1. Ch. 16: Energy and Chemical Change
Sec. 16.3: Thermochemical Equations

2. Objectives
Write thermochemical equation for chemical reactions and other processes.
Describe how energy is lost or gained during changes of state.
Calculate the heat absorbed or released in a chemical reaction.

3. Review
The change in energy is an important part of chemical reactions so chemists include ΔH as part of the chemical equation.

4. Thermochemical Equations
A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products and the energy change.
The energy change is usually expressed as the change in enthalpy, ΔH.

5. Thermochemical Equations
A subscript of ΔH will often give you information about the type of reaction or process taking place.
For example, ΔHcomb is the change in enthalpy for a combustion reaction or, simply, the enthalpy (heat) of combustion.

6. Thermochemical Equations
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
ΔHcomb = kJ/mol
This is the thermochemical equation for the combustion of glucose.
The enthalpy of combustion is kJ/mol.

7. Thermochemical Equations
The enthalpy of combustion (ΔHcomb) of a substance is defined as the enthalpy change for the complete burning of one mole of the substance.
That means, 2808 kJ of heat are released for every mole of glucose that is oxidized (or combusts). Two moles would release 5616 kJ:
2 mol glucose x kJ = kJ
1 mol
The zero superscript tells you that the reactions were carried out under standard conditions.
Standard conditions are one atmosphere of pressure and 298 K (25 degrees C).

8. Standard Enthalpies Standard enthalpy changes have the symbol ΔHo.
Standard conditions in thermochemistry are 1 atm pressure and 25 0C. **DO NOT CONFUSE THESE WITH THE STP CONDITIONS OF THE GAS LAWS.

9. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
Calculations
How much heat is evolved when 54.0 g of glucose (C6H12O6) is burned according to this equation?
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
ΔHcomb = kJ/mol
Since the ΔHcomb given is for 1 mole of glucose, you must first determine the # of moles you have.
54.0 g x 1 mole = moles glucose
180 g

10. Calculations (cont.)
You can now use the ΔHcomb value as a conversion factor:
0.300 mol glucose x kJ = -842 kJ
1 mole

11. Practice Problems
How much heat will be released when 6.44 g of sulfur reacts with O2 according to this equation:
2S + 3O2  2SO3 ΔH0 = kJ/mol
How much heat will be released when 11.8 g of iron react with O2 according to the equation:
3Fe + 2O2  Fe3O4 ΔH0 = kJ/mol

12. Changes of State
The heat required to vaporize one mole of a liquid is called its molar enthalpy (heat) of vaporization (ΔHvap).
The heat required to melt one mole of a solid substance is called its molar enthalpy (heat) of fusion (ΔHfus).
Both phase changes are endothermic & ΔH has a positive value.
Because vaporizing a liquid and melting a solid are endothermic processes, their ΔH values are positive.
These are standard values that can be looked up.

13. Changes of State
The vaporization of water and the melting of ice can be described by the following equations:
H2O(l) → H2O(g) ΔHvap = 40.7 kJ/mol
One mole of water requires 40.7 kJ to vaporize.
H2O(s) → H2O(l) ΔHfus = 6.01 kJ/mol
One mole of ice requires 6.01 kJ to melt.
The 1st equation indicates that 40.7 kJ of energy is absorbed when one mole of water is converted to one mole of water vapor.
The 2nd equation shows that when one mole of ice melts to form one mole of liquid water, 6.01 kJ of energy is absorbed.
What happens in the reverse process?

14. ΔHcond = - ΔHvap = -40.7 kJ/mol ΔHsolid = - ΔHfus = -6.01 kJ/mol
Changes of State
The same amounts of energy are released in the reverse processes (condensation and solidification (freezing)) as are absorbed in the processes of vaporization and melting.
Therefore, they have the same numerical values but are opposite in sign.
ΔHcond = - ΔHvap = kJ/mol
ΔHsolid = - ΔHfus = kJ/mol

15. Practice Problems
How much heat is released when 275 g of ammonia gas condenses at its boiling point? (ΔHvap = 23.3 kJ/mol)
If water at 00 C releases 52.9 J as it freezes, what is the mass of the water? (ΔHfus= 6.01 kJ/mol)
How much heat is required to melt 25 g of ice at its MP? (ΔHfus= 6.01 kJ/mol)

16. Combination Problems
At times, you will need to calculate the amount of heat that is absorbed or released when a temperature change AND a phase change occur in sequence.
Recall, the heat involved in a temperature change is calculated by using: ΔH = mCΔT. NOTE: In this expression, ΔH is found in joules or cal.
Heat involved in a phase change is calculated using dimensional analysis. NOTE: ΔH will be in kJ.

17. Example
How much heat is released when 37.5 g of water that is at C freezes? (ΔHfus= 6.01 kJ/mol)
Water cannot freeze at C. It must be at 0 0C. Therefore, we must first determine how much energy is released when it is cooled from C to 0 0C.
ΔH = mCΔT = (37.5 g)(4.184 J/g 0C)( C)
= J = kJ

18. Example
How much heat is released when 37.5 g of water that is at C freezes? (ΔHfus= 6.01 kJ/mol)
-3.14 kJ of heat are released when the sample is cooled to 0 0C.
Now we must calculate the energy released when the entire sample freezes. Recall that ΔHsolid= -ΔHfus= -6.01kJ/mol
37.5 g x 1 mole x kJ = kJ
18 g mole

19. Example
How much heat is released when 37.5 g of water that is at C freezes? (ΔHfus= 6.01 kJ/mol)
-3.14 kJ of heat are released when the sample is cooled to 0 0C.
-12.5 kJ of heat are released when the sample freezes.
The total heat released is or kJ.

20. Use Cw = 4.184 J/g0C; ΔHfus = 6.01 kJ/mol; ΔHvap = 40.7 kJ/mol.
Practice Problems
Use Cw = J/g0C; ΔHfus = 6.01 kJ/mol; ΔHvap = 40.7 kJ/mol.
How much heat is needed to melt 8 g of ice at 0 0C to water at 15 0C?
How much heat is needed to change 28.0 g of water at C to steam?