1. Thermochemistry
Chapter 16

2. Thermochemistry
The study of energy involved during chemical reactions.
Energy sources:
~ chemical
~ nuclear
~ solar
~ geothermal
~ wind/water

3. Heat: the energy of motion of molecules
Kinetic molecular theory: substances are composed of particles that are continually moving and colliding with other particles.
Kinetic energy: energy of motion
Potential energy: stored energy

5. Temperature:
transfer of heat to a substance because of faster molecular movement (as long as there is no phase change)
Measured with a thermometer

6. A temperature change is explained as a change in kinetic energy
Temperature depends on the quantity of heat (q) flowing out or in of the substance.

7. Heat (q) q=mc ∆t q=heat m=mass ∆t=change in temperature (tf-ti)
c=specific heat capacity (J/(g oC)
Specific heat capacity is the quantity of heat required to raise the temperature of a unit mass of a substance by one degree Celsius.

8. Table of Specific Heats

9. Law of conservation of energy
∆E universe = O
The total energy of the universe is constant, it is not created or destroyed, however it can be transferred from one substance to another.
∆E universe = ∆E system + ∆E surroundings

10. First Law of thermodynamics
Any change in energy of a system is equivalent by an opposite change in energy of the surroundings.
∆E system = - ∆E surroundings
According to this law, any energy released or absorbed by a system will have a transfer of heat, q.
So, q system = -q surroundings

11. Sample Problem
15 g of ice was added to 60.0g of water. The Ti of water was 26.5 oC, the final temperature of the mixture was 9.70 oC. How much heat was lost by the water?
q=mc ∆t
q=(60.0g) (4.18 J/g oC) ( oC)
q= J =-4.21 kJ

12. Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC.
q = m x c x (Tf - Ti) q = 250g x 4.18 J/g oC x (56oC - 20oC) q = J = 38 kJ

13. Calculate the specific heat capacity of copper given that 204
Calculate the specific heat capacity of copper given that J of energy raises the temperature of 15g of copper from 25oC to 60oC.
q = m x c x (Tf - Ti) q = J m = 15g Ti = 25 oC Tf = 60 oC
J = 15 g x c x (60oC - 25oC) J= 15 g x c x 35oC J= 525 g oC x c c = J ÷ 525 g oC = 0.39 J/ g oC

14. Watch flash about heat flow

15. Homework:
Handout

16. Enthalpy (H)
Total kinetic and potential energy (internal energy) of a system under constant pressure.
System: area where reaction takes place
Its particles average motion define its properties.
Surroundings: outside of the system

17. Open-system
both matter and energy can freely cross from the system to the surroundings and back.
Ex: an open test tube

18. Closed-System energy can cross the boundary, but matter cannot.
Ex: a sealed test tube

19. Isolated-System
neither matter nor energy can cross between the system and the surroundings.
Ex: The universe
there are no surroundings to exchange matter or energy with (as far as we know!)

20. ∆ H = Hproducts – Hreactants
The internal energy of a reactant or product cannot be measured, but their change in enthalpy (heat of reaction) can.
∆ H = Hproducts – Hreactants
A change in enthalpy occurs during phase changes, chemical reactions and nuclear reactions.
∆ H system = q surroundings

21. If energy is released out of the system then it is considered to be an exothermic enthalpy change (exit).
If energy is absorbed into the system then it is considered to be an endothermic enthalpy change (enter).

22. Endothermic Reactions

23. Endothermic Reactions
Method 2:
2 HgO (s)  2 Hg (l) + O2(g) ΔH= kJ
Method 3:
2 HgO (s) kJ  2 Hg (l) + O2(g)

24. Exothermic Reactions

25. Exothermic Reactions Method 2:
4Al(s)+ 3O2(g) 2 Al2O3(g) ΔH= kJ
Method 3:
4 Al(s) + 3O2(g) 2 Al2O3(g) kJ

26. Calorimeter
Instrument used to measure amount of energy involved in a chemical reaction.
It is equivalent to an isolated or closed system. (nothing may enter or exit the system)
The energy change is not measured within the system, but the energy transferred to its surroundings.

27. A basic calorimeter
Two styrofoam cups nestled within one another (insulation), then filled with a specific quantity of water.
A chemical reaction or phase change takes place inside and a thermometer is placed within to measure any change in temperature that occurs to the system.

