Second Law of Thermodynamics

Second Law of Thermodynamics

The second law of thermodynamics is a general principle which places constraints upon the direction of heat transfer and the attainable efficiencies of heat engines. In so doing, it goes beyond the limitations imposed by the first law of thermodynamics. It's implications may be visualized in terms of the waterfall analogy.

Second Law: Heat Engines

Second Law of Thermodynamics: It is impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W . Some amount of heat QC must be exhausted to a cold reservoir. This precludes a perfect heat engine.
This is sometimes called the "first form" of the second law, and is referred to as the Kelvin-Planck statement of the second law.

Second Law: Refrigerator

Second Law of Thermodynamics: It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object. This precludes a perfect refrigerator. The statements about refrigerators apply to air conditioners and heat pumps, which embody the same principles.

This is the "second form" or Clausius statement of the second law.

Second Law: Entropy

Second Law of Thermodynamics: In any cyclic process the entropy will either increase or remain the same.

Entropy:	a state variable whose change is defined for a reversible process at T where Q is the heat absorbed.
Entropy:	a measure of the amount of energy which is unavailable to do work.
Entropy:	a measure of the disorder of a system.
Entropy:	a measure of the multiplicity of a system.

Since entropy gives information about the evolution of an isolated system with time, it is said to give us the direction of "time's arrow" . If snapshots of a system at two different times shows one state which is more disordered, then it could be implied that this state came later in time. For an isolated system, the natural course of events takes the system to a more disordered (higher entropy) state.

First Law of Thermodynamics

The first law of thermodynamics is the application of the conservation of energy principle to heat and thermodynamic processes:

The first law makes use of the key concepts of internal energy, heat, andsystem work. It is used extensively in the discussion of heat engines. The standard unit for all these quantities would be the joule, although they are sometimes expressed in calories or BTU.


It is typical for chemistry texts to write the first law as ΔU=Q+W. It is the same law, of course - the thermodynamic expression of the conservation of energy principle. It is just that W is defined as the work done on the system instead of work done by the system. In the context of physics, the common scenario is one of adding heat to a volume of gas and using the expansion of that gas to do work, as in the pushing down of a piston in an internal combustion engine. In the context of chemical reactions and process, it may be more common to deal with situations where work is done on the system rather than by it.

Enthalpy

Four quantities called "thermodynamic potentials" are useful in the chemical thermodynamics of reactions and non-cyclic processes. They are internal energy, the enthalpy, the Helmholtz free energy and the Gibbs free energy. Enthalpy is defined by H = U + PV


where P and V are the pressure and volume, and U is internal energy. Enthalpy is then a precisely measurable state variable, since it is defined in terms of three other precisely definable state variables. It is somewhat parallel to the first law of thermodynamics for a constant pressure system Q = ΔU + PΔV since in this case Q=ΔH





It is a useful quantity for tracking chemical reactions. If as a result of an exothermic reaction some energy is released to a system, it has to show up in some measurable form in terms of the state variables. An increase in the enthalpy H = U + PV might be associated with an increase in internal energy which could be measured by calorimetry, or with work done by the system, or a combination of the two.



The internal energy U might be thought of as the energy required to create a system in the absence of changes in temperature or volume. But if the process changes the volume, as in a chemical reaction which produces a gaseous product, then work must be done to produce the change in volume. For a constant pressure process the work you must do to produce a volume change ΔV is PΔV. Then the term PV can be interpreted as the work you must do to "create room" for the system if you presume it started at zero volume.

System Work

When work is done by a thermodynamic system, it is ususlly a gas that is doing the work. The work done by a gas at constant pressure is:
For non-constant pressure, the work can be visualized as the area under the pressure-volume curve which represents the process taking place. The more general expression for work done is:

Work done by a system decreases the internal energy of the system, as indicated in the First Law of Thermodynamics. System work is a major focus in the discussion of heat engines.

