Other Gas Laws

7/25/2019 Other Gas Laws

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Other gas laws - Boyle's Law and Charles' Law

This page takes a simple look at Boyle's Law and Charles' Law, and is

suitable for 16 - 18 year old chemistry students doing a course the equivalent

of UK A level. The aim is simply to show how these laws relate to Kinetic

Theory (in a non-mathematical way), and to the ideal gas equation.

Before you waste time on this, be sure that you actually need to know about it.

Certainly in the UK exam system, it is pretty rare for chemistry students to be

expected to know either of these laws these days. They have been almost

completely replaced by the ideal gas equation.

Boyle's Law

Statement

For a fixed mass of gas at constant temperature, the volume is

inversely proportional to the pressure.

That means that, for example, if you double the pressure, you will halve the

volume. If you increase the pressure 10 times, the volume will decrease 10

times.

You can express this mathematically as

pV = constant

Is this consistent with pV = nRT ?

• You have a fixed mass of gas, so n (the number of moles) is constant.



• R is always constant - it is called the gas constant.

• Boyle's Law demands that temperature is constant as well.

That means that everything on the right-hand side of pV = nRT is constant,and so pV is constant - which is what we have just said is a result of Boyle's

Law.

Simple Kinetic Theory explanation

I'm not going to try to prove the relationship between pressure and volume

mathematically - I'm just showing that it is reasonable.

This is easiest to see if you think about the effect of decreasing the volume of

a fixed mass of gas at constant temperature.

Pressure is caused by gas molecules hitting the walls of the container.

With a smaller volume, the gas molecules will hit the walls more frequently,

and so the pressure increases.

You might argue that this isn't actually what Boyle's Law says - it wants you to

increase the pressure first and see what effect that has on the volume. But, in

fact, it amounts to the same thing.

If you want to increase the pressure of a fixed mass of gas without changing

the temperature, the only way you can do it is to squeeze it into a smaller

volume. That causes the molecules to hit the walls more often, and so the

pressure increases.

Charles' Law

Statement



For a fixed mass of gas at constant pressure, the volume is

directly proportional to the kelvin temperature.

That means, for example, that if you double the kelvin temperature from, say

to 300 K to 600 K, at constant pressure, the volume of a fixed mass of the gaswill double as well.

You can express this mathematically as

V = constant x T

Is this consistent with pV = nRT ?

• You have a fixed mass of gas, so n (the number of moles) is constant.

• R is the gas constant.

• Charles' Law demands that pressure is constant as well.

If you rearrange the pV = nRT equation by dividing both sides by p, you will

get

V = nR/p x T

But everything in the nR/p part of this is constant.

That means that V = constant x T, which is Charles' Law.

Simple Kinetic Theory explanation

Again, I'm not trying to prove the relationship between pressure and volume

mathematically - just that it is reasonable.



Suppose you have a fixed mass of gas in a container with a moveable barrier -

something like a gas syringe, for example. The barrier can move without any

sort of resistance.

The barrier will settle so that the pressure inside and outside is identical.

Now suppose you heat the gas, butnot the air outside.

The gas molecules will now be moving faster, and so will hit the barrier more

frequently, and harder. Meanwhile, the air molecules on the outside are hitting

it exactly as before.

Obviously, the barrier will be forced to the right, and the volume of the gas will

increase. That will go on until the pressure inside and outside is the same. In

other words, the pressure of the gas will be back to the same as the air again.

So we have fulfilled what Charles' Law says. We have a fixed mass of gas

(nothing has been added, and nothing has escaped). The pressure is the

same before and after (in each case, the same as the external air pressure).

And the volume increases when you increase the temperature of the gas.

What we haven't shown, of course, is that there is a "directly proportional"

relationship. It can be done, but it needs some maths.

REAL GASES



This page looks at how and why real gases differ from ideal gases,and takes a brief look at the van der Waals equation. If you have

come straight to this page via a search engine, it might be a good

idea to read the page about ideal gases first.

Real gases v ideal gases

Real gases and the molar volume

I want to use this to illustrate the slight differences between the

numerical properties of real and ideal gases at normal

temperatures and pressures.

If you have read the page about ideal gases, you will remember

that we used the ideal gas equation to work out a value for the

molar volume of an ideal gas at stp (standard temperature and

pressure).

If you know the density of a gas at a particular temperature and

pressure, it is very easy to work out its molar volume.

For example, at 273 K and 1 atmosphere pressure, the density of

helium is 0.1785 g dm-3.

That means that 0.1785 g of helium occupies 1 dm3 at stp. It is a

fairly simple sum to work out what 1 mole of helium, He, would

occupy.

1 mole of He weighs 4 g, and would occupy 4 / 0.1785 dm3 = 22.4

dm3.

That's the same (at least to 3 significant figures) as the ideal gas

value, suggesting that helium behaves as an ideal gas under these

http://www.chemguide.co.uk/physical/kt/idealgases.html#top





conditions.

Note:If your maths isn't very good, and you can't understand why I divided

4 by 0.1785 to get the answer, think of it like this:

Replace the awkward value of 0.1785 by something simple, like 2. If 2 g of

He occupied 1 dm3, what would 4 g occupy? It is obviously twice as much, 2

dm3 - but how did you get at that mathematically? You found out how many

times 2 would go into 4 - in other words, you divided 4 by 2. Do exactly the

same with the more complicated number.

If you do this for a random sample of other gases, you get these

values (to 3 significant figures) for the molar volume at stp (273 K

and 1 atmosphere pressure).

density (g dm-3) molar volume at stp

He 0.1785 22.4

N2 1.2506 22.4

O2 1.4290 22.4

CH4 0.717 22.3

CO2 1.977 22.3

C2H4 1.260 22.2

NH3 0.769 22.1

SO2 2.926 21.9



So although for simple calculation purposes we use the value 22.4

dm3 for all gases, you can see that it isn't exactly true. Even at

ordinary temperatures and pressures, real gases can deviate

slightly from the ideal value. The effect is much greater under moreextreme conditions, as we will see next.

Compression factors

For an ideal gas, pV = nRT. If pV and nRT are the same, and you

divide one by the other, then the answer will, of course, be 1. For

real gases, pV doesn't equal nRT, and so the value will be

something different.

The term pV / nRT is called thecompression factor. The graphs

below show how this varies for nitrogen as you change the

temperature and the pressure.

Note:These diagrams were generated using data produced fromPatrick

http://www.ceb.cam.ac.uk/thermo/




Barrie's program, and converted into graphs using Excel. The figures are

derived from the van der Waals equation - not because it is the best source,

but because it is the only one you are likely to come across at this level, and

I shall mention it below. If you wanted to play around with some of the other

equations, you would find that the results produce similarly shaped curves,

but the absolute sizes of the deviations would vary.

If nitrogen was an ideal gas under all conditions of temperature and

pressure, every one of these curves would be a horizontal straight

line showing a compression factor of 1. That's obviously not true!

Things to notice

• At low pressures of about 1 bar (100 kPa - just a bit less

than 1 atmosphere), the compression factor is close to 1.

Nitrogen approximates to ideal behaviour at ordinary

pressures.

• The non-ideal behaviour gets worse at lower temperatures.

For temperatures of 300 or 400 K, the compression factor is

close to 1 over quite a large pressure range. The nitrogen

becomes more ideal over a greater pressure range as the

temperature rises.

• The non-ideal behaviour gets worse at higher pressures.

• There must be at least two different effects causing these

deviations. There must be at least one effect causing the

pV / nRT ratio to be too low, especially at low temperatures.

And there must be at least one effect causing it to get too

high as pressure increases. We will explore those effects in

a while.

Other gases






Is the same behaviour shown by other gases? The next diagram

shows how the compression factors vary with pressure for a variety

of gases at a fixed temperature.

If you were to redo the set of original nitrogen graphs (at varyingtemperatures) for any of these other gases, you would find that

each of them will produce a set of curves similar to the nitrogen

ones. What varies is the temperature at which the different graph

shapes occur.

For example, if you look at the carbon dioxide graph at 273 K, it

looks similar to the nitrogen one at 100 K from the first set of

curves, although it doesn't increase so steeply at higher pressures.

It is easy to say that gases become less ideal at low temperatures,

but what counts as a low temperature varies from gas to gas. The

closer you get to to the temperature at which the gas would turn

into a liquid (or, in the case of carbon dioxide, a solid), the more

non-ideal the gas becomes.



What causes non-ideal behaviour?

In the compression factor expression, pV / nRT, everything on the

bottom of the expression is either known or can be measured

accurately. But that's not true of pressure and volume.

In the assumptions we make about ideal gases, there are two

statements which say things which can't be true of a real gas, and

these have an effect on both pressure and volume.

The volume problem

The kinetic theory assumes that, for an ideal gas, the volume taken

up by the molecules themselves is entirely negligible compared

with the volume of the container.

For a real gas, that assumption isn't true. The molecules

themselves do take up a proportion of the space in the container.

The space in the container available for things to move around in is

less than the measured volume of the container.

This problem gets worse the more the gas is compressed. If the

pressure is low, the volume taken up by the actual

moleculesisinsignificant compared with the total volume of the



container.

But as the gas gets more compressed, the proportion of the total

volume that the molecules themselves take up gets higher and

higher. You could imagine compressing it so much that the

molecules were actually all touching each other. At that point the

volume available for them to move around in is zero!

Suppose at some high pressure, you measure the volume of the

container as, say, 1000 cm3, but suppose the molecules

themselves occupy as much as 100 cm3 of it.

