion reactions are the only ones that will be considered here and we know that combustion reactions are all exothermic. Another way of saying that is that the delta H value or the enthalpy change is always going to be negative. We can use a device called a calorimeter.
A calorimeter is a well insulated reaction vessel in which heat changes can be measured and so that's something that we will look at doing in the classroom. To measure these heat changes we can use a formula, and again we should have been introduced to this back in stage one chemistry. The formula is Q is equal to m times C times delta T. What do all those actually mean?
So this table summarises what each of these quantities actually relate to. So firstly Q relates to the heat change, so that is the heat that's being transferred to our calorimeter. We measure that in the units of Joules. M is for the mass of water that is being heated so we are transferring this heat into water and we measure its mass in grams.
C refers to the specific heat capacity of water which is what we're trying to heat and we know that it has a specific heat capacity of 4.18 joules per gram per degree Celsius. This represents the amount of energy required to raise the temperature of one gram of water by one degree Celsius or even one Kelvin. Delta T is the temperature change of water. So whether it either increases or decreases in temperature.
If we are looking at combustion, it can be done in this way. And we would notice that this is always going to result in an increase in temperature. If we use a setup like the one in the previous slide, we can look at how we can solve for the heat energy. So example question one here says calculate the heat energy required to warm 250.0 mL of water water from 22.0 degrees Celsius to 98.0 degrees Celsius.
Assume the specific heat capacity of water is 4.18 joules per gram per degree Celsius. Let's go ahead and let's state what information we do know and one thing just to factor in is that the mass of water is equivalent to its volume. That's because water has a density of one gram per mil, meaning that if we measure one mil of water it's equivalent to a mass of one gram.
So that means that 250.0 mL of water is going to be equivalent to 250.0 g. Another thing we can calculate is the delta T value. So that's looking at the difference in the final and initial temperature. So it went up to 98.0 and it started off at 22.0. So we've had a temperature change of 76 degrees Celsius.
From here we can... plug this into our formula for measuring Q. So this is the formula here.
We'll substitute in our values and we know the specific heat capacity is given here. So we've got all of our values here which we need to just put into our calculator. We get an answer of 79,420 joules. We can convert this into more appropriate units which could be 79.4 kilojoules expressed to three significant figures.
Another way in which we can measure the heat being released from a combustion reaction is referred to as the molar enthalpy of combustion. So this is the heat released when one molar of our fuel undergoes complete combustion. So with this it also comes with a formula, so we just have to factor in how many moles of our fuel is undergone combustion, incorporate that into our formula to work out the heat released in kilojoules per mole.
The formula is given as such here, so delta H is equal to M times C times delta T. We divide it by 1000 to convert the units of joules to kilojoules, and we also divide it by the number of moles of our fuel that has undergone complete combustion. Another way of writing this is delta H is equal to Q divided by N, provided that Q is given in kilojoules and N obviously is given in moles.
We can see in this table here what each of these quantities mean and the units we use to measure them. So delta H being the molar enthalpy, N being the number of moles of our fuel undergoing combustion. In a previous slide we saw a diagram like this, which shows us how we can measure the energy change from a combustion reaction. So typically we house our fuel in what we call a spirit burner.
So this has our fuel here. It has a wick which essentially allows for the combustion in a fairly safe controlled manner. The heat that gets produced from our combustion reaction will hopefully go into directly heating. the water contained within our calorimeter.
So an example of a calorimeter could be a copper can or an aluminium can. We would need to have a thermometer to measure the change in temperature. It may even be used to help stir the water to ensure that the temperature is constant throughout. And another thing to factor in is we need to look at measuring the mass of our fuel burn.
So we could measure the mass of our spirit burner before combustion. measure it after combustion, and look at the difference between the final and initial mass to work out the mass of our fuel burn. When we carry out this technique, we make some assumptions.
They're not the right assumptions to make, however we do this just to simplify the calculation of our heat being transferred or the enthalpy change for our reaction. So these assumptions can consist of the fact that heat is being completely transferred, to heat the water, that the fuel undergoes complete combustion, and also that the maximum temperature is reached before the water starts returning to its starting temperature. Because we know that eventually, if something is heated, it will cool down and return to its original starting temperature.
