1. Chapter 6 Thermochemistry
6.1 The Nature of Energy

2. The Nature of Energy Energy- the capacity to do work or produce heat
Law of conservation of energy- energy can be converted but not created or destroyed. Energy of universe is constant

3. potential energy- (PE) due to position or composition
Types of Energy
potential energy- (PE) due to position or composition
ex. attractive or repulsive forces
kinetic energy- (KE) due to motion of the object
KE = ½mv2 :depends on mass and volume

4. Types of Energy
(a): PEA > PEB
(b):
ball A has rolled down the hill
has lost PE to friction and PE in ball B

5. Transfer of Energy
Temperature- measure of the average kinetic energy of the particles. It reflects random motion of particles in substance
Two Ways to Transfer Energy:
Heat- (q) transfer of energy between two objects because of a temperature difference
Work- (w) force acting over a distance

6. Pathway the specific conditions of energy transfer
energy change is independent of pathway because it is a state function
State function: a property of a system that depends only on its present state
work and heat depend on pathway so are not state functions
state function- depends only on current conditions, not past or future

7. Transfer of Energy Chemical Energy
System - part of the universe you are focused on
Surroundings- everything else in the universe
usually
system: what is inside the container. Reactants or products of a chemical reaction
surroundings: container ,room, etc.

8. Transfer of Energy Exothermic energy is produced in reaction
flows out of system
container feels hot to the touch
Endothermic
energy is consumed by the reaction
flows into the system
container feels cold to the touch

9. CH4(g) + 2O2 (g) CO2 + 2H2O(g) + energy (heat)
Transfer of Energy
CH4(g) + 2O2 (g) CO2 + 2H2O(g) + energy (heat)
Combustion of Methane Gas is exothermic

10. N2(g) + 2O2 (g) energy (heat) CO2 + 2H2O(g) +
Transfer of Energy
N2(g) + 2O2 (g) energy (heat) CO2 + 2H2O(g) +
Reaction between nitrogen and oxygen is endothermic

11. Transfer of Energy
the energy comes from the potential difference between the reactants and products
energy produced (or absorbed) by reaction must equal the energy absorbed (or produced) by surroundings
usually the molecules with higher potential energy have weaker bonds than molecules with lower potential energy

12. Thermodynamics
Thermodynamics- study of energy and its interconversions (transfers)
First Law of Thermodynamics Energy of universe is constant
(Law of conservation of energy)

13. Internal Energy (E) sum of potential and kinetic energy in system
can be changed by work, heat, or both
E = PE + KE
∆E = q + w
E of a system can
be changed by flow
of heat, work or both

14. change in internal energy
Signs
Signs are very important in thermodynamic quantities
Signs will always reflect the system’s point of view unless otherwise stated ∆E q w
change in internal energy
heat
work
Exothermic -
Endothermic +

15. Signs

16. Work common types of work Expansion- work done by gas
Compression- work done on a gas
P is external pressure – not internal like we normally refer to
expansion
+∆V -w
compression
-∆V +w

17. Work

18. Example 1
Find the ∆E for endothermic process where 15.6 kJ of heat flows and 1.4 kJ of work is done on system
Since it is endothermic, q is + and w is +

19. Since it is an expansion, ∆V is + and w is -
Example 2
Calculate the work of expansion of a gas from 46 L to 64 L at a constant pressure of 15 atm.
Since it is an expansion, ∆V is + and w is -

20. Example 3
A balloon was inflated from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of heat. Assuming the pressure is 1.0 atm, find the ∆E in Joules.
(1 L∙atm=101.3 J)
Since it is an expansion, ∆V is + and w is -

21. 6.2 Enthalpy and Calorimetry

22. Enthalpy Enthalpy is a property of a system definition: H = E + PV
since E, P and V are all state functions, then H is too
for the following, the process is at constant P and the only type of work allowed is PV work
∆E = qP + w = qP - P∆V  qP = ∆ E +P∆V
H = E + PV  ∆H = ∆E+P∆V
so, qP = ∆H at constant P
At constant P, ∆H of a
System is equal to the
Energy flow as heat