The ideal gas equation was worked out by doing calculations based

on Kinetic Theory assumptions. The V in pV is assumed to be the

volume which the molecules are free to move around in - but in this

case, it would only be 900 cm3, not 1000 cm3.

If you worked out the compression factor, pV / nRT, by putting the

total volume of the container into the formula, the answer is bound

to be higher than it ought to be. It doesn't allow for the volume

taken up by the molecules themselves.

Let's just repeat one of the earlier diagrams so that you can seethis effect in operation.



For an ideal gas, the compression factor would be 1 over the whole

pressure range. For a real gas like nitrogen, notice how the

compression factor tends to increase with pressure.

The value of the compression factor is too high at high pressuresfor a real gas. The reason for that is that the measured volume that

you put into the expression is too high because you aren't allowing

for the volume taken up by the molecules. That error gets relatively

worse the more compressed the gas becomes.

The pressure problem

Another key assumption of the Kinetic Theory for ideal gases is thatthere are no intermolecular forces between the molecules. That is

wrong for every real gas.

If there weren't any intermolecular forces then it would be

impossible to condense the gas as a liquid. Even helium, with the

weakest of all intermolecular forces, can be turned to a liquid if the



temperature is low enough.

So what effect do intermolecular forces have?

For a gas molecule in the middle of the gas, there is no net effect. It

will be attracted to some extent to all the other molecules around it,

but, on average, those attractions will cancel each other out.

Attractions from behind a molecule, tending to slow it down, will be

cancelled out by attractions from in front of it, tending to speed it

up.

Despite all the intermolecular forces it is experiencing, the molecule

picked out in green will just continue to move in the same direction

at the same speed.

That's different if the molecule is just about to hit the wall of the

container.



Now there aren't any gas molecules in front of it, and the net pull is

backwards into the body of the gas. The molecule will slow down just before it hits the wall.

If it slows down, it will hit the wall less hard, and so exert less

pressure.

The overall effect of this is to make the measured pressure less

than it would be if the gas was ideal. That means that if you put the

measured pressure into the expression pV / nRT, the value of the

compression factor will be less than it would be if the gas was ideal.

This is why, under some conditions, graphs of compression factors

drop below the ideal value of 1.

Look yet again at the nitrogen curves:



This effect is most important at low temperatures. Why is that?

At lower temperatures, the molecules are moving more slowly on

average. Any pull they feel back into the gas will have relatively

more effect on a slow moving particle than a faster one.

At higher temperatures, where the molecules are moving a lot

faster, any small pull back into the body of the gas is hardly going to

be noticeable. At high temperatures, the effect of intermolecular

forces is indeed negligible.

And there is one final effect concerning intermolecular forces which

is slightly more hidden away.

As pressure increases, the molecules are forced more closely

together. If they are closer, the intermolecular forces will become

more important. So, as pressure increases, you would expect more

lowering of the compression factor relative to the ideal case. The

molecules which slow down the one just about to hit the wall will be



closer to it, and so more effective.

Is that what happens? Yes, up to a point.

Look again at the nitrogen curve at 100 K. As the pressure

increases, at first the value of the compression factor falls.

But it soon starts to rise again. Why? Because at this point, the

effect of the size of the molecules starts to become more important

- and as the pressure is increased even more, this other effect

becomes dominant.

Which is the most ideal gas?

You are looking for a gas with the smallest possible molecules, and

the lowest possible intermolecular forces. That is helium.

A helium molecule consists of a single small atom, and the van der

Waals dispersion forces are as low as it is possible for them to be.

Note:If you aren't happy about the factors which affect the size ofvan derWaals dispersion forces, follow this link. Use the BACK button to come back

to this page afterwards.

Like helium, a hydrogen molecule also has two electrons,

and so the intermolecular forces are going to be small - but

not as small as helium. In the hydrogen molecule, you have

two atoms that you can distribute the charges over.

As molecules get larger, then dispersion forces will

increase, and you may get other intermolecular forces such

as dipole-dipole attractions as well. Gases made of

molecules such as these will be much less ideal.

http://www.chemguide.co.uk/atoms/bonding/vdw.html#top







The van der Waals Equation

The van der Waals equation was the first attempt to try to

produce an equation which related p, V, n and T for real

gases. It looks like this:

If you think this looks complicated, you should see some of

the more modern attempts! Incidentally, you may come

across the equation in a simpler form for 1 mole of a gas

rather than for n moles. Under those circumstances, every

single n in the equation disappears.

Look at the left-hand side in two stages. First the pressure

term.

Remember from above, that the measured pressure is less

than the ideal pressure for a real gas. van der Waals has

added a term to compensate for that.

In the volume term, van der Waals has subtracted the value

nb to allow for the space taken up by the molecules

themselves.

a and b are constants for any particular gas, but they vary

from gas to gas to allow for the different intermolecular

forces, and molecular sizes. That means that, unfortunately,you no longer have a single equation that you can use for

any gas.

Fortunately, however, the ideal gas equation works well

enough for most gases at ordinary pressures, as long as



the temperature is reasonably high.

DEAL GASES AND THE IDEAL GAS LAW

This page looks at the assumptions which are made in the

Kinetic Theory about ideal gases, and takes an introductory

look at the Ideal Gas Law: pV = nRT. This is intended only

as an introduction suitable for chemistry students at about

UK A level standard (for 16 - 18 year olds), and so there is

no attempt to derive the ideal gas law using physics-style

calculations.

Kinetic Theory assumptions about ideal gases

There is no such thing as an ideal gas, of course, but many

gases behave approximately as if they were ideal at

ordinary working temperatures and pressures. Real gases

are dealt with in more detail on another page.

The assumptions are:

• Gases are made up of molecules which are in

constant random motion in straight lines.

• The molecules behave as rigid spheres.

• Pressure is due to collisions between the molecules



and the walls of the container.

• All collisions, both between the molecules

themselves, and between the molecules and the wallsof the container, are perfectly elastic. (That means

that there is no loss of kinetic energy during the

collision.)

• The temperature of the gas is proportional to the

average kinetic energy of the molecules.

And then two absolutely key assumptions, because these

are the two most important ways in which real gases differ

from ideal gases:

• There are no (or entirely negligible) intermolecular

forces between the gas molecules.

• The volume occupied by the molecules themselves is

entirely negligible relative to the volume of the

container.

The Ideal Gas Equation

The ideal gas equation is:

pV = nRT

On the whole, this is an easy equation to remember and

use. The problems lie almost entirely in the units. I am

assuming below that you are working in strict SI units (as

you will be if you are doing a UK-based exam, for example).



Exploring the various terms

Pressure, p

Pressure is measured in pascals, Pa - sometimes

expressed as newtons per square metre, N m-2. These

mean exactly the same thing.

Be careful if you are given pressures in kPa (kilopascals).

For example, 150 kPa is 150,000 Pa. You must make that

conversion before you use the ideal gas equation.

Should you want to convert from other pressure

measurements:

• 1 atmosphere = 101,325 Pa

• 1 bar = 100 kPa = 100,000 Pa

Volume, V

This is the most likely place for you to go wrong when you

use this equation. That's because the SI unit of volume is

the cubic metre, m3 - not cm3 or dm3.

1 m3 = 1000 dm3 = 1,000,000 cm3

So if you are inserting values of volume into the equation,

you first have to convert them into cubic metres.

You would have to divide a volume in dm3 by 1000, or in



cm3 by a million.

Similarly, if you are working out a volume using the

equation, remember to covert the answer in cubic metresinto dm3 or cm3if you need to - this time by multiplying by a

1000 or a million.

If you get this wrong, you are going to end up with a silly

answer, out by a factor of a thousand or a million. So it is

usually fairly obvious if you have done something wrong,

and you can check back again.

Number of moles, n

This is easy, of course - it is just a number. You already

know that you work it out by dividing the mass in grams by

the mass of one mole in grams.

You will most often use the ideal gas equation by firstmaking the substitution to give:

I don't recommend that you remember the ideal gas

equation in this form, but you must be confident that you

can convert it into this form.

The gas constant, R

A value for R will be given you if you need it, or you can

look it up in a data source. The SI value for R is 8.31441 J



K-1 mol-1.

Note:You may come across other values for this with

different units. A commonly used one in the past was 82.053

cm3 atm K

-1 mol

-1. The units tell you that the volume would be

in cubic centimetres and the pressure in atmospheres.

Unfortunately the units in the SI version aren't so obviously

helpful.

The temperature, T

The temperature has to be in kelvin. Don't forget to add 273 if youare given a temperature in degrees Celsius.

Using the ideal gas equation

Calculations using the ideal gas equation are included in my

calculations book (see the link at the very bottom of the page), and

I can't repeat them here. There are, however, a couple of

calculations that I haven't done in the book which give a reasonableidea of how the ideal gas equation works.

The molar volume at stp

If you have done simple calculations from equations, you have

probably used the molar volume of a gas.

1 mole of any gas occupies 22.4 dm3 at stp (standard temperature

and pressure, taken as 0°C and 1 atmosphere pressure). You mayalso have used a value of 24.0 dm

3 at room temperature and

pressure (taken as about 20°C and 1 atmosphere).

These figures are actually only true for an ideal gas, and we'll have

a look at where they come from.



We can use the ideal gas equation to calculate the volume of 1

mole of an ideal gas at 0°C and 1 atmosphere pressure.

First, we have to get the units right.

0°C is 273 K. T = 273 K

1 atmosphere = 101325 Pa. p = 101325 Pa

We know that n = 1, because we are trying to calculate the volume

of 1 mole of gas.

And, finally, R = 8.31441 J K-1 mol-1.

