1. Part 3
Thermochemistry

2. Thermochemistry

3. Thermochemistry
Thermochemistry is the study of heat changes in chemical reactions.
- Exchange of Heat between System and Surroundings:
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

4. Examples of endothermic and exothermic reactions:
1) 2NH4SCN(s)+ Ba(OH)2*8 H2O(s) → Ba(SCN)2(aq) + 2NH3(aq) + 10H2O(l)
As a result, the temperature of the system drops from about 20 °C to -9 °C.
2) 2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(l ).
The reaction of powdered aluminum with Fe2O3 is highly exothermic. The reaction proceeds vigorously to form Al2O3 and molten iron

5. Endothermic and Exothermic Reactions
The change in enthalpy can be calculated by
ΔHreaction = ƩΔHproducts - ƩΔHreactants
if ΔHreaction is negative, then the reaction is exothermic reaction.
if ΔHreaction is positive, then the reaction is endothermic reaction.

7. Thermochemical reaction
- To write the thermochemical reaction equation, the reaction conditions and states must be written; e.g. C6H6 (l, 1atm, 273K) + O2 (g, 1atm, 273K).
- The thermochemical reaction equation must be balanced:
- ΔH for a reaction in the forward direction is equal in value, but opposite in sign to ΔH for the reveres reaction.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (s)
DH = kJ

8. Factors that affect the reaction heat:
1- Physical state of materials
ΔHreaction depends on the state of the reactants and the state of the products. This factor can be divided into:
A- Crystaline of the reactants

9. B- State of matter

11. 2- Number of moles ( amount of materials)

12. 3- Reaction conditions (P, V, T)

14. Example
Olive oils is completely burned in oxygen at ˚C according to:
∆H= kJ
Calculate the change in the internal energy ∆E (in kJ) for this combustion process.
Solution:

16. Enthalpy ( H) Hess’s Law
H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.
Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”
- The total enthalpy change depends only on the initial state of the reactants and the final state of the products.

22. Example
The enthalpy of reaction for the combustion of C to CO2 is –393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO2 is –283.0 kJ/mol CO:
(1) C(s) + O2(g) CO2(g) H1 = kJ
(2) CO(g) O2(g) CO2(g) H2 = kJ
Using these data, calculate the enthalpy for combustion of C to CO:
(3) C(s) O2(g) CO(g) H3 = ? kJ
Solution:

23. Example
The enthalpy of reaction for the combustion of C to CO2 is –393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO2 is –283.0 kJ/mol CO:
(1) C(s) + O2(g) CO2(g) H1 = kJ
(2) CO(g) O2(g) CO2(g) H2 = kJ
Using these data, calculate the enthalpy for combustion of C to CO:
(3) C(s) O2(g) CO(g) H3 = ? kJ
Solution:

24. Kirchhoff's law

29. Standard Enthalpies of Formation ( ΔHfo )
- An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.
- Standard of enthalpies formation (ΔHf0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.
- The superscript zero indicates that the corresponding process has been carried out under standard conditions.
- The standard enthalpy of formation of any element in its most stable form (allotrope) is zero.

30. Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure).
- To calculate the ΔH of the reaction:

31. Determination of ΔHf0
There are a number of ways in which to measure the enthalpy of formation of a compound; here are two. The most obvious is to simply carry out the formation reaction from the constituent elements in their standard states in a constant pressure calorimeter (will be explained later). For example, consider the combustion of graphite to form carbon dioxide
The heat released in this reaction is -ΔHf0 (CO2), since the standard enthalpy of formation of the reactants is zero. For this method to work, two conditions must be met:
1) the reaction goes to completion.
2) only one product is formed.

32. Thus, the reaction:
is not suitable for this method since it doesn’t readily go to completion and we get a complicated mixture of hydrocarbons. In order to get around this, note that it is often possible to burn something to completion (and measure ΔHcombustion, the heat released). Thus consider:
Using the equation:
ΔHcombustion = ΔH0f (CO2(g)) + 2ΔH0f (H2O(l)) - ΔH0f (CH4(g)).

33. The standard enthalpies of formation of carbon dioxide and water can be measured using the first method; hence, once we measure the heat of combustion, the only unknown is the standard enthalpy of formation of methane (CH4) and a little algebra gives:
ΔH0f (CH4(g)) = ΔH0f (CO2(g)) + 2ΔH0f (H2O(l)) - ΔHcombustion
= [ ( ) - ( )]kJ/mole
= kJ/mole

34. Standard Enthalpy of formation

35. Example
The enthalpy of reaction for the combustion of C to CO2 is –393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO2 is –283.0 kJ/mol CO:
(1) C(s) + O2(g) CO2(g) H1 = kJ
(2) CO(g) O2(g) CO2(g) H2 = kJ
Using these data, calculate the enthalpy for combustion of C to CO:
(3) C(s) O2(g) CO(g) H3 = ? kJ
Solution:

36. Example
The enthalpy of reaction for the combustion of C to CO2 is –393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO2 is –283.0 kJ/mol CO:
(1) C(s) + O2(g) CO2(g) H1 = kJ
Solution:

