1. Thermochemistry

2. Energy, Enthalpy and Thermochemistry
Energy: the ability to do work or transfer heat.
Kinetic Energy: energy due to the motion of the object (1/2 mv2)
Potential Energy: energy due to position or composition
Thermal energy: the energy associated with the random motion of atoms and molecules
Heat: the transfer of thermal energy between two objects due to a temperature difference
Work: a force acting over a distance.

3. Units of Energy Units of Energy: kg m2 1 J = 1  s2
The SI unit of energy is the joule (J):
An older, non-SI unit is still in widespread use: the calorie (cal).
1 cal = J
1 J = 1 
kg m2 s2
Energy of Matters:

4. Thermochemistry
Thermochemistry is the study of heat changes in chemical reactions
The system is the specific part of the universe that is of interest in the study.
The surrounding is everything in the universe
other than the system.

5. SURROUNDINGS
SYSTEM
open
closed
isolated
Exchange:
mass & energy
energy
nothing

6. First Law of Thermodynamics
The total energy of the universe is constant; energy can be transformed from one form to another, but cannot be created or destroyed.
In closed system
∆E= ∆U = q+w
q: the heat added to the system during the process
w: the work done on the system during the process

7. Internal Energy
The change in internal energy, E, is the final energy of the system minus the initial energy of the system:
E = Efinal − Einitial

8. Changes in Internal Energy
When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).
That is, E = q + w.
DE is the change in internal energy of a system
q is the heat exchange between the system and the surroundings
w is the work done on (or by) the system
w = -PDV when a gas expands against a constant external pressure

9. E, q, w, and Their Signs
q>0 heat flows into the system from the surroundings
q<0 an outflow of heat from the system to the surroundings
w>0 work is done on the system by the surroundings
w<0 the system does work on the surroundings

10. 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.
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
energy + 2HgO (s) Hg (l) + O2 (g)
energy + H2O (s) H2O (l)

11. Examples of endothermic and exothermic reactions
2 NH4SCN(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.
Endothermic reaction

12. The reaction of powdered aluminum with Fe2O3 is highly exothermic
The reaction of powdered aluminum with Fe2O3 is highly exothermic. The reaction proceeds vigorously to form Al2O3 and molten iron:
2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(l ).
Exothermic reaction

13. Work
Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas).

14. P-V Work
w = -PDV when a gas expands against a constant external pressure
w is the work done on (or by) the system

15. We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.
w = -PV

16. Enthalpy
If a process takes place at constant pressure and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy (H) of the system, in other word enthalpy (H)is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
Enthalpy is the internal energy plus the product of pressure and volume:
H = E + PV

17. When the system changes at constant pressure, the change in enthalpy, H, is
H = (E + PV)
This can be written
H = E + PV
Since E = q + w and w = -PV, we can substitute these into the enthalpy expression:
H = (q+w) − w
H = q
So, at constant pressure, the change in enthalpy is the heat gained or lost.

18. PV = nfRT – niRT = (nf – ni)RT = nRT
H = E + PV
E = H - PV
For Gas Materials
PV = nfRT – niRT = (nf – ni)RT = nRT
Where n is the numbers of moles of gas molecules

19. Olive oils is completely burned in oxygen at 100
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. n = nf – ni = (52+57) – 80 = 29 PV = nRT = 29 x x 373 = J = kJ E = H - PV = – = kJ

20. Enthalpy of Reaction
For a chemical reaction : ΔH= Hproducts- Hreactants
If Hreactants< Hproducts (endothermic)
If Hreactants> Hproducts (exothermic)
DH = heat given off or absorbed during a reaction
Hproducts < Hreactants
Hproducts > Hreactants
DH < 0 (negative)
DH > 0 (positive)

21. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles
H2O (s) H2O (l)
DH = 6.01 kJ
If you reverse a reaction, the sign of DH is changed
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must be multiplied by n
2H2O (s) H2O (l)
DH = 2 x 6.01 = 12.0 kJ

22. The states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ

23. The Truth about Enthalpy
Enthalpy is an extensive property.
H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.
H for a reaction depends on the state of the products and the state of the reactants.