Slotting all of this into the ideal gas equation and then rearranging it

gives:

And finally, because we are interested in the volume in cubicdecimetres, you have to remember to multiply this by 1000 to

convert from cubic metres into cubic decimetres.

The molar volume of an ideal gas is therefore 22.4 dm3 at stp.

And, of course, you could redo this calculation to find the volume of

1 mole of an ideal gas at room temperature and pressure - or any

other temperature and pressure.

Finding the relative formula mass of a gas from its density

This is about as tricky as it gets using the ideal gas equation.

The density of ethane is 1.264 g dm-3 at 20°C and 1 atmosphere.



Calculate the relative formula mass of ethane.

The density value means that 1 dm3 of ethane weighs 1.264 g.

Again, before we do anything else, get the awkward units sorted

out.

A pressure of 1 atmosphere is 101325 Pa.

The volume of 1 dm3 has to be converted to cubic metres, by

dividing by 1000. We have a volume of 0.001 m3.

The temperature is 293 K.

Now put all the numbers into the form of the ideal gas equation

which lets you work with masses, and rearrange it to work out the

mass of 1 mole.

The mass of 1 mole of anything is simply the relative formula mass

in grams.

So the relative formula mass of ethane is 30.4, to 3 sig figs.

Now, if you add up the relative formula mass of ethane, C2H6using

accurate values of relative atomic masses, you get an answer of

30.07 to 4 significant figures. Which is different from our answer -

so what's wrong?



There are two possibilities.

• The density value I have used may not be correct. I did the

sum again using a slightly different value quoted at a

different temperature from another source. This time I got an

answer of 30.3. So the density values may not be entirely

accurate, but they are both giving much the same sort of

answer.

• Ethane isn't an ideal gas. Well, of course it isn't an ideal gas

- there's no such thing! However, assuming that the density

values are close to correct, the error is within 1% of what you

would expect. So although ethane isn't exactly behaving like

an ideal gas, it isn't far off.

If you need to know about real gases, now is a good time to read

about them.

AN INTRODUCTION TO CHEMICAL ENERGETICS

This page deals with the basic ideas about energy changes

during chemical reactions, including simple energy diagrams

and the terms exothermic and endothermic.

Energy changes during chemical reactions

Obviously, lots of chemical reactions give out energy as heat.

Getting heat by burning a fuel is a simple example, but you

will probably have come across lots of others in the lab.

Other reactions need a continuous supply of heat to make

them work. Splitting calcium carbonate into calcium oxide and



carbon dioxide is a simple example of this.

Any chemical reaction will involve breaking some bonds and

making new ones. Energy is needed to break bonds, and is

given out when the new bonds are formed. It is very unlikely

that these two processes will involve exactly the same amount

of energy - and so some energy will either be absorbed or

released during a reaction. You will find this discussed in

more detail in the page aboutbond enthalpies.

Simple energy diagrams

A reaction in which heat energy is given off is said to

beexothermic.

A reaction in which heat energy is absorbed is said to

beendothermic.

You can show this on simple energy diagrams.

For an exothermic change:

Notice that in an exothermic change, the products have a

lower energy than the reactants. The energy that the system

http://www.chemguide.co.uk/physical/energetics/bondenthalpies.html

http://www.chemguide.co.uk/physical/energetics/bondenthalpies.html



loses is given out as heat. The surroundings warm up.

For an endothermic change:

This time the products have a higher energy than the

reactants. The system absorbs this extra energy as heat from

the surroundings.

Expressing exothermic and endothermic changes in numbers

Here is an exothermic reaction, showing the amount of heat

evolved:

This shows that 394 kJ of heat energy are evolved when

equation quantities of carbon and oxygen combine to give

carbon dioxide. The mol-1 (per mole) refers to the whole

equation in mole quantities.

How do you know that heat is evolved? That is shown by the

negative sign.



You always think of the energy change during a reaction from

the point of view of the reactants. The reactants (carbon and

oxygen) have lost energy during the reaction. When you burn

carbon in oxygen, that is the energy which is causing thesurroundings to get hotter.

And here is an endothermic change:

In this case, 178 kJ of heat are absorbed when 1 mole of

calcium carbonate reacts to give 1 mole of calcium oxide and

1 mole of carbon dioxide.

You can tell that energy is being absorbed because of the plus

sign. A simple energy diagram for the reaction looks like this:

The products have a higher energy than the reactants. Energy

has been gained by the system - hence the plus sign.

Whenever you write values for any energy change, you must

always write a plus or a minus sign in front of it. If you have an

endothermic change, and write, say, 178 kJ mol-1 instead of

+178 kJ mol-1, you risk losing a mark in an exam.



Energetic stability

You are likely to come across statements that say that

something is energetically more stable than something else.

For example, in the next page in this section you will find that I

have said that oxygen, O2, is more energetically stable than

ozone, O3. What does this mean?

If you plot the positions of oxygen and ozone on an energy

diagram, it looks like this:

The lower down the energy diagram something is, the more

energetically stable it is. If ozone converted into ordinary

oxygen, heat energy would be released, and the oxygen would

be in a more energetically stable form than it was before.

So why doesn't ozone immediately convert into the more

energetically stable oxygen?

Similarly, if you mix petrol (gasoline) and air at ordinary

temperatures (when you are filling up a car, for example), why

doesn't it immediately convert into carbon dioxide and water?

It would be much more energetically stable if it turned into

carbon dioxide and water - you can tell that, because lots of

heat is given out when petrol burns in air. But there is no



reaction when you mix the two.

For any reaction to happen, bonds have to be broken, and new

ones made. Breaking bonds takes energy. There is a minimum

amount of energy needed before a reaction can start

-activation energy. If the molecules don't, for example, hit each

other with enough energy, then nothing happens. We say that

the mixture is kinetically stable, even though it may

beenergetically unstable with respect to its possible products.

So a petrol and air mixture at ordinary temperatures doesn't

react, even though a lot of energy would be released if the

reaction took place. Petrol and air are energetically unstable

with respect to carbon dioxide and water - they are much

higher up the energy diagram. But a petrol and air mixture is

kinetically stable at ordinary temperatures, because the

activation energy barrier is too high.

If you expose the mixture to a flame or a spark, then you get a

major fire or explosion. The initial flame supplies activation

energy. The heat given out by the molecules that react first is

more than enough to supply the activation energy for the next

molecules to react - and so on.

The moral of all this is that you should be very careful using

the word "stable" in chemistry!

Note:You will find a bit more about activation energy on the

introductory page aboutrates of reaction.

VARIOUS ENTHALPY CHANGE

DEFINITIONS

http://www.chemguide.co.uk/physical/basicrates/introduction.html





This page explains what an enthalpy change is, and then gives a

definition and brief comment for three of the various kinds of

enthalpy change that you will come across. You will find some more

definitions on other pages in this section.

It isessential that youlearn the definitions. You aren't going to be

able to do any calculations successfully if you don't know exactly

what all the terms mean.

Enthalpy changes

Enthalpy change is the name given to the amount of heat evolvedor absorbed in a reaction carried out at constant pressure. It is

given the symbol ∆H, read as "delta H".

Note:The term "enthalpy change" only applies to reactions

done at constant pressure. That is actually how most lab

reactions are done - in tubes or flasks (or whatever) open to the

atmosphere, so that the pressure is constant at atmospheric

pressure.

The phrase "at constant pressure" is an essential part of the

definition but, apart from that, you are unlikely to need to worry

about it if you are doing a UK-based exam at the equivalent of A

level.

Standard enthalpy changes

Standard enthalpy changes refer to reactions done understandard

conditions, and with everything present in theirstandard states.

Standard states are sometimes referred to as "reference states".

Standard conditions



Standard conditions are:

• 298 K (25°C)

• a pressure of 1 bar (100 kPa).

• where solutions are involved, a concentration of 1 mol dm-3

Warning!Standard pressure was originally defined as 1

atmosphere (101.325 kPa), and you will still find that in older

books (including my calculations book). At the time of writing

(August 2010) there was at least one UK-based syllabus that

was still talking in terms of "1 atmosphere". It isessential to

check your syllabus to find out exactly what you need to learn.

Standard states

For a standard enthalpy change everything has to be present in its

standard state. That is the physical and chemical state that you

would expect to find it in under standard conditions.

That means that the standard state for water, for example, is liquid

water, H2O(l) -not steam or water vapour or ice.

Oxygen's standard state is the gas, O2(g) -not liquid oxygen or

oxygen atoms.

For elements which have allotropes (two different forms of the

element in the same physical state), the standard state is the most

energetically stable of the allotropes.

For example, carbon exists in the solid state as both diamond and

graphite. Graphite is energetically slightly more stable than

diamond, and so graphite is taken as the standard state of carbon.

Similarly, under standard conditions, oxygen can exist as O2(simply

called oxygen) or as O3 (called ozone - but it is just an allotrope of



oxygen). The O2 form is far more energetically stable than O3, so

the standard state for oxygen is the common O2(g).

The symbol for standard enthalpy changes

The symbol for a standard enthalpy change is ∆H°, read as "delta

H standard" or, perhaps more commonly, as "delta H nought".

Note:Technically, the "o" in the symbol should have a

horizontal line through it, extending out at each side. This is

such a bother to produce convincingly without the risk of

different computers producing unreliable results, that I shall usethe common practice of simplifying it to "o".

Standard enthalpy change of reaction, ∆H°r

Remember that an enthalpy change is the heat evolved or

absorbed when a reaction takes place at constant pressure.

The standard enthalpy change of a reaction is the enthalpy

change which occurs when equation quantities of materials

react under standard conditions, and with everything in its

standard state.

That needs exploring a bit.

