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Saturday, 22 June 2013

SIMPLE MEASUREMENTS OF ENTHALPY CHANGES

SIMPLE MEASUREMENTS OF ENTHALPY CHANGES This page is just a brief introduction to simple measurements of enthalpy changes of reaction which can be easily carried out in a lab. There are also pointers to other sources of information.
Two common enthalpy change measurements
Enthalpy changes involving solutions
The experiments
There are a whole range of different enthalpy changes that can be measured by reacting solutions (or a solution plus a solid) in a simple expanded polystyrene cup. A common example would be the measurement of the enthalpy change of neutralisation of, say, hydrochloric acid and sodium hydroxide solution.
The polystyrene cup serves to insulate the reaction mixture, and slows heat losses from the side and bottom. Heat is still lost from the surface of the liquid mixture, of course, and that can be reduced by using a polystyrene lid with a hole for a thermometer.
You can allow for heat losses during the reaction by plotting a cooling curve.
All of this is described in some detail at the beginning of Chapter 5 of my chemistry calculations book, and I can't repeat it on Chemguide.
There are other sources of error in these experiments - in particular in the accuracy of the thermometer used. Since the temperature rise probably won't be very great, you would need to use the most accurate thermometer possible in order to keep the percentage error low.
The calculations
The heat evolved or absorbed during a reaction is given by the expression:
Heat evolved or absorbed = mass x specific heat x temperature change
That can be written in symbols as
Heat evolved or absorbed = m s ΔT
You will find that the specific heat is sometimes given the symbol "s" and sometimes the symbol "c".
The specific heat of a substance is the amount of heat needed to increase the temperature of 1 gram of it by 1 K (which is the same as 1°C).
So for water (the value you are most likely to come across), the specific heat is 4.18 J g-1 K-1.
That means that it takes 4.18 joules to increase the temperature of 1 gram of water by 1 K (or 1°C).
In experiments of this sort, you will quite commonly have measured out a volume of solution rather than a mass, and it is very common to assume that the density of the solution is exactly the same as water (1 g cm-3). That means that 25 cm3 of solution would have a mass of 25 g.
That is an approximation, though, and it does introduce some error.
The assumption is also made that the specific heat of a solution is the same as the specific heat of water. That's another approximation, and it also introduces errors into the answer.
What is generally true is that these errors are relatively small compared with errors caused by heat losses, and so on.
Anyway . . .
To do the calculation, you would normally just work out the amount of heat evolved or absorbed in your particular reaction, and then scale it up to give an enthalpy change per mole.
You will find examples of this in Chapter 5 of my chemistry calculations book.
There is another example, though, which you can access for free: it was written for my first GCSE book for syllabuses which no longer exist.
It takes the form of a sample coursework assessment relating to the enthalpy changes of reaction when magnesium reacts with a variety of acids. Bear in mind that this was written for a level below A level (or its equivalents), but the calculations and a lot of the discussion are still relevant. This was a time when you had to produce a serious bit of work to get a good GCSE grade on a piece of coursework.
All the results in this pdf document are real readings that I took myself, and you will find it on the Longman website.


Note:  Publishers' websites tend to change without warning. If you find that this link doesn't work, please get in touch with me via the address on the about this site page.


Enthalpy changes of combustion
As normally measured in a lab at this level, these are far less accurate than the simple solution reactions above. It is impossible to eliminate heat losses, or to be sure that you have complete combustion.
You will find these experiments discussed in my chemistry calculations book, but you will also find that a Google search will throw up endless versions with practical details, and possibly sample calculations.
The examples in my book include a fairly simple way of getting around the heat loss problem.
I'm afraid that is as far as I am prepared to go on this topic without risking sales of my book, or breaching my contract with my publishers. Sorry!

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