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# Heat of Sublimation

The heat (or enthalpy) of sublimation is the amount of energy that must be added to a solid at constant pressure in order to turn it directly into a gas (without passing through the liquid phase).  The heat of sublimation is generally expressed as ΔHsub in units of Joules or kiloJoules per mole or kilogram of substance.

### Things to recall...

1. A "Δ" in front of any variable indicates that the following equation is a state function in which the value of an indicator measured in the initial state of a reaction is subtracted from the value of the same indicator measured in the final state of the reaction.  The resulting value is the change in the indicator at the end of the said reaction (Δindicator).
2. Enthalpy is the amount of energy that is required to induce a phase change (change in the state of matter).
3. Energy is measured in Joules or kiloJoules and is transferred through either through heat (q) or work (w).
4. The phases involved in sublimation are the solid phase and the gas phase.

### Introduction to Sublimation and Heat of Sublimation

Sublimation is the process of changing a solid into a gas without allowing the solid to pass through the liquid phase.  In order to sublime a substance, a certain amount of energy must be transferred to the substance via heat (q) or work (w).  The energy needed to sublime a substance is particular to the substance's identity and temperature and must be enough to do all of the following:

1. Excite the substance(s) so that it reaches its maximum heat (energy) capacity (q) in the solid state.
2. Sever all the atomic bonds in the substance(s).
3. Excite the unbonded atoms of the substance so that it reaches its minimum heat capacity in the gaseous state.

(See attached powerpoint show in the Files section at the bottom of the page.  Select the option to play slideshow.)

### The Equations

The heat (or enthalpy) of sublimation is the amount of energy that must be added to a solid at constant pressure in order to turn it directly into a gas (without passing through the liquid phase).  The heat of sublimation is generally expressed as ΔHsub in units of Joules or kiloJoules per mole or kilogram of substance.

### Things to recall...

1. A "Δ" in front of any variable indicates that the following equation is a state function in which the value of an indicator measured in the initial state of a reaction is subtracted from the value of the same indicator measured in the final state of the reaction.  The resulting value is the change in the indicator at the end of the said reaction (Δindicator).
2. Enthalpy is the amount of energy that is required to induce a phase change (change in the state of matter).
3. Energy is measured in Joules or kiloJoules and is transferred through either through heat (q) or work (w).
4. The phases involved in sublimation are the solid phase and the gas phase.

### Introduction to Sublimation and Heat of Sublimation

Sublimation is the process of changing a solid into a gas without allowing the solid to pass through the liquid phase.  In order to sublime a substance, a certain amount of energy must be transferred to the substance via heat (q) or work (w).  The energy needed to sublime a substance is particular to the substance's identity and temperature and must be enough to do all of the following:

1. Excite the substance(s) so that it reaches its maximum heat (energy) capacity (q) in the solid state.
2. Sever all the atomic bonds in the substance(s).
3. Excite the unbonded atoms of the substance so that it reaches its minimum heat capacity in the gaseous state.

(See attached powerpoint show in the Files section at the bottom of the page.  Select the option to play slideshow.)

### The Equations

#### ΔHsub = ΔEtot

The energies involved in sublimation steps 1 through 3 (see Sublimation ) compose the total amount of energy that is involved in sublimation.  This can be expressed by the equation,

$$\Delta H_{sub} = \Delta E_1+\Delta E_2+\Delta E_3$$

in which

$$\Delta E_1=\Delta E_{thermal_{solid}}$$
$$\Delta E_2=\Delta E_{bond_{from solid to liquid}}$$
$$\Delta E_3=\Delta E_{thermal_{solid}}+\Delta E_{bond_{from liquid to gas}}$$

Although a solid does not actually pass through the liquid phase during the process of sublimation, the fact that enthalpy is a state function allows us to add the various energies associated with the solid, liquid, and gas phases together.  Recall that for state functions, only the initial and final states of the substance are important.  Say for example that state A is the initial state and state B is the final state.  How a substance goes from state A to state B does not matter so much as what state A and what state B are.  Concerning the state function of enthalpy, the energies associated with enthalpies (whose associated states of matter are contiguous to one another) are additive.  Though in sublimation a solid does not pass through the liquid phase on its way to the gas phase, it takes the same amount of energy that it would to first melt (fuse) and then vaporize.