Here is a simple reaction between hydrogen and oxygen to make

water:

• First, notice that the symbol for a standard enthalpy change

of reaction is ∆H°r. For enthalpy changes of reaction, the "r"



(for reaction) is often missed off - it is just assumed.

• The "kJ mol-1" (kilojoules per mole) doesn't refer to any

particular substance in the equation. Instead it refers to the

quantities of all the substances given in the equation. In this

case, 572 kJ of heat is evolved when 2 moles of hydrogen

gas react with 1 mole of oxygen gas to form 2 moles of liquid

water.

• Notice that everything is in its standard state. In particular,

the water has to be formed as a liquid.

• And there is a hidden problem! The figure quoted is for the

reaction under standard conditions, but hydrogen and

oxygen don't react under standard conditions.

Whenever a standard enthalpy change is quoted, standard

conditions are assumed. If the reaction has to be done under

different conditions, a different enthalpy change would be

recorded. That has to be calculated back to what it would be

under standard conditions. Fortunately, you don't have to

know how to do that at this level.

Some important types of enthalpy change

Standard enthalpy change of formation, ∆H°f

The standard enthalpy change of formation of a compound

is the enthalpy change which occurs when one mole of the

compound is formed from its elements under standardconditions, and with everything in its standard state.

Note:When you are trying to learn these definitions, you can

make life easier for yourself by picking out the key bit, and

adding the other bits on afterwards. The key bit about this

definition is that you are forming 1 mole of a compound from its



elements. All the stuff about enthalpy change and standard

conditions and standard states is common to most of these

definitions.

The equation showing the standard enthalpy change of formation

for water is:

When you are writing one of these equations for enthalpy change

of formation, youmust end up with 1 mole of the compound. If that

needs you to write fractions on the left-hand side of the equation,

that is OK. (In fact, it is not just OK, it isessential, because

otherwise you will end up with more than 1 mole of compound, or

else the equation won't balance!)

The equation shows that 286 kJ of heat energy is given out when 1

mole of liquid water is formed from its elements under standard

conditions.

Standard enthalpy changes of formation can be written for any

compound, even if you can't make it directly from the elements. For

example, the standard enthalpy change of formation for liquid

benzene is +49 kJ mol-1. The equation is:

If carbon won't react with hydrogen to make benzene, what is the

point of this, and how does anybody know what the enthalpy

change is?

What the figure of +49 shows is the relative positions of benzene



and its elements on an energy diagram:

How do we know this if the reaction doesn't happen? It is actually

very simple to calculate it from other values which wecan measure

- for example, from enthalpy changes of combustion (coming up

next). We will come back to this again when we look at calculations

on another page.

Knowing the enthalpy changes of formation of compounds enables

you to calculate the enthalpy changes in a whole host of reactions

and, again, we will explore that in a bit more detail on another page.

And one final comment about enthalpy changes of formation:

The standard enthalpy change of formation of an element in its

standard state is zero. That's an important fact. The reason is

obvious . . .

For example, if you "make" one mole of hydrogen gas starting from

one mole of hydrogen gas you aren't changing it in any way, so you

wouldn't expect any enthalpy change. That is equally true of any

other element. The enthalpy change of formation of any element

has to be zero because of the way enthalpy change of formation is

defined.



Note:Some sources say that the enthalpy change of formation

of elements is taken as zeroby convention. That is simply

nonsense! The standard enthalpy change of formation of

elements is zero because of the way the enthalpy change is

defined.

If this confuses you, ignore it!

Standard enthalpy change of combustion, ∆H°c

The standard enthalpy change of combustion of a

compound is the enthalpy change which occurs when one

mole of the compound is burned completely in oxygen

under standard conditions, and with everything in its

standard state.

The enthalpy change of combustion will always have a negative

value, of course, because burning always releases heat.

Two examples:

Notice:

• Enthalpy of combustion equations will often contain

fractions, because you must start with only 1 mole of

whatever you are burning.

• If you are talking aboutstandard enthalpy changes of

combustion, everything must be in its standard state. One

important result of this is that any water you write amongst

the products must be there as liquid water.



Similarly, if you are burning something like ethanol, which is

a liquid under standard conditions, you must show it as a

liquid in any equation you use.

• Notice also that the equation and amount of heat evolved in

the hydrogen case is exactly the same as you have already

come across further up the page. At that time, it was

illustrating the enthalpy of formation of water. That can

happen in some simple cases. Talking about the enthalpy

change of formation of water is exactly the same as talking

about the enthalpy change of combustion of hydrogen.

HESS'S LAW AND ENTHALPY CHANGE CALCULATIONS

This page explains Hess's Law, and uses it to do some simple

enthalpy change calculations involving enthalpy changes of

reaction, formation and combustion.

Hess's Law

Stating Hess's Law

Hess's Law is the most important law in this part of chemistry.

Most calculations follow from it. It says . . .

The enthalpy change accompanying a chemical change is

independent of the route by which the chemical changeoccurs.



Explaining Hess's Law

Hess's Law is saying that if you convert reactants A into

products B, the overall enthalpy change will be exactly the

same whether you do it in one step or two steps or however

many steps.

If you look at the change on an enthalpy diagram, that is

actually fairly obvious.

This shows the enthalpy changes for an exothermic reaction

using two different ways of getting from reactants A to

products B. In one case, you do a direct conversion; in the

other, you use a two-step process involving some

intermediates.

In either case, the overall enthalpy change must be the same,

because it is governed by the relative positions of the

reactants and products on the enthalpy diagram.

If you go via the intermediates, you do have to put in someextra heat energy to start with, but you get it back again in the

second stage of the reaction sequence.

However many stages the reaction is done in, ultimately the

overall enthalpy change will be the same, because the

positions of the reactants and products on an enthalpy



diagram will always be the same.

Note:It is possibly confusing that I am switching between the

terms enthalpy and energy. Enthalpy change is simply a particular

measure of energy change. You will remember that the enthalpy

change is the heat evolved or absorbed during a reaction

happening at constant pressure.

I have labelled the vertical scale on this particular diagram as

enthalpy rather than energy, because we are specifically thinking

about enthalpy changes. I could have just kept to the more general

term "energy", but I prefer to be accurate.

You can do calculations by setting them out as enthalpy diagrams

as above, but there is a much simpler way of doing it which needs

virtually no thought.

You could set out the above diagram as:

Hess's Law says that the overall enthalpy change in these two

routes will be the same. That means that if you already know two ofthe values of enthalpy change for the three separate reactions

shown on this diagram (the three black arrows), you can easily

calculate the third - as you will see below.

The big advantage of doing it this way is that you don't have to

worry about the relative positions of everything on an enthalpy



diagram. It is completely irrelevant whether a particular enthalpy

change is positive or negative.

Warnings!

Although most calculations you will come across will fit into a

triangular diagram like the above, you may also come across other

slightly more complex cases needing more steps. That doesn't

make it any harder!

You need to take care in choosing your two routes. The pattern

willnot always look like the one above. You will see that in the

examples below.

Enthalpy change calculations using Hess's Law cycles

I can only give a brief introduction here, because this is covered in

careful, step-by-step detail in mychemistry calculations book.

Working out an enthalpy change of formation from enthalpychanges of combustion

If you have read an earlier page in this section, you may remember

that I mentioned that the standard enthalpy change of formation of

benzene was impossible to measure directly. That is because

carbon and hydrogen won't react to make benzene.

Important:If you don't know (without thinking about it too much)

exactly what is meant by standard enthalpy change of formation or

combustion, youmust get this sorted out now. Re-read the page

about enthalpy change definitions before you go any further -

andlearn them!

http://www.chemguide.co.uk/book.html

http://www.chemguide.co.uk/physical/energetics/definitions.html#top







Standard enthalpy changes of combustion, ∆H°c are relatively easy

to measure. For benzene, carbon and hydrogen, these are:

∆H°c (kJ mol-1)

C6H6(l) -3267

C(s) -394

H2(g) -286

First you have to design your cycle.

• Write down the enthalpy change you want to find as a simple

horizontal equation, and write ∆H over the top of the arrow.

(In diagrams of this sort, we often miss off the standard

symbol just to avoid clutter.)

• Then fit the other information you have onto the same

diagram to make a Hess's Law cycle, writing the known

enthalpy changes over the arrows for each of the other

changes.

• Finally, find two routes around the diagram, always going

with the flow of the various arrows. You must never have one

of your route arrows going in the opposite direction to one of

the equation arrows underneath it.

In this case, what we are trying to find is the standard enthalpy

change of formation of benzene, so that equation goes horizontally.



You will notice that I haven't bothered to include the oxygen that the

various things are burning in. The amount of oxygen isn't critical

because you just use an excess anyway, and including it really

confuses the diagram.

Why have I drawn a box around the carbon dioxide and water at the

bottom of the cycle? I tend to do this if I can't get all the arrows to

point to exactly the right things. In this case, there is no obvious

way of getting the arrow from the benzene to point atboth the

carbon dioxide and the water. Drawing the box isn't essential - I just

find that it helps me to see what is going on more easily.

Notice that you may have to multiply the figures you are using. Forexample, standard enthalpy changes of combustion start with 1

mole of the substance you are burning. In this case, the equations

need you to burn 6 moles of carbon, and 3 moles of hydrogen

molecules. Forgetting to do this is probably the most common

mistake you are likely to make.

How were the two routes chosen? Remember that you have to go

with the flow of the arrows. Choose your starting point as the corner

that only has arrows leaving from it. Choose your end point as thecorner which only has arrows arriving.

Now do the calculation:

Hess's Law says that the enthalpy changes on the two routes are



the same. That means that:

∆H - 3267 = 6(-394) + 3(-286)

Rearranging and solving:

∆H = 3267 + 6(-394) + 3(-286)

∆H = +45 kJ mol-1

Note:If you have a good memory, you might remember that I gave

a figure of +49 kJ mol-1 for the standard enthalpy change of

formation of benzene on an earlier page in this section. So why is

this answer different?

The main problem here is that I have taken values of the enthalpies

of combustion of hydrogen and carbon to 3 significant figures

(commonly done in calculations at this level). That introduces small

errors if you are just taking each figure once. However, here you

are multiplying the error in the carbon value by 6, and the error in

the hydrogen value by 3. If you are interested, you could rework the

calculation using a value of -393.5 for the carbon and -285.8 for the

hydrogen. That gives an answer of +48.6.

So why didn't I use more accurate values in the first place?

Because I wanted to illustrate this problem! Answers you get to

questions like this are often a bit out. The reason usually lies either

in rounding errors (as in this case), or the fact that the data may

have come from a different source or sources. Trying to get

consistent data can be a bit of a nightmare.

Working out an enthalpy change of reaction from enthalpychanges of formation

This is the commonest use of simple Hess's Law cycles that you

are likely to come across.

In this case, we are going to calculate the enthalpy change for the



reaction between ethene and hydrogen chloride gases to make

chloroethane gas from the standard enthalpy of formation values in

the table. If you have never come across this reaction before, it

makes no difference.

∆H°f (kJ mol-1)

C2H4(g) +52.2

HCl(g) -92.3

C2H5Cl(g) -109

Note:I'm not too happy about the value for chloroethane! The data

sources I normally use give a wide range of values. The one I have

chosen is an average value from theNIST Chemistry WebBook.

This uncertainty doesn't affect how you do the calculation in any

way, but the answer may not be exactly right - don't quote it as if

itwas right.

In the cycle below, this reaction has been written horizontally, and

the enthalpy of formation values added to complete the cycle.

Again, notice the box drawn around the elements at the bottom,

because it isn't possible to connect all the individual elements to

http://webbook.nist.gov/cgi/cbook.cgi?ID=C75003&Units=SI&Mask=1#Thermo-Gas





the compounds they are forming in any tidy way. Be careful to

count up all the atoms you need to use, and make sure they are

written as they occur in the elements in their standard state. You

mustn't, for example, write the hydrogens as 5H(g), because thestandard state for hydrogen is H2.

Note:In truth, if I am doing this type of enthalpy sum myself (with

nobody watching!), I tend to just write the word "elements" in the

bottom box to save the bother of working out exactly how many of

everything I need. I would be wary of doing that in an exam,

though.

And now the calculation. Just write down all the enthalpy changes

which make up the two routes, and equate them.

+52.2 - 92.3 + ∆H = -109

Rearranging and solving:

∆H = -52.2 + 92.3 - 109

∆H = -68.9 kJ mol-1

Note:I am afraid that this is as much as I feel I can give you on

this topic without risking sales of my book, or ending up in breach

of contract with my publishers. Unfortunately, it isn't enough for you

to be confident of being able to do these calculations every time.

Apart from anything else, you need lots of practice.

I have talked this through more gently in the book, with lots of

examples. If you chose to work through chapter 5 in the book, youwould be confident that you could do any chemical energetics

calculation that you were given.

Obviously I'm biased, but I strongly recommend that you either buy

the book, or get hold of a copy from your school or college or local

library. Don't just take my word for it - read the reviews on



the Amazon website.

BOND ENTHALPY (BOND ENERGY)

This page introduces bond enthalpies (bond energies) and looks at

some simple calculations involving them.

One of the most confusing things about this is the way the words

are used. These days, the term "bond enthalpy" is normally used,

but you will also find it described as "bond energy" - sometimes inthe same article. An even older term is "bond strength". So you can

take all these terms as being interchangeable.

As you will see below, though, "bond enthalpy" is used in several

different ways, and you might need to be careful about this.

Note:Bond enthalpies quoted from different sources often vary

by a few kilojoules, even if they are referring to exactly the same

thing. Don't worry if you come across slightly different values. Asyou will see later on this page, calculations involving bond

enthalpies hardly ever give accurate answers anyway.

You may even find differences in values between different pages

of Chemguide, or differences between Chemguide and my

calculations book. If so, I apologise, but I tend to use a lot of

different data sources which have varied over the years.

Explaining the terms

Bond dissociation enthalpy and mean bond enthalpy

http://www.amazon.co.uk/gp/product/0582411270?ie=UTF8&tag=chemguide&linkCode=as2&camp=1634&creative=6738&creativeASIN=0582411270





Simple diatomic molecules

A diatomic molecule is one that only contains two atoms. They

could be the same (for example, Cl2) or different (for example, HCl).

Thebond dissociation enthalpy is the energy needed to break

one mole of the bond to give separated atoms - everything being in

the gas state.

Important! The point about everything being in the gas state

isessential. You cannot use bond enthalpies to do calculations

directly from substances starting in the liquid or solid state.

As an example of bond dissociation enthalpy, to break up 1 mole of

gaseous hydrogen chloride molecules into separate gaseous

hydrogen and chlorine atoms takes 432 kJ. The bond dissociation

enthalpy for the H-Cl bond is +432 kJ mol-1.

More complicated molecules

What happens if the molecule has several bonds, rather than just

1?

Consider methane, CH4. It contains four identical C-H bonds, and it

seems reasonable that they should all have the same bond

enthalpy.

However, if you took methane to pieces one hydrogen at a time, it

needs a different amount of energy to break each of the four C-H

bonds. Every time you break a hydrogen off the carbon, the

environment of those left behind changes. And the strength of abond is affected by what else is around it.

In cases like this, the bond enthalpy quoted is an average value.

In the methane case, you can work out how much energy is

needed to break a mole of methane gas into gaseous carbon and



hydrogen atoms. That comes to +1662 kJ and involves breaking 4

moles of C-H bonds. The average bond energy is therefore

+1662/4 kJ, which is +415.5 kJ per mole of bonds.

That means that many bond enthalpies are actually quoted

asmean (or average) bond enthalpies, although it might not

actually say so. Mean bond enthalpies are sometimes referred to

as "bond enthalpy terms".

In fact, tables of bond enthalpies give average values in another

sense as well, particularly in organic chemistry. The bond enthalpy

of, say, the C-H bond varies depending on what is around it in the

molecule. So data tables use average values which will work well

enough in most cases.

That means that if you use the C-H value in some calculation, you

can't be sure that it exactly fits the molecule you are working with.

So don't expect calculations using mean bond enthalpies to give

very reliable answers.

You may well have to know the difference between a bond

dissociation enthalpy and a mean bond enthalpy, and you should

be aware that the wordmean (or average) is used in two slightly

different senses. But for calculation purposes, it isn't something you

need to worry about. Just use the values you are given.

Important:The rest of this page assumes that you have already

read the page about Hess's Law and enthalpy change

calculations. If you have come straight to the current page from a

search engine, you won't make sense of the way the calculations

are set out unless you first read the Hess's Law page.

http://www.chemguide.co.uk/physical/energetics/sums.html








Finding enthalpy changes of reaction from bond

enthalpies

I can only give a brief introduction here, because this is covered in

careful, step-by-step detail in mychemistry calculations book.

Cases where everything present is gaseous

Remember that you can only use bond enthalpies directly if

everything you are working with is in the gas state.

Using the same method as for other enthalpy sums

We are going to estimate the enthalpy change of reaction for the

reaction between carbon monoxide and steam. This is a part of the

manufacturing process for hydrogen.

The bond enthalpies are:

bond enthalpy (kJ mol-1)

C-O in carbon monoxide +1077

C-O in carbon dioxide +805

O-H +464

H-H +436

So let's do the sum. Here is the cycle - make sure that you

understand exactly why it is the way it is.





And now equate the two routes, and solve the equation to find the

enthalpy change of reaction.

∆H + 2(805) + 436 = 1077 + 2(464)

∆H = 1077 + 2(464) - 2(805) - 436

∆H = -41 kJ mol-1

Using a short-cut method for simple cases

You could do any bond enthalpy sum by the method above - taking

the molecules completely to pieces and then remaking the bonds. If

you are happy doing it that way, just go on doing it that way.

However, if you are prepared to give it some thought, you can save

a bit of time - although only in very simple cases where the

changes in a molecule are very small.

For example, chlorine reacts with ethane to give chloroethane and

hydrogen chloride gases.

(All of these are gases. I have left the state symbols out this time to



avoid cluttering the diagram.)

It is always a good idea to draw full structural formulae when you

are doing bond enthalpy calculations. It makes it much easier to

count up how many of each type of bond you have to break and

make.

If you look at the equation carefully, you can see what I mean by a

"simple case". Hardly anything has changed in this reaction. You

could work out how much energy is needed to break every bond,

and how much is given out in making the new ones, but quite a lot

of the time, you are just remaking the same bond.

All that has actually changed is that you have broken a C-H bond

and a Cl-Cl bond, and made a new C-Cl bond and a new H-Cl

bond. So you can just work those out.


C-H +413

Cl-Cl +243

C-Cl +346

H-Cl +432

Work out the energy needed to break C-H and Cl-Cl:

+413 + 243 = +656 kJ mol-1

Work out the energy released when you make C-Cl and H-Cl:

-346 - 432 = -778 kJ mol-1

So the net change is +656 - 778 = -122 kJ mol-1



Note:Even if you choose not to use this method, it might be a

good idea to be aware of it. It is possible to imagine an examiner

setting a question which assumes that you will use this method,

and therefore doesn't give a particular bond enthalpy value that

you would need if you did it by the longer method. For example, inthe case above, you don't actually need to know the C-C bond

enthalpy.