THE CONTACT PROCESS

THE CONTACT PROCESS This page describes the Contact Process for the manufacture of sulphuric acid, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process.
A brief summary of the Contact Process The Contact Process: makes sulphur dioxide; convers the sulphur dioxide into sulphur trioxide (the reversible reaction at the heart of the process); converts the sulphur trioxide into concentrated sulphuric acid. Making the sulphur dioxide This can either be made by burning sulphur in an excess of air: . . . or by heating sulphide ores like pyrite in an excess of air: In either case, an excess of air is used so that the sulphur dioxide produced is already mixed with oxygen for the next stage. Converting the sulphur dioxide into sulphur trioxide This is a reversible reaction, and the formation of the sulphur trioxide is exothermic. A flow scheme for this part of the process looks like this: The reasons for all these conditions will be explored in detail further down the page. Converting the sulphur trioxide into sulphuric acid This can't be done by simply adding water to the sulphur trioxide - the reaction is so uncontrollable that it creates a fog of sulphuric acid. Instead, the sulphur trioxide is first dissolved in concentrated sulphuric acid: The product is known as fuming sulphuric acid or oleum. This can then be reacted safely with water to produce concentrated sulphuric acid - twice as much as you originally used to make the fuming sulphuric acid. Explaining the conditions The proportions of sulphur dioxide and oxygen The mixture of sulphur dioxide and oxygen going into the reactor is in equal proportions by volume. Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. That means that the gases are going into the reactor in the ratio of 1 molecule of sulphur dioxide to 1 of oxygen. That is an excess of oxygen relative to the proportions demanded by the equation. According to Le Chatelier's Principle, Increasing the concentration of oxygen in the mixture causes the position of equilibrium to shift towards the right. Since the oxygen comes from the air, this is a very cheap way of increasing the conversion of sulphur dioxide into sulphur trioxide. Why not use an even higher proportion of oxygen? This is easy to see if you take an extreme case. Suppose you have a million molecules of oxygen to every molecule of sulphur dioxide. The equilibrium is going to be tipped very strongly towards sulphur trioxide - virtually every molecule of sulphur dioxide will be converted into sulphur trioxide. Great! But you aren't going to produce much sulphur trioxide every day. The vast majority of what you are passing over the catalyst is oxygen which has nothing to react with. By increasing the proportion of oxygen you can increase the percentage of the sulphur dioxide converted, but at the same time decrease the total amount of sulphur trioxide made each day. The 1 : 1 mixture turns out to give you the best possible overall yield of sulphur trioxide. The temperature Equilibrium considerations You need to shift the position of the equilibrium as far as possible to the right in order to produce the maximum possible amount of sulphur trioxide in the equilibrium mixture. The forward reaction (the production of sulphur trioxide) is exothermic. According to Le Chatelier's Principle, this will be favoured if you lower the temperature. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat. In order to get as much sulphur trioxide as possible in the equilibrium mixture, you need as low a temperature as possible. However, 400 - 450°C isn't a low temperature! Rate considerations The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to produce as much sulphur trioxide as possible per day. It makes no sense to try to achieve an equilibrium mixture which contains a very high proportion of sulphur trioxide if it takes several years for the reaction to reach that equilibrium. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor. The compromise 400 - 450°C is a compromise temperature producing a fairly high proportion of sulphur trioxide in the equilibrium mixture, but in a very short time. The pressure Equilibrium considerations Notice that there are 3 molecules on the left-hand side of the equation, but only 2 on the right. According to Le Chatelier's Principle, if you increase the pressure the system will respond by favouring the reaction which produces fewer molecules. That will cause the pressure to fall again. In order to get as much sulphur trioxide as possible in the equilibrium mixture, you need as high a pressure as possible. High pressures also increase the rate of the reaction. However, the reaction is done at pressures close to atmospheric pressure! Economic considerations Even at these relatively low pressures, there is a 99.5% conversion of sulphur dioxide into sulphur trioxide. The very small improvement that you could achieve by increasing the pressure isn't worth the expense of producing those high pressures. The catalyst Equilibrium considerations The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't produce any greater percentage of sulphur trioxide in the equilibrium mixture. Its only function is to speed up the reaction. Rate considerations In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor.

THE HABER PROCESS

This page describes the Haber Process for the manufacture of ammonia from nitrogen and hydrogen, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process.
A brief summary of the Haber Process
The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic.

A flow scheme for the Haber Process looks like this:

Some notes on the conditions
The catalyst
The catalyst is actually slightly more complicated than pure iron. It has potassium hydroxide added to it as a promoter - a substance that increases its efficiency.
The pressure
The pressure varies from one manufacturing plant to another, but is always high. You can't go far wrong in an exam quoting 200 atmospheres.
Recycling
At each pass of the gases through the reactor, only about 15% of the nitrogen and hydrogen converts to ammonia. (This figure also varies from plant to plant.) By continual recycling of the unreacted nitrogen and hydrogen, the overall conversion is about 98%.