##### ΔEthermal(state of matter)

A change in thermal energy is indicated by a change in temperature (in Kelvin) of a substance at any particular state of matter.  Change in thermal energy is expressed by the equation

$$\Delta E_{thermal_{(of a particular state of matter)}}=(C_p)*\Delta T$$

in which

$$C_p=heat capacity_{(of a particular state of matter)}$$
$$C_p=(specific heat capacity_{(of a particular state of matter)})*mass_{(substance)}$$
$$\Delta T = T_{(final)}-T_{(initial)}$$
$$\Delta T = T_{(at phase change, going from state_1 to state_2)}-T_{(initial)}$$

For more information on heat capacity and specific heat capacity, see heat capacity.  Specific heat capacities of common substances can easily be found online or in a reference or text book.

##### ΔEbond(going from state 1 to state 2)

Bond energy is the amount of energy that a group of atoms must absorb so that it can undergo a phase change (going from a state of lower energy to a state of higher energy).  It is measured

$$\Delta E_{bond}=\Delta H_{substance_{phase change}}*\Delta mass_{(substance)}$$

in which $$\Delta H_{substance_{phase change}}$$ is the enthalpy associated with a specific substance at a specific phase change.  Common types of enthalpies include the heat of fusion (melting) and the heat of vaporization. Recall that fusion is the phase change that occurs between the solid state and the liquid state, and vaporization is the phase change that occurs between the liquid state and the gas state. Note that if the substance has more than one type of bond (or intramolecular force), then the substance must absorb enough energy to break all the different types of bonds before the substance can sublime.1

### Graphical Representations of the Heat of Sublimation

Note that although the graph below indicates the inclusion of the liquid phase, the graph is merely a representation of how much energy is needed to sublime a solid substance.  Recall that sublimation does not include the liquid phase and that the fact that enthalpy is a state function allows us to add the enthalpies of fusion and vaporization together to find the enthalpy of sublimation.

### ΔHsub > ΔHvap

Though both enthalpies involve the changing a substance into its gaseous state, the change in energy associated with sublimation is generally greater than that of vaporization.  This is because of the initial state of the substances and the amount of initial energy that each substance has.  Particles in a solid have less energy than those of a liquid, meaning it is takes more energy to excite a solid to its gaseous phase that it does to excite a liquid to its gaseous phase.

Another way to look this phenomena is to take a look at the different energies involved with the heat of sublimation:  ΔEthermal (s), ΔEbond(s-l), ΔEthermal(l), and ΔEbond(l-g).  Already we know that ΔEbond=ΔH(phase change)*Δm(changed substance) and ΔEbond(l-g)=ΔH(l-g)*Δm(gas created).  ΔH(l-g) is essentially ΔH(vaporization), meaning that ΔHvap is actually a component of ΔHsub.

### Where does the added energy go?

Energy can be observed in many different ways.  As shown above, ΔEtot can be expressed as ΔEthermal + ΔEbondAnother way in which  ΔEtot can be expressed is change in potential energy, ΔPE, plus change in kinetic energy, ΔKE.  Potential energy is the energy associated with random movement, whereas kinetic energy is the energy associated with velocity (movement with direction).  ΔEtot = ΔEthermal + ΔEbond and ΔEtot = ΔPE + ΔKE are related by the equations

ΔPE = (0.5)ΔEthermal + ΔEbond
ΔKE = (0.5)ΔEthermal

for substances in the solid and liquid states.  Note that ΔEthermal is divided equally between ΔPE and ΔKE for substances in the solid and liquid states.  This is because the intermolecular and intramolecular forces that exist between the atoms of the substance (i.e. atomic bond, van der Waals forces, etc) have not yet been dissociated and prevent the atomic particles from moving freely about the atmosphere (with velocity).  Potential energy is just a way to have energy, and it generally describes the random movement that occurs when atoms are forced to be close to one another.  Likewise, kinetic energy is just another way to have energy, which describes an atom's vigorous struggle to move and to break away from the group of atoms.  The thermal energy that is added to the substance is thus divided equally between the potential and the kinetic energies  because all aspects of the atoms' movement must be excited equally

However, once the intermolecular and intramolecular forces which restrict the atoms' movement are dissociated (when enough energy has been added), potential energy no longer exists (for monatomic gases) because the atoms of the substance are no longer forced to vibrate and be in contact with other atoms.  When a group of atoms is in the gaseous state, it's atoms can devote all their energies into moving away from one another (kinetic energy).