29. Assumptions
It is an isolated or closed system and there is no heat transfer between the calorimeter and its surroundings.
The amount of heat absorbed or released by the calorimeter itself is too small to influence calculations.
Any dilute solutions involved in the reaction are treated as if they are water.

30. Bomb calorimeter Cannot use basic design for combustion reactions.
Used in research for H for fuels, oils, food, explosives…
Larger and more sophisticated
The reaction container is strong enough to with stand an explosion, hence the name “bomb”

31. Have fixed components, like volume of water, thermometer…
Heavily insulated or vacuum insulated so no convection or conduction can occur affecting the enthalpy of the system.
Use the equation: q=CΔt

32. A Bomb Calorimeter

33. Problems
Pg 638 # 5-7
Handout Questions 1-10

34. Standard Molar Enthalpy of Formation
Quantity of energy released (-) or absorbed (+) when one mole of a compound is formed directly from its elements at standard temperature and pressure.
We use a table to find them.
Unit for ΔHf: kJ/mol
Watch your states!

35. Calculating enthalpy changes
Amount of a substance reacting matters, so can use q= nΔH.
Remember n=amount of moles.
If you are given a mass (g) and molar mass (g/mol), then you can solve for n by dividing mass by molar mass. (review from chem 11 stoichiometry section)

36. Sample Problem: Show the formation reaction of methanol.
If I had 10.5 g of CH3OH(l), how much energy would be released?
C(s) + 2H2(g) + 1/2O2(g)  CH3OH(l) ΔHf = -239 kJ/mol
mm= (12.01) +(4 x 1.01) + (16.00) = g/mol
q = nΔH = (10.5 g / g/mol) (-239 kJ/mol) = KJ

37. On Your Own:
I have g of NaOH(s) formed at standard temperature and pressure, how much energy will be released?
Answer: kJ

38. Standard Molar Enthalpy of Combustion
Energy changes involved with combustion reactions of one mole of a substance.
Remember that these reactions are only measured once cooled to 25oC
Combustion is a reaction with oxygen as a reactant

39. Practice Problem: Write the combustion reaction for methane (CH4)
ΔHcomb for CH4 is kJ/mol
How much energy is released when 12.8 g of CH4 is combusted?
CH4(g) + 2O2 (g)  2H2O(l) + CO2 (g) ΔHcomb = kJ/mol
q=nΔH = (12.8 g/ g/mol) ( kJ/mol) = kJ

40. Combustion reaction: Always form water and carbon dioxide.
Ex: CH4 + 2O2  CO2 + 2H2O
Remember your alkanes: CnH2n+2
Meth: C=1, Eth: C=2, Prop: C=3, But: C=4, Pent: C=5, Hex: C=6…

41. Practice
Pg 643 # 15-17
Pg 645 # 19,21,23

42. Enthalpy and changes of state
A change in matter without any change in chemical composition of the system
*** Always involves energy changes, but not temperature changes. There is constant temperature during a phase change. Why?

43. Because…
No change in kinetic energy, however we know that energy is entering the system because the bonds that are holding the molecules together are being broken or altered. This increases Ep of the molecules

44. Just think!
We know that there are energy changes during phase changes, because when we sweat our body is trying to cool itself down using evaporation!

45. Phase changes: (surroundings)
(s) to (l)
fusion
endo T
(l) to (g)
vaporization
(g) to (l)
condensation
exo
(l) to (s)
solidifying
(g) to (s)
sublimation
(s) to (g)

46. Latent Heat of Phase Change ΔH melt = - ΔH freez
Molar Heat of Fusion or melting
The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point.
Molar Heat of Freezing or solidifying
The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.

47. ΔH vap = - ΔH cond Molar Heat of Vaporization
The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point.
Molar Heat of Condensation
The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.

48. Molar Enthalpy for changes of state
You can find tables of these values listed
Unit: kJ/mol

49. Latent Heat – Sample Problem
Problem: The molar heat of fusion of water is
6.009 kJ/mol. How much energy is needed to convert 60g of ice at 0C to liquid water at 0C?
Mass
of ice
Molar
Mass of
water
Heat of
fusion

50. Chemical reactions:
If a reaction is exothermic, energy is released to the surroundings (temp increases) and chemical potential energy of the system decreases
∆ H is a negative number

51. If a reaction is endothermic, energy is absorbed from the surroundings (temp. decreases) and the chemical potential energy of the system increases
Δ H is a positive number

52. It requires times more energy for a chemical reaction to take place than a phase change because a chemical reaction requires stronger ionic and covalent bonds to be broken, but a phase change only requires intermolecular bonds to be broken.