23. Enthalpy
For reactions carried out at constant
P, the flow of heat is a measure of
change in H of a system
heat of reaction and change in enthalpy are used interchangeably for a reaction at constant P
∆H = Hproducts - Hreactants
endo: + ∆H exo: - ∆H

24. Calorimetry science of measuring heat
calorimeter- device used to experimentally find the heat associated with a chemical reaction
substances respond differently when heated ( to raise T for two substances by 1 degree, they require different amount of heat)

25. Heat Capacity
(C) how much heat it takes to raise a substance’s T by one °C or K
the amount of energy depends on the amount of substance

26. Heat Capacity Specific heat capacity Molar heat capacity
(s) heat capacity per gram
in J/°C*g or J/K*g
Molar heat capacity
heat capacity per mole
in J/°C*mol or J/K*mol

27. Constant-Pressure Calorimetry
uses simplest calorimeter (like coffee-cup calorimeter) since it is open to air
used to find changes in enthalpy (heats of reaction) for reactions occurring in a solution since qP = ∆H
heat of reaction is an extensive property, so we usually write them per mole so they are easier to use

28. Constant-Pressure Calorimetry
when 2 reactants are mixed and T increases, the chemical reaction must be releasing heat so is exothermic
the released energy from the reaction increases the motion of molecules, which in turn increases the T

29. Constant-Pressure Calorimetry
If we assume that the calorimeter did not leak energy or absorb any itself (that all the energy was used to increase the T), we can find the energy released by the reaction:
E released by rxn = E absorbed by soln
∆H = qP = sP x m x ∆T

30. Constant-Volume Calorimetry
uses a bomb calorimeter
weighed reactants are placed inside the rigid, steel container and ignited
water surrounds the reactant container so the T of it and other parts are measured before and after reaction

31. Constant-Volume Calorimetry
Here, the ∆V = 0 so -P∆V = w = 0
∆E = q + w = qV for constant volume
E released by rxn = ∆T x Ccalorimeter

32. Example 1
When 1 mol of CH4 is burned at constant P, 890 kJ of heat is released. Find ∆H for burning of 5.8 g of CH4 at constant P.
890 kJ is released per mole of CH4

33. Example 2
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0°C in a coffee-cup calorimeter, solid BaSO4 forms and the T increases to 28.1°C. The specific heat capacity of the solution is 4.18 J/g*°C and the density is 1.0 g/mL. Find the enthalpy change per mole of BaSO4 formed.

34. Example 2 Write the net ionic equation for the reaction:
Ba2+ (aq) + SO42- (aq)  BaSO4(s)
Is the energy released or absorbed? What does that mean about ∆H and q?
exothermic: -∆H and –qP
How can we calculate ∆H or heat?
heat = q = sP x m x ∆T
How can we find the m?
use density and volume

35. Example 2 Find the mass: Find the change in T:
Calculate the heat created:

36. Example 2
since it is a one-to-one ratio and the moles of reactants are the same, there is no limiting reactant
1.0 mol of solid BaSO4 is made so
∆H= -2.6x104 J/mol = -26 kJ/mol

37. Example 3
Compare the energy released in the combustion of H2 and CH4 carried out in a bomb calorimeter with a heat capacity of 11.3 kJ/°C. The combustion of 1.50 g of methane produced a T change of 7.3°C while the combustion of 1.15 g of hydrogen produced a T change of 14.3°C. Find the energy of combustion per gram for each.

38. Example 3 methane: CH4 hydrogen: H2
The energy released by H2 is about 2.5 times the energy released by CH4

39. 6.3 Hess’ Law

40. Hess’ Law
since H is a state function, the change in H is independent of pathway
Hess’ Law- when going from a set of reactants to a set of products, the ∆H is the same whether it happens in one step or a series of steps

41. Example 1

42. Example 1
N2(g) + 2O2(g)  2NO2(g) ∆H = 68 kJ OR
N2(g) + O2(g)  2NO(g) ∆H = 180 kJ
2NO(g) + O2(g)  2NO2(g)∆H = -112 kJ

43. If a reaction is reversed, the sign of ∆H must be reversed as well.
Rules
If a reaction is reversed, the sign of ∆H must be reversed as well.
because the sign tells us the direction of heat flow at constant P
The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction.
If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way.
because ∆H is an extensive property

44. Example 2
Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to diamond.
Cgraphite (s)  Cdiamond (s) ∆H=?

45. Example 2 (1) Cgraphite(s) + O2(g)  CO2(g) ∆H=-394kJ/mol
(2) Cdiamond(s) + O2(g)  CO2(g) ∆H=-396kJ/mol
to get the desired equation, we must reverse 2nd equation:
-(2) CO2(g)  Cdiamond(s) + O2(g) ∆H=396kJ/mol
Cgraphite (s)  Cdiamond (s) ∆H=
∆H=2 kJ/mol

46. H2(g) + ½O2(g)  H2O(l) ∆H3=-286kJ
Example 3
Find ∆H for the synthesis of B2H6, diborane:
2B(s) + 3H2(g)  B2H6(g)
Given:
2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ
B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g) ∆H2=-2035kJ
H2(g) + ½O2(g)  H2O(l) ∆H3=-286kJ
H2O(l)  H2O(g) ∆H4=44 kJ

47. Example 3 Need 3 H2 (g) so 3 x (3) Need 3 H2O to cancel so 3 x (4)
(1) 2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ
-(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g) ∆H2=-(-2035kJ)
3x(3) 3H2(g) + 3/2O2(g)  3H2O(l) 3∆H3=3(-286kJ)
3x(4) 3H2O(l)  3H2O(g) 3∆H4=3(44 kJ)
2B(s) + 3H2(g)  B2H6(g)
∆H = (-2035) + 3(-286) + 3(44) = 36kJ

48. 6.4 Standard Enthalpies of Formation

49. Standard Enthalpy of Formation
∆Hf°
change in enthalpy that accompanies the formation of one mole of a compound from its elements in standard states
° means that the process happened under standard conditions so we can compare more easily

50. Standard States For a COMPOUND: For an ELEMENT:
for gas: P = 1 atm
pure liquid or solid state
in solution: concentration is 1 M
For an ELEMENT:
form that it exists in at 1 atm and 25°C
O: O2(g) K: K(s) Br: Br2(l)

51. Writing Formation Equations
always write equation where 1 mole of compound is formed (even if you must use non-integer coefficients)
NO2(g):
½N2(g) + O2(g)  NO2(g)
∆Hf°= 34 kJ/mol
CH3OH(l):
C(s) + H2(g) + O2(g) CH3OH(l)
∆Hf°= -239 kJ/mol

53. Using Standard Enthalpies of Formation
where
n = number of moles of products/reactants
∑ means “sum of”
∆Hf° is the standard enthalpy of formation for reactants or products
∆Hf° for any element in standard state is zero so elements are not included in the summation

54. Using Standard Enthalpies of Formation
since ∆H is a state function, we can use any pathway to calculate it
one convenient pathway is to break reactants into elements and then recombine them into products

55. Using Standard Enthalpies of Formation

56. Using Standard Enthalpies of Formation

57. Example 1
Calculate the standard enthalpy change for the reaction that occurs when ammonia is burned in air to make nitrogen dioxide and water
4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(l)
break them apart into elements and then recombine them into products

58. Example 1

59. Example 1 can be solved using Hess’ Law:
4NH3(g)  2N2(g) + 6H2(g) -4∆Hf°NH3
7O2(g)  7O2(g) 0
2N2(g) + 4O2(g)  4NO2(g) 4 ∆Hf°NO2
6H2(g) + 3O2(g)  6H2O(l) 6 ∆Hf°H2O

60. Example 1 can also be solved using enthalpy of formation equation:
values are in Appendix 4

61. Example 2
Calculate the standard enthalpy change for the following reaction:
2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s)

62. Example 3
Compare the standard enthalpy of combustion per gram of methanol with per gram of gasoline (it is C8H18).
Write equations:
2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(l)
2C8H18(l) + 25O2(g)16CO2(g) + 18H2O(l)

63. Example 3
Calculate the enthalpy of combustion per mole:

64. Example 3 Convert to per gram using molar mass:
so octane is about 2x more effective