Cases where you have a liquid present

I have to keep on saying this! Remember that you can only use

bond enthalpies directly if everything you are working with is in the

gas state.

If you have one or more liquids present, you need an extra energy

term to work out the enthalpy change when you convert from liquid

to gas, or vice versa. That term is theenthalpy change of

vaporisation, and is given the symbol ∆Hvap or ∆Hv.

This is the enthalpy change when 1 mole of the liquid converts to

gas at its boiling point with a pressure of 1 bar (100 kPa).

(Older sources might quote 1 atmosphere rather than 1 bar.)

For water, the enthalpy change of vaporisation is +41 kJ mol-1. That

means that it take 41 kJ to change 1 mole of water into steam. If 1

mole of steam condenses into water, the enthalpy change would be

-41 kJ. Changing from liquid to gas needs heat; changing gas back

to liquid releases exactly the same amount of heat.

To see how this fits into bond enthalpy calculations, we willestimate the enthalpy change of combustion of methane - in other

words, the enthalpy change for this reaction:



Notice that the product is liquid water. You cannot apply bond

enthalpies to this. You must first convert it into steam. To do this

you have to supply 41 kJ mol-1.

The bond enthalpies you need are:


C-H +413

O=O +498

C=O in carbon dioxide +805

O-H +464

The cycle looks like this:

This obviously looks more confusing than the cycles we've looked

at before, but apart from the extra enthalpy change of vaporisation

stage, it isn't really any more difficult. Before you go on, make sure



that you can see why every single number and arrow on this

diagram is there.

In particular, make sure that you can see why the first 4 appears in

the expression "4(+464)". That is an easy thing to get wrong. (In

fact, when I first drew this diagram, I carelessly wrote 2 instead of 4

at that point!)

That's the hard bit done - now the calculation:

∆H + 2(805) + 2(41) + 4(464) = 4(413) + 2(498)

∆H = 4(413) + 2(498) - 2(805) - 2(41) - 4(464)

∆H = -900 kJ mol-1

The measured enthalpy change of combustion is -890 kJ mol-1, and

so this answer agrees to within about 1%. As bond enthalpy

calculations go, that's a pretty good estimate.

Note:Because this is all covered in more detail in mycalculations

book, I am afraid that this is as far as I am prepared to go with

this topic. The book will give you a lot more examples, includingsome variations such as calculating bond enthalpies from

enthalpies of formation, and vice versa.

LATTICE ENTHALPY (LATTICE ENERGY)

This page introduces lattice enthalpies (lattice energies) and Born-

Haber cycles.

Lattice enthalpy and lattice energy are commonly used as if they

mean exactly the same thing - you will often find both terms used

within the same textbook article or web site, including on university







sites.

In fact, there is a difference between them which relates to the

conditions under which they are calculated. However, the difference

is small, and negligible compared with the differing values for lattice

enthalpy that you will find from different data sources.

Unless you go on to do chemistry at degree level, the difference

between the two terms isn't likely to worry you.

Note:While I have been writing this section, the different values for the

same piece of data from different data sources has driven me crazy,

because there is no easy way of knowing which is the most recent ormost accurate data.

In the Born-Haber cycles below, I have used numbers which give a

consistent answer, but please don't assume that they are necessarily the

most accurate ones. If you are doing a course for 16 - 18 year olds,

none of this really matters - you just use the numbers you are given.

If you use mychemistry calculations book, you will find a slightly

different set of numbers. These came from the Chemistry Data Book

edited by Stark and Wallace, published by John Murray. Values from this

now fairly old book often differ slightly from more recent sources.

Don't worry about this. It doesn't affect the principles in any way. Just

don't assume that any bit of data you are given (even by me) is

necessarily "right"!

What is lattice enthalpy?

Two different ways of defining lattice enthalpy

There are two different ways of defining lattice enthalpy which

directly contradict each other, and you will find both in common

use. In fact, there is a simple way of sorting this out, but many





sources don't use it.

I will explain how you can do this in a moment, but first let's look at

how the problem arises.

Lattice enthalpy is a measure of the strength of the forces between

the ions in an ionic solid. The greater the lattice enthalpy, the

stronger the forces.

Those forces are only completely broken when the ions are present

as gaseous ions, scattered so far apart that there is negligible

attraction between them. You can show this on a simple enthalpy

diagram.

For sodium chloride, the solid is more stable than the gaseous ions

by 787 kJ mol-1, and that is a measure of the strength of the

attractions between the ions in the solid. Remember that energy (in

this case heat energy) is given out when bonds are made, and is

needed to break bonds.

So lattice enthalpy could be described in either of two ways.

• You could describe it as the enthalpy change when 1 mole of

sodium chloride (or whatever) was formed from its scattered

gaseous ions. In other words, you are looking at a downward

arrow on the diagram.



In the sodium chloride case, that would be -787 kJ mol-1.

• Or, you could describe it as the enthalpy change when 1

mole of sodium chloride (or whatever) is broken up to form

its scattered gaseous ions. In other words, you are looking at

an upward arrow on the diagram.

In the sodium chloride case, that would be +787 kJ mol-1.

Both refer to the same enthalpy diagram, but one looks at it from

the point of view of making the lattice, and the other from the point

of view of breaking it up.

Unfortunately, both of these are often described as "latticeenthalpy".

This is an absurdly confusing situation which is easily resolved. I

suggest that you never use the term "lattice enthalpy" without

qualifying it.

• You should talk about "lattice dissociation enthalpy" if you

want to talk about the amount of energy needed to split up a

lattice into its scattered gaseous ions.

For NaCl, the lattice dissociation enthalpy is +787 kJ mol-1.

• You should talk about "lattice formation enthalpy" if you want

to talk about the amount of energy released when a lattice is

formed from its scattered gaseous ions.

For NaCl, the lattice formation enthalpy is -787 kJ mol-1

.

That immediately removes any possibility of confusion.

So . . .



The lattice dissociation enthalpy is the enthalpy change

needed to convert 1 mole of solid crystal into its scattered

gaseous ions. Lattice dissociation enthalpies are always

positive.

The lattice formation enthalpy is the enthalpy change when

1 mole of solid crystal is formed from its scattered gaseous

ions. Lattice formation enthalpies are always negative.

Note:Find out which of these versions your syllabus is likely to want

you to know (even if they just call it "lattice enthalpy") and concentrate

on that one, but be aware of the confusion!

Incidentally, if you are ever uncertain about which version is being used,

you can tell from the sign of the enthalpy change being discussed. If the

sign is positive, for example, it must refer to breaking bonds, and

therefore to a lattice dissociation enthalpy.

Factors affecting lattice enthalpy

The two main factors affecting lattice enthalpy are the charges on

the ions and the ionic radii (which affects the distance between the

ions).

The charges on the ions

Sodium chloride and magnesium oxide have exactly the same

arrangements of ions in the crystal lattice, but the lattice enthalpies

are very different.



Note:In this diagram, and similar diagrams below, I am not interested

in whether the lattice enthalpy is defined as a positive or a negative

number - I am just interested in their relative sizes. Strictly speaking,

because I haven't added a sign to the vertical axis, the values are for

lattice dissociation enthalpies. If you prefer lattice formation enthalpies,

just mentally put a negative sign in front of each number.

You can see that the lattice enthalpy of magnesium oxide is much

greater than that of sodium chloride. That's because in magnesium

oxide, 2+ ions are attracting 2- ions; in sodium chloride, the

attraction is only between 1+ and 1- ions.

The radius of the ions

The lattice enthalpy of magnesium oxide is also increased relative

to sodium chloride because magnesium ions are smaller than

sodium ions, and oxide ions are smaller than chloride ions.



That means that the ions are closer together in the lattice, and that

increases the strength of the attractions.

You can also see this effect of ion size on lattice enthalpy as you go

down a Group in the Periodic Table.

For example, as you go down Group 7 of the Periodic Table from

fluorine to iodine, you would expect the lattice enthalpies of their

sodium salts to fall as the negative ions get bigger - and that is the

case:

Attractions are governed by the distances between the centres of

the oppositely charged ions, and that distance is obviously greater

as the negative ion gets bigger.

And you can see exactly the same effect if as you go down Group

1. The next bar chart shows the lattice enthalpies of the Group 1

chlorides.



Note:To save anyone the bother of getting in touch with me to point it

out, it's not strictly fair to include caesium chloride in this list. Caesiumchloride has a different packing arrangement of ions in its crystal, and

that has a small effect on the lattice enthalpy. The effect is small enough

that it doesn't actually affect the trend.

Calculating lattice enthalpy

It is impossible to measure the enthalpy change starting from asolid crystal and converting it into its scattered gaseous ions. It is

even more difficult to imagine how you could do the reverse - start

with scattered gaseous ions and measure the enthalpy change

when these convert to a solid crystal.

Instead, lattice enthalpies always have to be calculated, and there

are two entirely different ways in which this can be done.

You can can use a Hess's Law cycle (in this case called a Born-

Haber cycle) involving enthalpy changes whichcan be measured.

Lattice enthalpies calculated in this way are described as

experimental values.

Or you can do physics-style calculations working out how much

energy would be released, for example, when ions considered as



point charges come together to make a lattice. These are described

as theoretical values. In fact, in this case, what you are actually

calculating are properly described as latticeenergies.

Note:If you aren't confident aboutHess's Law cycles, it isessential that

you follow this link before you go on.