53. Manipulating the Enthalpy Change Term
The value of Δ H given as kJ/ mol refers to kJ per 1 mole of reactant or product as written in the equation.
N2(g) + 3H2(g) 2NH3(g)       Δ H = kJ
92.4 kJ of energy is released for every 1 mole of N2(g), 3 moles of H2(g), 2 moles of NH3(g) produced.

54. How much energy is released if only 1 mole of ammonia (NH3) gas is produced?
92.4 kJ of energy is released in the production of 2 moles of ammonia gas
Half as much energy will be released if only half the amount of ammonia gas is produced
½ x 92.4 = 46.2 kJ of energy will be released in the production of 1 mole of ammonia.

55. How much energy is released if 10 moles of nitrogen (N2) gas and 30 moles of hydrogen (H2) gas is used in the reaction?
92.4 kJ of energy is released for every 1 mole of N2(g)
10 times as much energy will be released if 10 times the amount of reactants are used
10 x 92.4 = 924kJ of energy will be released

56. Problems Pg
# 26,28-29
Handout #16-18

57. Heat Curve: Temperature vs. Time

58. Total energy change of a system
When a given amount of energy is available to heat a substance, sometimes there is more then one change of state involved as the substance goes from an initial to a final temperature.

59. This requires energy calculations for changing the temperature in a certain state and then calculations for a change of state.
To calculate the energy involved in a change of temperature of a substance without a change of state, we use q=mc ∆t

60. To calculate the energy involved in a change of state we would use:
q = nΔHrxn

61. Sample Problem #1
Calculate the total energy needed or released during the cooling of 1000 g of liquid iron (c= 0.82 J/g oC) to solid iron (c=0.52 J/g oC), where the initial temperature is 1700 oC to 80 oC.
The enthalpy of solidification is -15 KJ/mol.
The melting point of iron is 1535 oC.

62. Heating Curve
1700 q1 q2
1535
Temp
(oC) q3 80
Time (min)

63. ΔE = cooling liquid + phase change from l to s + solid cooling
= q1 + q2 + q3
= q (cooling (l)) + n Δ Hsolid+q(cooling (s))
= mc Δt + n Δ Hsolid + mc Δt
= [(1000g)(0.82 J/g C)( )] + [(1000g/55.85 g/mol)(-15 KJ/mol) ] + [ (1000g)( 0.52 J/g C) ( )]
= J + ( KJ) + ( J) = KJ

64. Sample Problem #2
How much energy is required to raise the temperature of 50.0 g of water (ice) at °C to a temperature of 175°C?
c H2O(l) = 4.18 J/g C
c H2O(g)and (s) = 2.01 J/g C
∆Hmelt = 6.02 kJ/mol
∆Hvap = 40.7 kJ/mol
Melting point = 0 oC
Boiling point = 100 oC
First, start by drawing your own heat curve.

65. the energy required to change the temperature of the ice from -15
the energy required to change the temperature of the ice from -15.0°C to 0°C, q1; the energy involved in the change of state as the ice melts at 0°C, q2; the energy involved to heat the liquid water from 0°C to 100°C, q3; the energy involved in the change of state as the water boils, q4; and finally the energy required to heat water vapour at 100°C to 175°C, q5.

66. Heat Curve oC
100 oC
Temp (oC)
0 oC
- 15 oC
Time (min)

67. q1 = mcΔt = 50.0 g x (2.06 J/g•°C) x 15.0°C = 1540 J or 1.54 kJ
q2 = nΔHfus = 50.0 g / (18.01 g/mol) x 6.03 kJ/mol = 16.7 kJ
q3 = mcΔt = 50.0 g x (4.18 J/g•°C) x 100.0°C = J or 20.9 kJ
q4 = nΔHvap = 50.0 g/(18.01 g/mol) x 40.8 kJ/mol = 113 kJ
q5 = mcΔt = 50.0 g x (2.02 J/g•°C) x 75.0°C = J or 7.58 kJ
qtotal = q1 + q2 + q3 + q4 + q5 = 1.54 kJ kJ kJ kJ kJ = 160 kJ

68. Practice:
Pg 655 # 30-34
Handout #1 (finish it, #19&20)