### Practical Applications of the Heat of Sublimation

The heat of sublimation can be useful in determining the effectiveness of medicines.  Medicine is often administered in pill (solid) form, and the substances which they contain can sublime over time if the pill absorbs too much energy over time.  Often times you may see the phrase "avoid excessive heat"2 on the bottles of common painkillers (e.g. Advil).  This is because in high temperature conditions, the pills can absorb heat energy, and sublimation can occur3

### Practice Problems

1. If the heat of fusion for H2O is 333.5 kJ/kg, the specific heat capacity of H2O(l) is 4.18 kJ/(kg*K), the heat of vaporization for H2O is 2257 kJ/kg, then calculate the heat of sublimation for 1.00 kg of H2O(s) with the initial temperature, 273K (Hint:  273K is the solid-liquid phase change temperature and 373K is the liquid-gas phase change temperature).
2. Using the information given in question one, calculate the heat of sublimation for 1.00 mole H2O when the initial temperature of the solid is 273K.  (Hint: molar mass of H2O is ~18.0 g/mol or 0.018 kg/mol)
3. Using the information given in question one, calculate the heat of sublimation for 1.00 kg H2O when the initial temperature is 200K.  The specific heat capacity for H2O(s) is 2.05 kJ/(kg*K).
4. If the heat of fusion for Au is 1.24 kJ/mol, the specific heat capacity of Au(l) is 25.4 J/(mol*K), the heat of vaporization for Au is 1701 kJ/kg, then calculate the heat of sublimation for 1.00 mol of Au(s) with the initial temperature, 1336K (Hint:  1336K is the solid-liquid phase change temperature, and 3081K is the liquid-solid phase change temperature).
5. If the heat of sublimation for Cu at 3081K is 313.3245 kJ/mol, the specific heat capacity of Cu(l) is .0245 kJ/(mol*K), the heat of vaporization for Cu is 300.3 kJ/mol, then calculate the heat of fusion at 1356K for 1.00 mol of Cu(s) with the temperature (Hint:  1356K is the solid-liquid phase change temperature, and 3081K is the liquid-gas phase change temperature).

### Practice Problem Solutions

1. ΔHsub for 1kg H2O (at Ti=273K)= (333.5 kJ/kg)(1.0 kg) + (4.18 kJ/kg*K)(373-273K) + (2257 kJ/kg)(1.0 kg) = 30008.5 kJ/kg
2. ΔHsub for 1mol H2O (at Ti=273K)= (30008.5 kJ/kg)(0.018 kg/mol) =  54.153 kJ/mol
3. ΔHsub for 1kg H2O (at Ti=200K)= 30008.5 kJ/kg + (2.05 kJ/K*kg)(1.0kg)(273-200K) = 3158.15 kJ/kg
4. ΔHsub for 1mol Au (at Ti=1336K)= (1.24 kJ/mol)(1mol) + (.0254 kJ/mol*K)(3081-1336K) + (1701 kJ/kg)(0.197kg/mol) = 380.66 kJ/mol
5. ΔHfus for Cu (at T=1356K) = 337.8735 kJ/mol - (0.0245 kJ/mol*K)(2839-1356K) - (300.3 kJ/mol)(1mol) = 1.24 kJ/mol

### Footnotes

1. Dmitry Bedrov, Oleg Borodin, Grant D. Smith, Thomas D. Sewell, Dana M. Dattelbaum, and Lewis L. Steven. "A molecular dynamics simulation study of crystalline 1,3,5-triamino-2,4,6-trinitrobenzene as a function of pressure and temperature." THE JOURNAL OF CHEMICAL PHYSICS 131, 2009.

2. Advil bottle. 24 Ibuprofen tablets, 200mg. EXP 12/08.

3. Pascal Taulelle, Georges Sitja, Gerard Pepe, Eric Garcia, Christian Hoff, and Stephane Veesler. "Measuring Enthalpy of Sublimation for Active Pharmaceutical Ingredients: Validate Crystal Energy and Predict Crystal Habit." Crystal Growth & Design (2009): 4706–4709. Print.

### External References

1. Advil bottle. 24 Ibuprofen tablets, 200mg. EXP 12/08.
2. Dmitry Bedrov, Oleg Borodin, Grant D. Smith, Thomas D. Sewell, Dana M. Dattelbaum, and Lewis L. Steven. "A molecular dynamics simulation study of crystalline 1,3,5-triamino-2,4,6-trinitrobenzene as a function of pressure and temperature." The Journal of Chemical Physics 131 (2009): 1-4.  Print.
3. Pascal Taulelle, Georges Sitja, Gerard Pepe, Eric Garcia, Christian Hoff, and Stephane Veesler.  "Measuring Enthalpy of Sublimation for Active Pharmaceutical Ingredients: Validate Crystal Energy and Predict Crystal Habit."  Crystal Growth & Design (2009):  4706–4709.  Print.
4. Petrucci, Ralph H., William S. Harwood, F. G. Herring, and Jeffry D. Madura. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, NJ: Pearson Prentice Hall, 2007.  242-248.
5. Potter, Wendell.  "Applying Models to Thermal Phenomena."  College Physics:  A Models Approach, Part 1.  Hayden McNeil Publishing:  Plymouth, MI, 2010.  7-20.