To look at how we can calculate molar enthalpy of combustion, we've got a second example question here. A spirit burner containing methanol was ignited under a 150.0 ml volume of water contained in an aluminium can which is our calorimeter and allowed to burn until the temperature increased by 18.2 degrees celsius. The mass of the spirit burner decreased from 136.12 grams to 135.51 grams.
For part a of this question we need to calculate the experimental molar enthalpy of combustion for methanol. First thing we're going to do is just state the particular quantities that we know. So firstly we've got the mass of water which we know is equivalent to the volume of water.
That's equal to 150.0 grams. We can also work out that the temperature change was 18.2 degrees Celsius. We can also work out the mass of methanol that had undergone combustion.
So by taking the final mass and the initial mass of the spirit burner. we can see that we have a mass of 0.610 grams of methanol undergoing combustion. We can also work out the molar mass of methanol because we will need to use these two to work out the number of moles of methanol that had undergone combustion. And we have a molar mass of 32.042 grams per mole.
So here we can see we've calculated the number of moles of methanol. We've got the working out there and we should have a value close to 0.0190 moles. But just keep in mind you may need to keep that answer in your calculator for a further calculation. So using the information from the previous slide, we can now calculate this experimental molar enthalpy of combustion for methanol.
So we can do that by using this formula here. So delta H equals M times C times delta C divided by 1000 times N. So we should have all of these values now ready to use, so we'll substitute those in. And if we plug that in correctly, and again ensuring that the number of moles is exactly what we get in our calculator, we should end up with a value of negative 599 kilojoules per mole to three significant figures.
A second trial conducted yielded a value of negative 547 kilojoules per mole. The accepted value for the enthalpy of combustion for methanol is negative 715 kilojoules per mole. So for part b we need to suggest sources of random and systematic error that could contribute to these discrepancies and this is going to tie in with those assumptions that we made earlier. So in terms of the random errors, one example of a random error is the evaporation of fuel from the wick. Many fuels are quite volatile meaning that they can evaporate quite readily.
What this means is less mass of your alcohol has essentially undergone combustion. Some of that has essentially been lost into the atmosphere. Other sources include variations in human measurement of things like the volume, so volume of water used, the mass of our spirit burner, as well as measuring the temperature changes that occur.
There may also be variations in the height of the wick from the bottom of the can and then this can influence the amount of heat being transferred into our calorimeter. In terms of the systematic errors, we can see that not all heat will actually be transferred to the water. We know that some can be transferred to the can itself, or even the surrounding area. air so that could be from the flame produced from the spirit burner traveling to the surrounding air even if the water is being heated we could get heat loss from the water to its surroundings so to the surrounding air to the calorimeter as well we are likely to encounter some incomplete combustion and we know that incomplete combustion can result in less heat energy being released And finally, the maximum temperature hasn't been reached before it starts returning to the starting temperature. So that means it's already starting to cool down, and that would be because we're getting a loss of heat from water to the surroundings before it can reach its maximum temperature.
For part c, using the accepted value, calculate the theoretical heat release when 1.00 grams of methanol undergoes complete combustion. we need to use this value here. And to do this, we're just going to use a rearrangement of one of the formulas I introduced before.
Before we do that, let's just work out the number of moles of methanol that undergo complete combustion. So plug in our numbers there, we should get an answer of about 0.0312 moles. So using this formula here where delta H equals Q over N, We're going to rearrange that to solve for Q, which is the heat being released from this reaction.
The delta H value, as we said before, is here, minus 715 kilojoules per mole. So we'll multiply that by the number of moles of methanol, and we end up with an answer of 22.3 kilojoules of energy being released. And that's to three significant figures.
From this example, hopefully you can see how significantly different the calculated value for the enthalpy of combustion of a fuel is from the accepted value for the enthalpy of combustion. In the next video we're going to see how we can increase the accuracy of our results and also look at a rearrangement for the formula for the molar enthalpy of combustion and how we can use that to calculate temperature changes and also look at how we can compare fuels using different comparisons. I'll see you guys in the next video.