Explaining the conditions
The proportions of nitrogen and hydrogen
The mixture of nitrogen and hydrogen going into the reactor is in the ratio of 1 volume of nitrogen to 3 volumes of hydrogen.
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. That means that the gases are going into the reactor in the ratio of 1 molecule of nitrogen to 3 of hydrogen.
That is the proportion demanded by the equation.
In some reactions you might choose to use an excess of one of the reactants. You would do this if it is particularly important to use up as much as possible of the other reactant - if, for example, it was much more expensive. That doesn't apply in this case.
There is always a down-side to using anything other than the equation proportions. If you have an excess of one reactant there will be molecules passing through the reactor which can't possibly react because there isn't anything for them to react with. This wastes reactor space - particularly space on the surface of the catalyst.

The temperature
Equilibrium considerations
You need to shift the position of the equilibrium as far as possible to the right in order to produce the maximum possible amount of ammonia in the equilibrium mixture.
The forward reaction (the production of ammonia) is exothermic.

According to Le Chatelier's Principle, this will be favoured if you lower the temperature. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat.
In order to get as much ammonia as possible in the equilibrium mixture, you need as low a temperature as possible. However, 400 - 450°C isn't a low temperature!
Rate considerations
The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to produce as much ammonia as possible per day. It makes no sense to try to achieve an equilibrium mixture which contains a very high proportion of ammonia if it takes several years for the reaction to reach that equilibrium.
You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor.
The compromise
400 - 450°C is a compromise temperature producing a reasonably high proportion of ammonia in the equilibrium mixture (even if it is only 15%), but in a very short time.

The pressure
Equilibrium considerations

Notice that there are 4 molecules on the left-hand side of the equation, but only 2 on the right.
According to Le Chatelier's Principle, if you increase the pressure the system will respond by favouring the reaction which produces fewer molecules. That will cause the pressure to fall again.
In order to get as much ammonia as possible in the equilibrium mixture, you need as high a pressure as possible. 200 atmospheres is a high pressure, but not amazingly high.
Rate considerations
Increasing the pressure brings the molecules closer together. In this particular instance, it will increase their chances of hitting and sticking to the surface of the catalyst where they can react. The higher the pressure the better in terms of the rate of a gas reaction.
Economic considerations
Very high pressures are very expensive to produce on two counts.
You have to build extremely strong pipes and containment vessels to withstand the very high pressure. That increases your capital costs when the plant is built.
High pressures cost a lot to produce and maintain. That means that the running costs of your plant are very high.
The compromise
200 atmospheres is a compromise pressure chosen on economic grounds. If the pressure used is too high, the cost of generating it exceeds the price you can get for the extra ammonia produced.

The catalyst
Equilibrium considerations
The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't produce any greater percentage of ammonia in the equilibrium mixture. Its only function is to speed up the reaction.
Rate considerations
In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor.

Separating the ammonia

When the gases leave the reactor they are hot and at a very high pressure. Ammonia is easily liquefied under pressure as long as it isn't too hot, and so the temperature of the mixture is lowered enough for the ammonia to turn to a liquid. The nitrogen and hydrogen remain as gases even under these high pressures, and can be recycled.

LE CHATELIER'S PRINCIPLE

LE CHATELIER'S PRINCIPLE This page looks at Le Chatelier's Principle and explains how to apply it to reactions in a state of dynamic equilibrium. It covers changes to the position of equilibrium if you change concentration, pressure or temperature. It also explains very briefly why catalysts have no effect on the position of equilibrium.
It is important in understanding everything on this page to realise that Le Chatelier's Principle is no more than a useful guide to help you work out what happens when you change the conditions in a reaction in dynamic equilibrium. It doesn't explain anything. I'll keep coming back to that point! Using Le Chatelier's PrincipleA statement of Le Chatelier's Principle If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. Using Le Chatelier's Principle with a change of concentration Suppose you have an equilibrium established between four substances A, B, C and D.
What would happen if you changed the conditions by increasing the concentration of A? According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of A decreases again - by reacting it with B and turning it into C + D. The position of equilibrium moves to the right. This is a useful way of converting the maximum possible amount of B into C and D. You might use it if, for example, B was a relatively expensive material whereas A was cheap and plentiful. What would happen if you changed the conditions by decreasing the concentration of A? According to Le Chatelier, the position of equilibrium will move so that the concentration of A increases again. That means that more C and D will react to replace the A that has been removed. The position of equilibrium moves to the left. This is esssentially what happens if you remove one of the products of the reaction as soon as it is formed. If, for example, you removed C as soon as it was formed, the position of equilibrium would move to the right to replace it. If you kept on removing it, the equilibrium position would keep on moving rightwards - turning this into a one-way reaction. Important This isn't in any way an explanation of why the position of equilibrium moves in the ways described. All Le Chatelier's Principle gives you is a quick way of working out what happens.
Using Le Chatelier's Principle with a change of pressure This only applies to reactions involving gases: What would happen if you changed the conditions by increasing the pressure? According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the pressure is reduced again. Pressure is caused by gas molecules hitting the sides of their container. The more molecules you have in the container, the higher the pressure will be. The system can reduce the pressure by reacting in such a way as to produce fewer molecules. In this case, there are 3 molecules on the left-hand side of the equation, but only 2 on the right. By forming more C and D, the system causes the pressure to reduce. Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer molecules. What would happen if you changed the conditions by decreasing the pressure? The equilibrium will move in such a way that the pressure increases again. It can do that by producing more molecules. In this case, the position of equilibrium will move towards the left-hand side of the reaction. What happens if there are the same number of molecules on both sides of the equilibrium reaction? In this case, increasing the pressure has no effect whatsoever on the position of the equilibrium. Because you have the same numbers of molecules on both sides, the equilibrium can't move in any way that will reduce the pressure again. Important Again, this isn't an explanation of why the position of equilibrium moves in the ways described. You will find a rather mathematical treatment of the explanation by following the link below.
Using Le Chatelier's Principle with a change of temperature For this, you need to know whether heat is given out or absorbed during the reaction. Assume that our forward reaction is exothermic (heat is evolved): This shows that 250 kJ is evolved (hence the negative sign) when 1 mole of A reacts completely with 2 moles of B. For reversible reactions, the value is always given as if the reaction was one-way in the forward direction. The back reaction (the conversion of C and D into A and B) would be endothermic by exactly the same amount. What would happen if you changed the conditions by increasing the temperature? According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the temperature is reduced again. Suppose the system is in equilibrium at 300°C, and you increase the temperature to 500°C. How can the reaction counteract the change you have made? How can it cool itself down again? To cool down, it needs to absorb the extra heat that you have just put in. In the case we are looking at, the back reactionabsorbs heat. The position of equilibrium therefore moves to the left. The new equilibrium mixture contains more A and B, and less C and D. If you were aiming to make as much C and D as possible, increasing the temperature on a reversible reaction where the forward reaction is exothermic isn't a good idea! What would happen if you changed the conditions by decreasing the temperature? The equilibrium will move in such a way that the temperature increases again. Suppose the system is in equilibrium at 500°C and you reduce the temperature to 400°C. The reaction will tend to heat itself up again to return to the original temperature. It can do that by favouring the exothermic reaction. The position of equilibrium will move to the right. More A and B are converted into C and D at the lower temperature. Summary Increasing the temperature of a system in dynamic equilibrium favours the endothermic reaction. The system counteracts the change you have made by absorbing the extra heat. Decreasing the temperature of a system in dynamic equilibrium favours the exothermic reaction. The system counteracts the change you have made by producing more heat. Important Again, this isn't in any way an explanation of why the position of equilibrium moves in the ways described. It is only a way of helping you to work out what happens.
Le Chatelier's Principle and catalysts Catalysts have sneaked onto this page under false pretences, because adding a catalyst makes absolutely no difference to the position of equilibrium, and Le Chatelier's Principle doesn't apply to them. This is because a catalyst speeds up the forward and back reaction to the same extent. Because adding a catalyst doesn't affect the relative rates of the two reactions, it can't affect the position of equilibrium. So why use a catalyst? For a dynamic equilibrium to be set up, the rates of the forward reaction and the back reaction have to become equal. This doesn't happen instantly. For a very slow reaction, it could take years! A catalyst speeds up the rate at which a reaction reaches dynamic equilibrium.