### Contributors

• Kasey Nakajima (UCD)

The ChemWiki has 9242 Modules.

UC Davis ChemWiki by University of California, Davis is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

### Graphical Representations of the Heat of Sublimation

Note that although the graph below indicates the inclusion of the liquid phase, the graph is merely a representation of how much energy is needed to sublime a solid substance.  Recall that sublimation does not include the liquid phase and that the fact that enthalpy is a state function allows us to add the enthalpies of fusion and vaporization together to find the enthalpy of sublimation.

### ΔHsub > ΔHvap

Though both enthalpies involve the changing a substance into its gaseous state, the change in energy associated with sublimation is generally greater than that of vaporization.  This is because of the initial state of the substances and the amount of initial energy that each substance has.  Particles in a solid have less energy than those of a liquid, meaning it is takes more energy to excite a solid to its gaseous phase that it does to excite a liquid to its gaseous phase.

Another way to look this phenomena is to take a look at the different energies involved with the heat of sublimation:  ΔEthermal (s), ΔEbond(s-l), ΔEthermal(l), and ΔEbond(l-g).  Already we know that ΔEbond=ΔH(phase change)*Δm(changed substance) and ΔEbond(l-g)=ΔH(l-g)*Δm(gas created).  ΔH(l-g) is essentially ΔH(vaporization), meaning that ΔHvap is actually a component of ΔHsub.

### Where does the added energy go?

Energy can be observed in many different ways.  As shown above, ΔEtot can be expressed as ΔEthermal + ΔEbondAnother way in which  ΔEtot can be expressed is change in potential energy, ΔPE, plus change in kinetic energy, ΔKE.  Potential energy is the energy associated with random movement, whereas kinetic energy is the energy associated with velocity (movement with direction).  ΔEtot = ΔEthermal + ΔEbond and ΔEtot = ΔPE + ΔKE are related by the equations

ΔPE = (0.5)ΔEthermal + ΔEbond
ΔKE = (0.5)ΔEthermal

for substances in the solid and liquid states.  Note that ΔEthermal is divided equally between ΔPE and ΔKE for substances in the solid and liquid states.  This is because the intermolecular and intramolecular forces that exist between the atoms of the substance (i.e. atomic bond, van der Waals forces, etc) have not yet been dissociated and prevent the atomic particles from moving freely about the atmosphere (with velocity).  Potential energy is just a way to have energy, and it generally describes the random movement that occurs when atoms are forced to be close to one another.  Likewise, kinetic energy is just another way to have energy, which describes an atom's vigorous struggle to move and to break away from the group of atoms.  The thermal energy that is added to the substance is thus divided equally between the potential and the kinetic energies  because all aspects of the atoms' movement must be excited equally

However, once the intermolecular and intramolecular forces which restrict the atoms' movement are dissociated (when enough energy has been added), potential energy no longer exists (for monatomic gases) because the atoms of the substance are no longer forced to vibrate and be in contact with other atoms.  When a group of atoms is in the gaseous state, it's atoms can devote all their energies into moving away from one another (kinetic energy).

### Practical Applications of the Heat of Sublimation

The heat of sublimation can be useful in determining the effectiveness of medicines.  Medicine is often administered in pill (solid) form, and the substances which they contain can sublime over time if the pill absorbs too much energy over time.  Often times you may see the phrase "avoid excessive heat"2 on the bottles of common painkillers (e.g. Advil).  This is because in high temperature conditions, the pills can absorb heat energy, and sublimation can occur3

### Practice Problems

1. If the heat of fusion for H2O is 333.5 kJ/kg, the specific heat capacity of H2O(l) is 4.18 kJ/(kg*K), the heat of vaporization for H2O is 2257 kJ/kg, then calculate the heat of sublimation for 1.00 kg of H2O(s) with the initial temperature, 273K (Hint:  273K is the solid-liquid phase change temperature and 373K is the liquid-gas phase change temperature).
2. Using the information given in question one, calculate the heat of sublimation for 1.00 mole H2O when the initial temperature of the solid is 273K.  (Hint: molar mass of H2O is ~18.0 g/mol or 0.018 kg/mol)
3. Using the information given in question one, calculate the heat of sublimation for 1.00 kg H2O when the initial temperature is 200K.  The specific heat capacity for H2O(s) is 2.05 kJ/(kg*K).
4. If the heat of fusion for Au is 1.24 kJ/mol, the specific heat capacity of Au(l) is 25.4 J/(mol*K), the heat of vaporization for Au is 1701 kJ/kg, then calculate the heat of sublimation for 1.00 mol of Au(s) with the initial temperature, 1336K (Hint:  1336K is the solid-liquid phase change temperature, and 3081K is the liquid-solid phase change temperature).
5. If the heat of sublimation for Cu at 3081K is 313.3245 kJ/mol, the specific heat capacity of Cu(l) is .0245 kJ/(mol*K), the heat of vaporization for Cu is 300.3 kJ/mol, then calculate the heat of fusion at 1356K for 1.00 mol of Cu(s) with the temperature (Hint:  1356K is the solid-liquid phase change temperature, and 3081K is the liquid-gas phase change temperature).