24. P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burns in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ
mol P4
g P4 x kJ
mol P4 x
266 g P4
= ______________ kJ

25. Heat Capacity and Specific Heat
The amount of energy (heat (q)) required to raise the temperature of a substance by 1 K (1C) is its heat capacity.
We define specific heat capacity (or simply specific heat) as the amount of energy (heat (q)) required to raise the temperature of 1 g of a substance by 1 K.
Examples

26. mass  temperature change
Specific heat, then, is
Specific heat =
heat transferred
mass  temperature change
CS = q
m  T
DT = Tfinal - Tinitial

27. How much heat is given off when an 869 g iron bar cools from 940C to 50C?
Cs of Fe = J/g • 0C
DT = Tfinal – Tinitial = ___ 0C – ___ 0C = ____ 0C
q = mxCsxDT
= ____ g x ____ J/g • 0C x ___ 0C
= _____ J

28. How much heat is needed to warm 250 g of water from 22 °C to 98 °C
How much heat is needed to warm 250 g of water from 22 °C to 98 °C? (b) What is the molar heat capacity of water?
(a) T= 98 oC – 22 oC = 76 oC = 76 K
q = CS x m x T
= (4.18 J/g-K)(250 g)(76 K)= 7.9 x 104 J
(b) 1 mol H2O = 18 g H2O
Cm = (4.18 J/g-K)(18 g/mol) = 75.2 J/mol-K
CS = q
m  T

29. Calorimetry
Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow
A constant-pressure calorimetry is used in determining the change in enthalpy equals the heat.
∆H=q

30. Constant Pressure Calorimetry
By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one 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 H for the reaction with this equation:
q = m  CS  T

31. (Constant-Volume Calorimetry)
Bomb Calorimetry
(Constant-Volume Calorimetry)
Reactions can be carried out in a sealed “bomb” such as this one.
The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.

32. Total heat capacity of the calorimeter Ccal qrnx = - Ccal x ∆T
CH6N2 is used as a liquid rocked fuel. The combustion of it with O2 produces N2, CO2 and H2O. When 4 g of CH6N2 is combusted in a bomb calorimeter, the temperature of the calorimeter increase from 25 ˚C to 39.5 ˚C . Ccal of calorimeter is KJ/ ˚C. calculate the heat of reaction for the combustion of a mole of CH6N2.
∆T = 14.5 ˚C
qrnx = - Ccal x ∆T = -(7.794 KJ/ ˚C) x (14.5 ˚C ) = -113 KJ
= -113 KJ / (4 g / 46 g mol-1) = -1.3 x 103 KJ /mol

33. When a student mixes 50 mL of 1 M HCl and 50 mL of NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increase from 21 oC to 27.5 oC. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses a negligible quantity of heat, that its density is 1 g/mL and that its specific heat is 4.18 J/g-K.
Total mass = 100 mL x 1 g/mL = 100 g
ΔT = oC – 21 oC = 6.5 oC = 6.5 K
qrnx = -Cs x m x ΔT = J/g-K x 100 g x 6.5 K = 2.7 x 103 J = -2.7 kJ
ΔH = qp = kJ.
No of HCl moles = 0.05 L x 1 mol / L = 0.05 mol
ΔH = -2.7 kJ/0.05 mol = -54 kJ/mol

34. 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.

36. Consider the combustion reaction of methane to form CO2 and liquid H2O
CH4(g) + 2O2(g)→ CO2(g) + 2H2O(l)
∆H1 =-890KJ/mol
This reaction can be thought of as occurring in two steps:
CH4(g) + 2O2(g)→ CO2(g) + 2H2O(g)
∆H2 = -802 kJ/mol
2H2O(g)→2H2O(l)
∆H3 =-88KJ/mol
∆H1 = ∆H2 + ∆H3

37. The enthalpy of reaction for the combustion of C to CO2 is –393
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

38. Standard Enthalpies of Formation f
(DH0) f
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 (DH0) 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. f

39. The standard enthalpy of formation of any element in its most stable form (allotrope) is zero.
DH0 (O2) = 0 f
DH0 (O3) = 142 kJ/mol
DH0 (C, graphite) = 0
DH0 (C, diamond) = 1.90 kJ/mol
Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure).

40. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)

41. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)

42. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)

43. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

44. Calculation of H We can use Hess’s law in this way:
H = nHf°products – mHf° reactants
where n and m are the stoichiometric coefficients.

45. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3( kJ) + 4( kJ)] – [1( kJ) + 5(0 kJ)]
= [( kJ) + ( kJ)] – [( kJ) + (0 kJ)]
= ( kJ) – ( kJ) = kJ

46. Example