AN INTRODUCTION TO CHEMICAL EQUILIBRIA

AN INTRODUCTION TO CHEMICAL EQUILIBRIA This page looks at the basic ideas underpinning the idea of a chemical equilibrium. It talks about reversible reactions and how they behave if the system is closed. This leads to the idea of a dynamic equilibrium, and what the common term "position of equilibrium" means. Reversible reactionsA reversible reaction is one which can be made to go in either direction depending on the conditions. If you pass steam over hot iron the steam reacts with the iron to produce a black, magnetic oxide of iron called triiron tetroxide, Fe3O4. The hydrogen produced in the reaction is swept away by the stream of steam. Under different conditions, the products of this reaction will also react together. Hydrogen passed over hot triiron tetroxide reduces it to iron. Steam is also produced. This time the steam produced in the reaction is swept away by the stream of hydrogen. These reactions are reversible, but under the conditions normally used, they become one-way reactions. The products aren't left in contact with each other, so the reverse reaction can't happen. Reversible reactions happening in a closed system A closed system is one in which no substances are either added to the system or lost from it. Energy can, however, be transferred in or out at will. In the example we've been looking at, you would have to imagine iron being heated in steam in a closed container. Heat is being added to the system, but none of the substances in the reaction can escape. The system is closed. As the triiron tetroxide and hydrogen start to be formed, they will also react again to give the original iron and steam. So, if you analysed the mixture after a while, what would you find? You would find that you had established what is known as a dynamic equilibrium. To explain what that means, we are going to use a much simpler example . . . Dynamic equilibria Getting a visual feel for a dynamic equilibrium Imagine a substance which can exist in two forms - a blue form or an orange form - and that each form can react to give the other one. We are going to let them react in a closed system. Neither form can escape. Assume that the blue form turns into the orange one much faster than the other way round. In any given time, these are the chances of the two changes happening: You can simulate this very easily with some coloured paper cut up into small pieces (a different colour on each side), and a dice. The following are the real results of a "reaction" I did myself. I started with 16 blue squares and looked at each one in turn and decided whether it should change colour by throwing a dice. A blue square was turned into an orange square (the bit of paper was turned over!) if I threw a 4, 5 or 6 An orange square was turned into a blue square only if I threw a 6 while I was looking at that particular square. Once I had looked at all 16 squares, I started the process all over again - but obviously with a different starting pattern. The diagrams show the results of doing this 11 times (plus the original 16 blue squares). You can see that the "reaction" is continuing all the time. The exact pattern of orange and blue is constantly changing. However, the overall numbers of orange and of blue squares remain remarkably constant - most commonly, 12 orange ones to 4 blue ones.
Explaining the term "dynamic equilibrium" The reaction has reached equilibrium in the sense that there is no further change in the numbers of blue and orange squares. However, the reaction is still continuing. For every orange square that turns blue, somewhere in the mixture it is replaced by a blue square turning orange. This is known as a dynamic equilibrium. The word dynamic shows that the reaction is still continuing. You can show dynamic equilibrium in an equation for a reaction by the use of special arrows. In the present case, you would write it as: It is important to realise that this doesn't just mean that the reaction is reversible. It means that you have a reversible reaction in a state of dynamic equilibrium. The "forward reaction" and the "back reaction" The change from left to right in the equation (in this case from blue to orange as it is written) is known as the forward reaction. The change from right to left is the back reaction. Position of equilibrium In the example we've used, the equilibrium mixture contained more orange squares than blue ones. Position of equilibrium is a way of expressing this. You can say things like: "The position of equilibrium lies towards the orange." "The position of equilibrium lies towards the right-hand side." If the conditions of the experiment change (by altering the relative chances of the forward and back reactions happening), the composition of the equilibrium mixture will also change. For example, if changing the conditions produced more blue in the equilibrium mixture, you would say "The position of equilibrium has moved to the left" or "The position of equilibrium has moved towards the blue".
Reaching equilibrium from the other side What happens if you started the reaction with orange squares rather than blue ones, but kept the chances of each change happening the same as in the first example? This is the result of my "reaction". Once again, you can see that exactly the same position of equilibrium is being established as when we started with the blue squares. You get exactly the same equilibrium mixture irrespective of which side of the equation you start from - provided the conditions are the same in both cases.
A more formal look at dynamic equilibria Thinking about reaction rates This is the equation for a general reaction which has reached dynamic equilibrium: How did it get to that state? Let's assume that we started with A and B. At the beginning of the reaction, the concentrations of A and B were at their maximum. That means that the rate of the reaction was at its fastest. As A and B react, their concentrations fall. That means that they are less likely to collide and react, and so the rate of the forward reaction falls as time goes on. In the beginning, there isn't any C and D, so there can't be any reaction between them. As time goes on, though, their concentrations in the mixture increase and they are more likely to collide and react. With time, the rate of the reaction between C and D increases: Eventually, the rates of the two reactions will become equal. A and B will be converting into C and D at exactly the same rate as C and D convert back into A and B again. At this point there won't be any further change in the amounts of A, B, C and D in the mixture. As fast as something is being removed, it is being replaced again by the reverse reaction. We have reached a position of dynamic equilibrium. A summary A dynamic equilibrium occurs when you have a reversible reaction in a closed system. Nothing can be added to the system or taken away from it apart from energy. At equilibrium, the quantities of everything present in the mixture remain constant, although the reactions are still continuing. This is because the rates of the forward and the back reactions are equal. If you change the conditions in a way which changes the relative rates of the forward and back reactions you will change the position of equilibrium - in other words, change the proportions of the various substances present in the equilibrium mixture.