### Practice Problem Solutions

1. ΔHsub for 1kg H2O (at Ti=273K)= (333.5 kJ/kg)(1.0 kg) + (4.18 kJ/kg*K)(373-273K) + (2257 kJ/kg)(1.0 kg) = 30008.5 kJ/kg
2. ΔHsub for 1mol H2O (at Ti=273K)= (30008.5 kJ/kg)(0.018 kg/mol) =  54.153 kJ/mol
3. ΔHsub for 1kg H2O (at Ti=200K)= 30008.5 kJ/kg + (2.05 kJ/K*kg)(1.0kg)(273-200K) = 3158.15 kJ/kg
4. ΔHsub for 1mol Au (at Ti=1336K)= (1.24 kJ/mol)(1mol) + (.0254 kJ/mol*K)(3081-1336K) + (1701 kJ/kg)(0.197kg/mol) = 380.66 kJ/mol
5. ΔHfus for Cu (at T=1356K) = 337.8735 kJ/mol - (0.0245 kJ/mol*K)(2839-1356K) - (300.3 kJ/mol)(1mol) = 1.24 kJ/mol

### Footnotes

1. Dmitry Bedrov, Oleg Borodin, Grant D. Smith, Thomas D. Sewell, Dana M. Dattelbaum, and Lewis L. Steven. "A molecular dynamics simulation study of crystalline 1,3,5-triamino-2,4,6-trinitrobenzene as a function of pressure and temperature." THE JOURNAL OF CHEMICAL PHYSICS 131, 2009.

2. Advil bottle. 24 Ibuprofen tablets, 200mg. EXP 12/08.

3. Pascal Taulelle, Georges Sitja, Gerard Pepe, Eric Garcia, Christian Hoff, and Stephane Veesler. "Measuring Enthalpy of Sublimation for Active Pharmaceutical Ingredients: Validate Crystal Energy and Predict Crystal Habit." Crystal Growth & Design (2009): 4706–4709. Print.

### External References

1. Advil bottle. 24 Ibuprofen tablets, 200mg. EXP 12/08.
2. Dmitry Bedrov, Oleg Borodin, Grant D. Smith, Thomas D. Sewell, Dana M. Dattelbaum, and Lewis L. Steven. "A molecular dynamics simulation study of crystalline 1,3,5-triamino-2,4,6-trinitrobenzene as a function of pressure and temperature." The Journal of Chemical Physics 131 (2009): 1-4.  Print.
3. Pascal Taulelle, Georges Sitja, Gerard Pepe, Eric Garcia, Christian Hoff, and Stephane Veesler.  "Measuring Enthalpy of Sublimation for Active Pharmaceutical Ingredients: Validate Crystal Energy and Predict Crystal Habit."  Crystal Growth & Design (2009):  4706–4709.  Print.
4. Petrucci, Ralph H., William S. Harwood, F. G. Herring, and Jeffry D. Madura. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, NJ: Pearson Prentice Hall, 2007.  242-248.
5. Potter, Wendell.  "Applying Models to Thermal Phenomena."  College Physics:  A Models Approach, Part 1.  Hayden McNeil Publishing:  Plymouth, MI, 2010.  7-20.

### Contributors

• Kasey Nakajima (UCD)

The ChemWiki has 9242 Modules.

UC Davis ChemWiki by University of California, Davis is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License

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chemwiki.docx
final graphs (added into word document, modified in Microsoft Word, printed and rescanned into computer--could not upload excel graph into module directly as a graphic)
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IMG.jpg
sublimation vs. fusion and vaporization
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sublimation vs. vaporization
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sublimation
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IMG_NEW_0001.jpg
graph
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sublimation.ppsx
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sublimation.ppt
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sublimation.pptx
powerpoint 2007
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created graph in excel, transferred and printed with Word
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