Experimental values - Born-Haber cycles

Standard atomisation enthalpies

Before we start talking about Born-Haber cycles, there is an extra

term which we need to define. That isatomisation enthalpy, ∆H°a.

The standard atomisation enthalpy is the enthalpy change

when 1 mole of gaseous atoms is formed from the element

in its standard state. Enthalpy change of atomisation is

always positive.

You are always going to have to supply energy to break an element

into its separate gaseous atoms.

All of the following equations represent changes involving

atomisation enthalpy:

Notice particularly that the "mol-1" is per mole of atoms formed -

NOT per mole of element that you start with. You will quite

commonly have to write fractions into the left-hand side of the





equation. Getting this wrong is a common mistake.

Born-Haber cycles

I am going to start by drawing a Born-Haber cycle for sodium

chloride, and then talk it through carefully afterwards. You will see

that I have arbitrarily decided to draw this for lattice formation

enthalpy. If you wanted to draw it for lattice dissociation enthalpy,

the red arrow would be reversed - pointing upwards.



Focus to start with on the higher of the two thicker horizontal lines.

We are starting here with the elements sodium and chlorine in their

standard states. Notice that we only need half a mole of chlorine

gas in order to end up with 1 mole of NaCl.

The arrow pointing down from this to the lower thick line represents

the enthalpy change of formation of sodium chloride.

The Born-Haber cycle now imagines this formation of sodium

chloride as happening in a whole set of small changes, most of

which we know the enthalpy changes for - except, of course, for the

lattice enthalpy that we want to calculate.

• The +107 is the atomisation enthalpy of sodium. We have to

produce gaseous atoms so that we can use the next stage in

the cycle.

• The +496 is the first ionisation energy of sodium. Remember

that first ionisation energies go from gaseous atoms to

gaseous singly charged positive ions.

• The +122 is the atomisation enthalpy of chlorine. Again, we

have to produce gaseous atoms so that we can use the nextstage in the cycle.

• The -349 is the first electron affinity of chlorine. Remember

that first electron affinities go from gaseous atoms to

gaseous singly charged negative ions.

• And finally, we have the positive and negative gaseous ions

that we can convert into the solid sodium chloride using the

lattice formation enthalpy.

Note:If you have forgotten aboutionisation energies orelectron

affinities follow these links before you go on.

http://www.chemguide.co.uk/atoms/properties/ies.html#top


http://www.chemguide.co.uk/atoms/properties/eas.html#top







Now we can use Hess's Law and find two different routes around

the diagram which we can equate.

As I have drawn it, the two routes are obvious. The diagram is set

up to provide two different routes between the thick lines.

So, here is the cycle again, with the calculation directly underneath

it . . .

-411 = +107 + 496 + 122 - 349 + LE

LE = -411 - 107 - 496 - 122 + 349

LE = -787 kJ mol-1



Note:Notice that in the calculation, we aren't making any assumptions

about the sign of the lattice enthalpy (despite the fact that it is obviously

negative because the arrow is pointing downwards). In the first line of

the calculation, I have just written "+ LE", and have left it to the

calculation to work out that it is a negative answer.

How would this be different if you had drawn a lattice dissociation

enthalpy in your diagram? (Perhaps because that is what your

syllabus wants.)

Your diagram would now look like this:

The only difference in the diagram is the direction the lattice

enthalpy arrow is pointing. It does, of course, mean that you have



to find two new routes. You can't use the original one, because that

would go against the flow of the lattice enthalpy arrow.

This time both routes would start from the elements in their

standard states, and finish at the gaseous ions.

-411 + LE = +107 + 496 + 122 - 349

LE = +107 + 496 + 122 - 349 + 411

LE = +787 kJ mol-1

Once again, the cycle sorts out the sign of the lattice enthalpy for

you.

Note:You will find more examples of calculations involving Born-Haber

cycles in mychemistry calculations book. This includes rather more

complicated cycles involving, for example, oxides.

If you compare the figures in the book with the figures for NaCl above,

you will find slight differences - the main culprit being the electron affinity

of chlorine, although there are other small differences as well. Don't

worry about this - the values in the book come from an older data

source. In an exam, you will just use the values you are given, so it isn't

a problem.

Theoretical values for lattice energy

Let's assume that a compound is fully ionic. Let's also assume that

the ions are point charges - in other words that the charge is

concentrated at the centre of the ion. By doing physics-style

calculations, it is possible to calculate a theoretical value for what

you would expect the lattice energy to be.

And no - I am not being careless about this! Calculations of this

sort end up with values of latticeenergy, and not latticeenthalpy. If

http://www.chemguide.co.uk/book.html#top

http://www.chemguide.co.uk/book.html#top



you know how to do it, you can then fairly easily convert between

the two.

There are several different equations, of various degrees of

complication, for calculating lattice energy in this way. You won't be

expected to be able to do these calculations at this level, but you

might be expected to comment on the results of them.

There are two possibilities:

• There is reasonable agreement between the experimental

value (calculated from a Born-Haber cycle) and the

theoretical value.

Sodium chloride is a case like this - the theoretical and

experimental values agree to within a few percent. That

means that for sodium chloride, the assumptions about the

solid being ionic are fairly good.

• The experimental and theoretical values don't agree.

A commonly quoted example of this is silver chloride, AgCl.

Depending on where you get your data from, the theoreticalvalue for lattice enthalpy for AgCl is anywhere from about 50

to 150 kJ mol-1 less than the value that comes from a Born-

Haber cycle.

In other words, treating the AgCl as 100% ionic

underestimates its lattice enthalpy by quite a lot.

The explanation is that silver chloride actually has a

significant amount of covalent bonding between the silver

and the chlorine, because there isn't enough

electronegativity difference between the two to allow for

complete transfer of an electron from the silver to the

chlorine.



Comparing experimental (Born-Haber cycle) and theoretical values

for lattice enthalpy is a good way of judging how purely ionic a

crystal is.

Note:If you have forgotten aboutelectronegativity it might pay you to

revise it now by following this link.

Why is magnesium chloride MgCl2?

This section may well go beyond what your syllabus

requires. Before you spend time on it, check your syllabus

(and past exam papers as well if possible) to make sure.

The question arises as to why, from an energetics point of

view, magnesium chloride is MgCl2 rather than MgCl or

MgCl3 (or any other formula you might like to choose).

It turns out that MgCl2 is the formula of the compound which

has the most negative enthalpy change of formation - inother words, it is the most stable one relative to the

elements magnesium and chlorine.

Let's look at this in terms of Born-Haber cycles.

In the cycles this time, we are interested in working out what

the enthalpy change of formation would be for the imaginary

compounds MgCl and MgCl3.

That means that we will have to use theoretical values of

their lattice enthalpies. We can't use experimental ones,

because these compounds obviously don't exist!

I'm taking theoretical values for lattice enthalpies for these

http://www.chemguide.co.uk/atoms/bonding/electroneg.html#top





compounds that I found on the web. I can't confirm these,

but all the other values used by that source were accurate.

The exact values don't matter too much anyway, because

the results are so dramatically clear-cut.

We will start with the compound MgCl, because that cycle is

just like the NaCl one we have already looked at.

The Born-Haber cycle for MgCl

Find two routes around this without going against the flow of



any arrows. That's easy:

∆Hf = +148 + 738 + 122 - 349 - 753

∆Hf = -94 kJ mol-1

So the compound MgCl is definitely energetically more

stable than its elements.

I have drawn this cycle very roughly to scale, but that is

going to become more and more difficult as we look at the

other two possible formulae. So I am going to rewrite it as atable.

You can see from the diagram that the enthalpy change of

formation can be found just by adding up all the other

numbers in the cycle, and we can do this just as well in a

table.

kJ

atomisation enthalpy of Mg +148

1st IE of Mg +738

atomisation enthalpy of Cl +122

electron affinity of Cl -349

lattice enthalpy -753

calculated ∆Hf -94



The Born-Haber cycle for MgCl2

The equation for the enthalpy change of formation this time

is

So how does that change the numbers in the Born-Haber

cycle?

• You need to add in the second ionisation energy of

magnesium, because you are making a 2+ ion.

• You need to multiply the atomisation enthalpy of

chlorine by 2, because you need 2 moles of gaseous

chlorine atoms.

• You need to multiply the electron affinity of chlorine by

2, because you are making 2 moles of chloride ions.

•

You obviously need a different value for latticeenthalpy.

kJ


1st IE of Mg +738

2nd IE of Mg +1451

atomisation enthalpy of Cl (x 2) +244

electron affinity of Cl (x 2) -698




calculated ∆Hf -643

You can see that much more energy is released when you

make MgCl2 than when you make MgCl. Why is that?

You need to put in more energy to ionise the magnesium to

give a 2+ ion, but a lot more energy is released as lattice

enthalpy. That is because there are stronger ionic

attractions between 1- ions and 2+ ions than between the 1-

and 1+ ions in MgCl.

So what about MgCl3? The lattice energy here would be

even greater.

The Born-Haber cycle for MgCl3

The equation for the enthalpy change of formation this time

is

So how does that change the numbers in the Born-Haber

cycle this time?

• You need to add in the third ionisation energy of

magnesium, because you are making a 3+ ion.

• You need to multiply the atomisation enthalpy of

chlorine by 3, because you need 3 moles of gaseous

chlorine atoms.

• You need to multiply the electron affinity of chlorine by



3, because you are making 3 moles of chloride ions.

• You again need a different value for lattice enthalpy.

kJ


1st IE of Mg +738

2nd IE of Mg +1451

3rd IE of Mg +7733

atomisation enthalpy of Cl (x 3) +366

electron affinity of Cl (x 3) -1047


calculated ∆Hf +3949

This time, the compound is hugely energetically unstable,

both with respect to its elements, and also to other

compounds that could be formed. You would need to supply

nearly 4000 kJ to get 1 mole of MgCl3 to form!

Look carefully at the reason for this. The lattice enthalpy is

the highest for all these possible compounds, but it isn'thigh enough to make up for the very large third ionisation

energy of magnesium.

Why is the third ionisation energy so big? The first two

electrons to be removed from magnesium come from the 3s



level. The third one comes from the 2p. That is closer to the

nucleus, and lacks a layer of screening as well - and so

much more energy is needed to remove it.

The 3s electrons are screened from the nucleus by the 1

level and 2 level electrons. The 2p electrons are only

screened by the 1 level (plus a bit of help from the 2s

electrons).

Conclusion

Magnesium chloride is MgCl2 because this is the

combination of magnesium and chlorine which produces

the most energetically stable compound - the one with the

most negative enthalpy change of formation.

ENTHALPIES OF SOLUTION AND HYDRATION

This page looks at the relationship between enthalpies o

sol!tion" h#$ration enthalpies an$ latti%e enthalpies&

Note:You really ought to have read the pages aboutHess's Law

cycles andlattice enthalpies before you continue with this page.

Enthalpy change of solution

Defining enthalpy change of solution

The enthalpy change of solution is the enthalpy change

when 1 mole of an ionic substance dissolves in water to

give a solution of infinite dilution.



http://www.chemguide.co.uk/physical/energetics/lattice.html



http://www.chemguide.co.uk/physical/energetics/lattice.html



Enthalpies of solution may be either positive or negative - in other

words, some ionic substances dissolved endothermically (for

example, NaCl); others dissolve exothermically (for example

NaOH).

An infinitely dilute solution is one where there is a sufficiently large

excess of water that adding any more doesn't cause any further

heat to be absorbed or evolved.

So, when 1 mole of sodium chloride crystals are dissolved in an

excess of water, the enthalpy change of solution is found to be +3.9

kJ mol-1. The change is slightly endothermic, and so the

temperature of the solution will be slightly lower than that of the

original water.

Thinking about dissolving as an energy cycle

Why is heat sometimes evolved and sometimes absorbed when a

substance dissolves in water? To answer that it is useful to think

about the various enthalpy changes that are involved in the

process.

You can think of an imaginary process where the crystal lattice is

first broken up into its separate gaseous ions, and then those ions

have water molecules wrapped around them. That is how they exist

in the final solution.

The heat energy needed to break up 1 mole of the crystal lattice is

thelattice dissociation enthalpy.

The heat energy released when new bonds are made between the

ions and water molecules is known as thehydration enthalpy of

the ion.

The hydration enthalpy is the enthalpy change when 1



mole of gaseous ions dissolve in sufficient water to give an

infinitely dilute solution. Hydration enthalpies are always

negative.

Factors affecting the size of hydration enthalpy

Hydration enthalpy is a measure of the energy released when

attractions are set up between positive or negative ions and water

molecules.

With positive ions, there may only be loose attractions between the

slightly negative oxygen atoms in the water molecules and the

positive ions, or there may be formal dative covalent (co-ordinate

covalent) bonds.

With negative ions, hydrogen bonds are formed between lone pairs

of electrons on the negative ions and the slightly positive hydrogens

in water molecules.

Note:You will find the attractions between water molecules and

positive ions discussed on the page about dative covalent bonding.You will find the attractions between negative ions and water

molecules discussed on the page abouthydrogen bonding.

The size of the hydration enthalpy is governed by the amount of

attraction between the ions and the water molecules.

• The attractions are stronger the smaller the ion. For

example, hydration enthalpies fall as you go down a group inthe Periodic Table. The small lithium ion has by far the

highest hydration enthalpy in Group1, and the small fluoride

ion has by far the highest hydration enthalpy in Group 7. In

both groups, hydration enthalpy falls as the ions get bigger.

• The attractions are stronger the more highly charged the ion.

http://www.chemguide.co.uk/atoms/bonding/dative.html


http://www.chemguide.co.uk/atoms/bonding/hbond.html






For example, the hydration enthalpies of Group 2 ions (like

Mg2+) are much higher than those of Group 1 ions (like Na

+).

Estimating enthalpies of solution from lattice enthalpies and

hydration enthalpies

The hydration enthalpies for calcium and chloride ions are given by

the equations:

The following cycle is for calcium chloride, and includes a lattice

dissociation enthalpy of +2258 kJ mol-1.

We have to use double the hydration enthalpy of the chloride ion

because we are hydrating 2 moles of chloride ions. Make sure you

understand exactly how the cycle works.

So . . .

∆Hsol = +2258 - 1650 + 2(-364)



∆Hsol = -120 kJ mol-1

Whether an enthalpy of solution turns out to be negative or positive

depends on the relative sizes of the lattice enthalpy and the

hydration enthalpies. In this particular case, the negative hydration

enthalpies more than made up for the positive lattice dissociation

enthalpy.

Important:This diagram is basically just to show you how to do

these calculations, but I have no confidence whatsoever in the

accuracy of the data I have used. The measured value for the

enthalpy of solution for anhydrous calcium chloride (the value which

we are trying to calculate here) is about -80 kJ mol-1. That bears little

relationship to the value calculated here!

I have no idea what the source of this discrepancy is. One or more of

the figures I am using is obviously inaccurate. Trying to find reliable

values for energy terms like lattice enthalpies or hydration enthalpies

has been a total nightmare throughout the whole of this energetics

section. Virtually every textbook I have available (and I have quite a

few!) gives different values. Virtually every textbook that you can

access via Google Books has different values. Virtually every

website that you look at seems to have its own combination of values

which may or may not agree with any of the books.

Despite the fact that I am now totally fed up with this whole topic, it

shouldn't affect you as a student or a teacher working towards an

exam the equivalent of UK A level. You have to work with whatever

values your examiners give you. What is important is that you

understand what you are doing. As long as the answer you come up

with is consistent with the data you are given, that is actually all that

matters to you.

ENTHALPY CHANGE OF

NEUTRALISATION



This page looks briefly at enthalpy changes of neutralisation. In

common with my experience with most of the other pages in this

section, searches for reliable data throw up various values for the

same reaction. Don't worry too much about this. It doesn't actuallyaffect the arguments.

Enthalpy change of neutralisation

Defining standard enthalpy change of neutralisation

The standard enthalpy change of neutralisation is the

enthalpy change when solutions of an acid and an alkali

react together under standard conditions to produce 1 mole

of water.

Notice that enthalpy change of neutralisation is always

measured per mole of water formed.

Enthalpy changes of neutralisation are always negative - heat is

given out when an acid and and alkali react. For reactions involving

strong acids and alkalis, the values are always very closely similar,

with values between -57 and -58 kJ mol-1.

That varies slightly depending on the acid-alkali combination (and

also on what source you look it up in!).

Why do strong acids reacting with strong alkalis give closelysimilar values?

We make the assumption that strong acids and strong alkalis are

fully ionised in solution, and that the ions behave independently of

each other. For example, dilute hydrochloric acid contains hydrogen

ions and chloride ions in solution. Sodium hydroxide solution



consists of sodium ions and hydroxide ions in solution.

The equation for any strong acid being neutralised by a strong

alkali is essentially just a reaction between hydrogen ions and

hydroxide ions to make water. The other ions present (sodium and

chloride, for example) are just spectator ions, taking no part in the

reaction.

The full equation for the reaction between hydrochloric acid and

sodium hydroxide solution is:

. . . but what is actually happening is:

If the reaction is the same in each case of a strong acid and a

strong alkali, it isn't surprising that the enthalpy change is similar.

Note:Actually, of course, the enthalpy changes should be the same,

not similar, if the assumptions we are making are exactly true! The

small differences between strong acid-strong base combinations are

almost invariably glossed over at this level. In fact, I can't remember

ever seeing this discussed in any source - textbook or web. It isn't

uncommon to find a list of enthalpy changes of neutralisation showing

some variability in the strong acid-strong alkali cases, and then a few

lines later on, this is ignored completely with a statement that in these

cases, the enthalpy changes of neutralisation are the same,

because . . .

I have decide not to waste time trying to sort out the exact reasons for

the problem, because I suspect it will take ages and ages, and it is

never going to get asked at this level anyway.

Why do weak acids or weak alkalis give different values?



In a weak acid, such as ethanoic acid, at ordinary concentrations,

something like 99% of the acid isn't actually ionised. That means

that the enthalpy change of neutralisation will include other

enthalpy terms involved in ionising the acid as well as the reactionbetween the hydrogen ions and hydroxide ions.

And in a weak alkali like ammonia solution, the ammonia is also

present mainly as ammonia molecules in solution. Again, there will

be other enthalpy changes involved apart from the simple formation

of water from hydrogen ions and hydroxide ions.

For reactions involving ethanoic acid or ammonia, the measured

enthalpy change of neutralisation is a few kilojoules less exothermic

than with strong acids and bases.

For example, one source which gives the enthalpy change of

neutralisation of sodium hydroxide solution with HCl as -57.9 kJ

mol-1, gives a value of -56.1 kJ mol

-1 for sodium hydroxide solution

being neutralised by ethanoic acid.

For very weak acids, like hydrogen cyanide solution, the enthalpy

change of neutralisation may be much less. A different source

gives the value for hydrogen cyanide solution being neutralised by

potassium hydroxide solution as -11.7 kJ mol-1, for example.

Documents

Other Gas Laws