37. Bond enthalpy ( Bond energy)
A useful way to calculate enthalpy changes in chemical reactions is through the concept of bond energies. Bond energy is defined as the energy required to break a chemical bond. It is usually measured at 273K.
Enthalpy can be measured by
ΔHreaction = Ʃ nΔH(bonds broken) - Ʃ nΔH(bonds formed)
Consider the following reaction
one C-H bond is broken. Experimentally, ΔH for this reaction is 102 kcal/mole

38. A survey of such reactions will show that the heat required to break a single C-H bond is in the range kcal/mole. We can thus assign 98 kcal/mole as an average bond energy for the C-H bond. Similar tends are observed in the bond strength of other types of bonds, and the results are summarized in the following table:

39. How can this be used? Consider the hydrogenation of ethlyene:
At the molecular level, we break one H-H and one C=C bond, and form one C-C and two C-H bonds. The energy change is just the net energy left in the molecule in such a process. From the table, the bond breaking steps take =248 kcal/mole. The bond formation will give off 80+2(98)=276 kcal/mole. Hence the net energy change in the system is = -28 kcal/mol.

54. Enthalpy of solution (ΔHsoln)
ΔHsoln can be calculated using
ΔHsoln = ΔH1 + ΔH2 + ΔH3
ΔH1 is the separation of solute molecules.
ΔH2 is the separation of solvent molecules.
ΔH3 is the formation of solute-solvent interactions.

55. ΔHsoln may be positive sign (energy absorbed)
negative sign (energy released)

56. Example:
ΔHsoln is small positive value which indicate that the dissolving process requires a small amount of energy.

59. Determination the heat of chemical reaction
Since we cannot know the exact enthalpy of the reactants and products, we measure ΔH through calorimetry, the measurement of heat flow. Two calorimetry can be used:
1- Constant-pressure calorimetry
2- Constant-volume calorimetry

60. Constant-pressure calorimetry
qreaction= − qcalorimeter
A constant-pressure calorimetry is used in determining the change in enthalpy equals the heat (∆H=qreaction). By carrying out a reaction in aqueous solution in a simple calorimeter

61. Constant-pressure calorimetry
which can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter. Because the specific heat for water is well known (4.184 J/g.K), we can measure ΔH for the reaction with this equation:
qp = m x CS x ΔT

62. Example
A 155 g piece of a certain metal was heated to 120 oC and then plunged in a constant pressure coffee-cup calorimeter, containing 250 g of water at 22.oC. The temperature of the system became 28 oC. Assuming that there was no heat energy transferred to the surrounding and knowing that the specific heat of water is j/g.oC. Calculate (in j/g.oC) the specific heat of this metal?
Solution:

63. Example
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0 oC is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25oC in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1oC. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.18 J/g.K and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed
Solution:

64. Constant-Volume Calorimetry (Bomb Calorimetry)
The heat absorbed (or released) by the water is a very good approximation of the energy change for the reaction. Once the sample is completely combusted, the heat released in the reaction transfers to the water and the calorimeter. The temperature change of the water is measured with a thermometer. Because water and calorimetry absorb heat, then the total heat given off in the reaction will be equal to the heat gained by the water and the calorimeter:
qrxn= - qcal

65. Constant-Volume Calorimetry (Bomb Calorimetry)
the heat gained by the calorimeter is the sum of the heat gained by the water, as well as the calorimeter itself. This can be expressed as follows:
qcal = qwater + qbomb
qcal = mwater x Cswater x ΔT + Cbomb x ΔT
where mwater is the water mass, Cswater denotes the specific heat capacity of the water, and Cbomb is the heat capacity of the calorimeter.

66. Constant-Volume Calorimetry (Bomb Calorimetry)
A bomb calorimeter is used to measure the heat of reaction at constant volume (qv) which is equal to the change in internal energy, ΔU , of a reaction. Thus, the total heat given off by the reaction is related to the change in internal energy (ΔU), not the change in enthalpy (ΔH) which is measured under conditions of constant pressure. The enthalpy change (ΔH) can be calculated according to the formula:
ΔH = qv + ΔngRT
which Δng  is the change in the number of moles of gases in the reaction.

67. Example
1.150 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42°C to 27.64°C and the heat capacity of the calorimeter is 4.90 kJ/°C, then determine the heat of combustion of sucrose, (C12H22O12), in kJ per mole of (C12H22O12).
Solution:
Using equation (2) calculate qcalorimeter:
qcalorimeter = (4.90 kJ/°C) x ( )°C = (4.90 x 4.22) kJ = 20.7 kJ 
Plug into equation (1):
qrxn=−qcalorimeter=−20.7kJ
But the question asks for kJ/mol C12H22O12, so this needs to be converted:
qrxn=−20.7kJ1.150gC12H22O12=−18.0kJgC12H22O12
Per Mole C12H22O12:
qrxn=−18.0kJgC12H22O12X342.3gC12H22O121molC12H22O12=−6.16X103kJmolC12H22O12

68. Example
To compare the energies of combustion of methane and hydrogen, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/oC. When a 1.50g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3oC. When a 1.15g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3 oC. Calculate the energy of combustion (per mole) for hydrogen and methane.
Solution:

69. Example
A sample of biphenyl (C6H5)2 weighing g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid C6H5COOH weighing g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol–1 (that is, ΔU = –3226 kJ mol–1.) Use this information to determine the standard enthalpy of combustion of biphenyl.
Solution: