Thermodynamics: Energy Accountancy and Thermochemistry
In this section we will discuss about one of the main principles of thermodynamics which is energy accountancy. In this section, we will focus on heat transactions which is the application of first law of thermodynamics. Besides that, we will also discuss about enthalpy where the reaction happens at constant pressure. Furthermore, we will also discuss about heat capacity and adiabatic system.
The main principle in heat transactions is the change in the internal energy (dU) is the sum of the change in heat energy (dq) plus the change in work (dw). However, work can come in many forms such as mechanical, electrical, and chemical to name a few. Hence, dU can be expressed as the sum of many different works.
All the works stated above require change in volume.
If on the other hand the work is prevented from example (e.g. by preventing gas expansion), then any change in internal energy must be expressed as a change in heat. In adiabatic bomb calorimeter this is just what happens. The bomb is a strong metal container and a water bath and the bomb is both things which is not just metal container. The perfect insulation is produced not by adding thermal insulation but by having another water bath around it that is maintained at the same temperature as the inner water bath. If there is not temperature difference between the two baths then there is no loss of heat (recall the zeroth law of thermodynamics). Obviously, you need to put heat or take it out of the outer water bath to keep it at the same temperature. Therefore, from the calorimeter by knowing the calorimeter constant (the heat required to raise it 1°C), then the energy released can be calculated as shown below.
Where C is the calorimeter constant.
From the equation above, it takes a certain amount of energy to heat something up. Clearly, it takes twice as much energy to heat 2 kg of Cu as it does of 1 kg of Cu. Besides that, it also takes a different amount of heat to heat 1 kg of Cu or 1 kg of diamond or 1 kg of air. This resistance to heating is called the heat capacity and when something melts or boils, energy goes in but the temperature does not rise, therefore at a first order phase transition to the heat capacity is infinite.
Heat capacity is defined as the change in internal energy with temperature; at constant volume. Thus, it can be written mathematically as:
As you notice from the equation, the subscript V means at constant volume and this equation is a partial derivative equation, which means everything else is kept constant. Clearly, this depends on how much you have, so it is an extensive property. As an additional information, Cv is "the heat capacity at constant volume" and is an extensive properties, and Cv,m is "the molar heat capacity at constant volume" and is an intensive property. By knowing the molar heat capacity at constant volume, the change in internal energy can be calculated and because it is at constant volume (no work) the you usually use Δqv instead of just Δq.
As mentioned before, the molar heat capacity is an intensive property, it does not depend on how much matter you have. The molar heat capacity of a monoatomic gas 3R/2 and for diatomic and linear triatomic molecule is 5R/2, where R is gas constant (8.31451 J K−1 mol−1). This number is came from the degrees of freedom and for example diatomic has rotation and vibration as well as x, y, z motion.
In a simple way a derivative is nothing more than for a gradient of a line and in our problem we have a surface as shown below instead of a line. Then, the big question is how to solve it. In short, it can be solved depends on which way you look at the surface as to what the gradient is.
As shown above, we hold the volume as a constant, so we can say a gradient of a specific line on that surface.
In earlier part, it was mentioned that if you prevent expansion, system does no work, so dU = dq, therefore you can measure the internal energy. Furthermore, you can reverse this argument and say if the system cannot expand then the increase in internal energy is equal to the heat supplied. Hence, if you know the heat input you do not need to worry about if the temperature changes (as in an adiabatic system).
Now, the question what happens if it can expand? The change in internal energy is less than the heat supplied, as some energy goes to expansion. If this is done at constant pressure then the heat supplied is equal to the enthalpy. Mathematically, enthalpy can be written as:
Then, as we concern about the change and enthalpy is also a state function, it is more appropriate to rewrite the equation as:
The enthalpy can be equal to the heat supplied at constant pressure if the system is free to expand. Therefore, if liquid boils it increases its volume and pushes against a constant pressure, the atmosphere. The heat supplied is the latent heat of vaporisation (enthalpy of vaporisation).
Moreover, enthalpy is used extensively in chemistry because most reaction take place under constant pressure, usually atmospheric pressure in open system. For example, if we supply 25 kJ of energy to heat an open beaker of water, it happens at constant pressure. Therefore, the enthalpy of the water goes up by 25 kJ and we can write as ΔH = +25 kJ, not dH as dH is for a tiny change, ΔH for a large (measurable) change.
The changes in enthalpy is commonly measured by a differential scanning calorimetry (DSC). This measurement uses two heaters and maintains the temperature of a sample and a reference. The power input into the heaters gives the enthalpy of the transition. This machines are standard pieces of equipment in every chemistry department.
As we discussed early on about enthalpy phase transition, we will try to discuss briefly for some phase transitions. The phase transition for a solid-solid or solid-liquid or liquid-liquid transition has volume changes very small, so pV term can be ignored to give dU = dH and it can be measured by a bomb calorimeter (DSC is easier). However, this statement is not true for liquid or solid to gas transition as now the volume change is large.
In a solid or liquid to gas reaction the volume change is very large and usually 700-1000 times. Thus, we can use the ideal gas equation and ignore the volume of the solid or liquid. Therefore, the enthalpy for solid or liquid to gas transition can be written as:
Where nG is the number of moles of gas produced.
The enthalpy of a system increases if you add heat energy to it and if you increase the temperature its enthalpy must also be increase. This relationship is given by the capacity at constant pressure as shown below.
Cp and CV are extensive properties which they depend on how much you have as the molar heat capacities Cp,m and CV,m are intensive properties because these are per mole (the amount of material is defined as one mole). For small changes in temperature the capacity does not change much, so you can determine the heat supplied by knowing the heat capacity at constant pressure and the change in temperature.
Because the pressure is constant, so ΔH = Δq.
However for larger changes, the heat capacity is not constant and varies as:
and for example below is the Cp function of O2.
The heat capacities of perfect gases are related by a simple equation:
This means that the heat capacity at constant pressure is always bigger than the heat capacity at constant volume and in gases the heat capacity at constant volume is small.
Now, we move to another cases in thermodynamics as the big question what happens when a gas expands? Let takes the example of fridge. When gas inside the tube is compressed it heats up, this heat is released at the back of the fridge in the narrow pipes. Then the gas is allowed into bigger tubes inside the fridge, it expands and cools. This cooling phenomena because the internal energy decreases as the gas expands and in doing so does work. If the pipe was perfectly insulated the temperature of the gas would fall even more. Thus, it raise a question what would be the change in internal energy?
Now, we are going to make use of the fact that the internal energy is a state function and this means it does not matter how we go from one state to the other (just the difference is important). Therefore, in the real system the gas expands and cools at the same time but we will say first the gas expands the it cools (this makes it a lot easier). By keeping temperature constant, work done by an expanding gas, at constant pressure is:
As it is a perfect gas we can use the ideal gas equation as shown below.
Then, we keep the volume constant as temperature changes and applying from the heat capacity at constant volume:
After that from the definition of internal energy and remembering its an adiabatic system (no heat flow) means dq = 0, so both equations can be put together. Both equation must be equal as it is adiabatic, no heat flow, so there is no where else to get the energy from.
If the volume starts at Vi and ends up at Vf and the temperature starts at Ti and ends at Tf,so we can integrate the equation to give:
Then, to simplify the equation by manipulating it and make a constant c where c = CV/nR, the equation become as shown below.
There is another relationship between CV and Cp that might be useful for later calculation for adiabatic system and the relationship is shown below.
Then, for an ideal gas as pV = nRT,
besides that for an ideal and adiabatic system,
Then, by equating both equation, the relationship between p and V in an ideal and adiabatic system is:
For adiabatic changes pV curves are known as adiabats and adiabats falls more steeply than an isotherm. That is because in an isotherm heat must flow into the system to maintain the temperature as V increases.
Until this far, we have discussed what enthalpy is and how it relates to internal energy, work, and heat. Hence, using this information we will apply thermodynamics to chemistry. To begin with let have a quick revision that exothermic reaction is when ΔH<0 and="" endothermic="" is="" nbsp="" span="" when=""> ΔH > 0, and the reason why this way around is due to internal energy must decrease.
The standard enthalpy changes is measured at the standard state of a substance at a specified temperature is its pure form at 1 bar (100 kPa), so for Fe at 500 K it is pure iron at 500 K and 1 bar, it means we need to have a "clean" starting point and define everything from there. Another examples are the standard enthalpies of vaporisation and fusion is defined at 1 bar and 298.15 K.
When the enthalpy of vaporisation is added to the enthalpy of fusion we can calculate the enthalpy of sublimation because it is a state function.
Besides that, the enthalpy of the forward and reverse reaction differ only in sign and again because it is a state function.
The enthalpy of a reaction can be described with a chemical equations plus thermodynamics information, and this description is called thermochemical equations. For example, the thermochemical equation for the combustion of methane is shown below.
As you can see the thermochemical equation above, all the compounds are in its standard state at 298.15 K and the enthalpy is the standard enthalpy for combustion.
Hess's law is an important result of thermodynamics as it states that standard enthalpies of individual reaction can be combined to obtain the enthalpy of another reaction. The basic reason of Hess's law is simply as the enthalpy is a state function. These individual reactions (steps) may not be chemically possible, the only requirement is that they should balance.
The combustion reaction of methane that was shown early on can be written by relating this enthalpy to the standard molar enthalpies of formation of the pure substances.
In general, it can be written as:
In the same way, you can use elements in their pure forms to assemble chemical reactions. For example the standard enthalpy of formation of benzene is given by:
As you notice, carbon's standard state is not diamond but graphite, as graphite can turn into diamond after a certain long period. This is an exception for the concept of standard state especially for elements that have allotropic structure as the standard state is the most stable allotrope. The other exception is phosphorus take to be white phosphorus which is the most reproducible form.
Another application of Hess's law is to calculate the enthalpy of combustion of propene with the data:
The enthalpy of combustion of propene can be described as the diagram shown below:
From the diagram above we can calculate the enthalpy for combustion of propene as:
In all our cases earlier on about enthalpy, we used the standard reaction enthalpy and it is always at 298 K. However, if the temperature is raised the enthalpy changes, so if a reaction occurs at a higher temperature then there are changes that should be taken into account for accurate results. In practice the changes for reactants and products are almost the same in many cases. Hence, the big question is how do we calculate the change?
To solve the problem we need to refer back to heat capacity at constant pressure, then we can calculate the change in enthalpy. The important thing is that it is the difference in heat capacities that is important.
This equation is called Kirchoff's law. The heat capacities may be different but their variation with temperature is usually similar, so the correction is small.
For example, the standard enthalpy of formation of butane is -126.15 kJ mol−1
and the heat capacities are:
Cp (C(graphite)) = 8.527 J K−1 mol−1
Cp (H2) = 28.824 J K−1 mol−1
Cp (C4H10) = 97.45 J K−1 mol−1
Hence, we can calculate the enthalpy of formation of butane at 400 K. Firstly, we need to calculate ΔCp:
Lastly, we can input all the data into Kirchoff's law, but remember to convert the enthalpy into J mol−1.
The main principle in heat transactions is the change in the internal energy (dU) is the sum of the change in heat energy (dq) plus the change in work (dw). However, work can come in many forms such as mechanical, electrical, and chemical to name a few. Hence, dU can be expressed as the sum of many different works.
 |
| The cross-section of a bomb calorimeter |
From the equation above, it takes a certain amount of energy to heat something up. Clearly, it takes twice as much energy to heat 2 kg of Cu as it does of 1 kg of Cu. Besides that, it also takes a different amount of heat to heat 1 kg of Cu or 1 kg of diamond or 1 kg of air. This resistance to heating is called the heat capacity and when something melts or boils, energy goes in but the temperature does not rise, therefore at a first order phase transition to the heat capacity is infinite.
 |
| The curve of heat capacity at constant volume against temperature |
In a simple way a derivative is nothing more than for a gradient of a line and in our problem we have a surface as shown below instead of a line. Then, the big question is how to solve it. In short, it can be solved depends on which way you look at the surface as to what the gradient is.
 |
| Left: The internal energy of system varies with volume and temperature. Right: The internal energy of system varies with temperature (gradient = heat capacity) |
In earlier part, it was mentioned that if you prevent expansion, system does no work, so dU = dq, therefore you can measure the internal energy. Furthermore, you can reverse this argument and say if the system cannot expand then the increase in internal energy is equal to the heat supplied. Hence, if you know the heat input you do not need to worry about if the temperature changes (as in an adiabatic system).
Now, the question what happens if it can expand? The change in internal energy is less than the heat supplied, as some energy goes to expansion. If this is done at constant pressure then the heat supplied is equal to the enthalpy. Mathematically, enthalpy can be written as:
 |
| The schematic diagram of differential scanning calorimetry (DSC) |
As we discussed early on about enthalpy phase transition, we will try to discuss briefly for some phase transitions. The phase transition for a solid-solid or solid-liquid or liquid-liquid transition has volume changes very small, so pV term can be ignored to give dU = dH and it can be measured by a bomb calorimeter (DSC is easier). However, this statement is not true for liquid or solid to gas transition as now the volume change is large.
In a solid or liquid to gas reaction the volume change is very large and usually 700-1000 times. Thus, we can use the ideal gas equation and ignore the volume of the solid or liquid. Therefore, the enthalpy for solid or liquid to gas transition can be written as:
The enthalpy of a system increases if you add heat energy to it and if you increase the temperature its enthalpy must also be increase. This relationship is given by the capacity at constant pressure as shown below.
The heat capacities of perfect gases are related by a simple equation:
Now, we move to another cases in thermodynamics as the big question what happens when a gas expands? Let takes the example of fridge. When gas inside the tube is compressed it heats up, this heat is released at the back of the fridge in the narrow pipes. Then the gas is allowed into bigger tubes inside the fridge, it expands and cools. This cooling phenomena because the internal energy decreases as the gas expands and in doing so does work. If the pipe was perfectly insulated the temperature of the gas would fall even more. Thus, it raise a question what would be the change in internal energy?
Now, we are going to make use of the fact that the internal energy is a state function and this means it does not matter how we go from one state to the other (just the difference is important). Therefore, in the real system the gas expands and cools at the same time but we will say first the gas expands the it cools (this makes it a lot easier). By keeping temperature constant, work done by an expanding gas, at constant pressure is:
For adiabatic changes pV curves are known as adiabats and adiabats falls more steeply than an isotherm. That is because in an isotherm heat must flow into the system to maintain the temperature as V increases.
 |
| An adiabat depicts the variation of pressure with volume when a gas expands adiabatically |
The standard enthalpy changes is measured at the standard state of a substance at a specified temperature is its pure form at 1 bar (100 kPa), so for Fe at 500 K it is pure iron at 500 K and 1 bar, it means we need to have a "clean" starting point and define everything from there. Another examples are the standard enthalpies of vaporisation and fusion is defined at 1 bar and 298.15 K.
When the enthalpy of vaporisation is added to the enthalpy of fusion we can calculate the enthalpy of sublimation because it is a state function.
 |
| Germain Henri Hess |
The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided.
Hess's Law
The combustion reaction of methane that was shown early on can be written by relating this enthalpy to the standard molar enthalpies of formation of the pure substances.
In the same way, you can use elements in their pure forms to assemble chemical reactions. For example the standard enthalpy of formation of benzene is given by:
Another application of Hess's law is to calculate the enthalpy of combustion of propene with the data:
- The enthalpy of hydrogenation of propene = -124 kJ mol mol−1
- The standard reaction enthalpy for the combustion of propane = -2220 kJ mol−1
- The standard reaction enthalpy for the combustion of hydrogen = -286 kJ mol−1
The enthalpy of combustion of propene can be described as the diagram shown below:
In all our cases earlier on about enthalpy, we used the standard reaction enthalpy and it is always at 298 K. However, if the temperature is raised the enthalpy changes, so if a reaction occurs at a higher temperature then there are changes that should be taken into account for accurate results. In practice the changes for reactants and products are almost the same in many cases. Hence, the big question is how do we calculate the change?
 |
| The variation of enthalpies as temperature changing |
Cp (C(graphite)) = 8.527 J K−1 mol−1
Cp (H2) = 28.824 J K−1 mol−1
Cp (C4H10) = 97.45 J K−1 mol−1
Hence, we can calculate the enthalpy of formation of butane at 400 K. Firstly, we need to calculate ΔCp:
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