1. Energy
Chapters 3,4,5,6

2. Thermodynamics: The branch of science that deals with the study of energy and its transformations Thermochemistry : The study of heat changes that occur in a chemical reaction

3. What is Energy ? It is the capacity of a system to do work
Can cause a change in matter (change of state)
Plays a significant role in chemical reactions
It is neither created nor destroyed, it can only transform
Energy changes are measured in joules (J)
1 joule = 1 N m
1N = 1 kg x m/s2

4. Energy in Reactions
Most chemical reactions either release or absorb energy usually in the form of heat. Energy is always conserved amongst two parts:
i) the system – the chemical reaction
ii) the surroundings – everything outside the system
The total energy is : system + surroundings

5. 1. Exothermic Reactions Occurs when there is a release of energy
When atoms combine to form a compound
Example: combustion burning of a substance
Energy (in kJ) written on the products side of equations
System surroundings
2H2 (g) + O2 (g)  2 H2O (g) kJ
heat

6. 2. Endothermic Reactions
Occurs when energy is absorbed during the reactions
When a compound breaks apart into individual atoms (energy required to break bonds)
Energy (in kJ) written on reactants side of equations
Surroundings system
H2O (s) kJ  H2O (l)
2H2O (l) kJ  H2 (g) + ½ O2 (g)
heat

7. Experimental Observations for Energy Change
Exothermic
Heat loss by system to surroundings
Container feels warm
Endothermic
Heat gained by system from surroundings
Container feels cool
Measuring heat
Exothermic reaction: heat given off and temperature of water rises
Endothermic reaction: heat taken in and temperature of water drops

8. Enthalpy (H)
Def’n: is a measure of the energy content of a system (At a constant pressure)
Molar enthalpy: measures energy for exo/endo of one mole of a substance
The heat content of a substance. Heat cannot be measured directly so we measure the change of enthalpy (delta H).

9. Enthalpy changes (ΔH)
Def’n : is a measure of the heat change in a reaction
HR = enthalpy of the reactants
HP = enthalpy of the products
Reactants (HR)  Products (HP) .

10. If ΔH > 0 , the reaction is endothermic (heat is absorbed).
Energy lies on the left of the equation
Writing the equation:
2NH3 (g) KJ  N2(g) + 3H2 (g)
2NH3 (g)  N2(g) + 3H2 (g) ΔH = kJ/mol

11. If ΔH < 0, the reaction is exothermic (heat is released by system).
Energy lies on the right of the equation
Writing the equation:
N2 (g) + 3H2 (g)  2NH3 (g) KJ
N2 (g) + 3H2 (g)  2NH3 (g) ΔH = kJ/mol

12. Writing the equation:
N2 (g) + 3H2 (g)  2NH3 (g) kJ
N2 (g) + 3H2 (g)  2NH3 (g) ΔH = kJ/mol

13. Forms of Energy Kinetic: Energy due to motion
Potential: Stored energy due to position of the atoms within a molecule. It is stored in the bonds that hold atoms together.

14. Vibrational Energy : the atoms move towards and away from the center of mass
Rotational Energy : the molecule rotates around its center of mass
Translational Energy: the molecule moves from place to place

15. kinetic energy
kinetic energy
kinetic energy

16. Reaction Pathway Shows the change in energy during a chemical reaction
Activation energy ?
ΔH=

17. 2H2(l) + O2(l)  2H2O(g) + energy
A. Exothermic Reaction
reaction that releases energy
products have lower Potential Energy than reactants
energy released
2H2(l) + O2(l)  2H2O(g) + energy

18. B. Endothermic Reaction
reaction that absorbs energy
reactants have lower Potential Energy than products
energy absorbed
2Al2O3 + energy  4Al + 3O2

19. Types of Systems
Open system : is a system that freely exchanges energy and matter with its surroundings.
Closed system: is a system that exchanges only energy with its surroundings, not matter.
Isolated system: isolated system does not exchange energy or matter with its surroundings.

20. Calorimetry and the calorimeter
Calorimetry is used to experimentally determine the quantities of heat involved in transformation
Calorimeter: instrument that experimentally determines the heat energy (ΔH) absorbed or released during a reaction.
Reaction in a sealed container representing a closed system
Determines the enthalpy of a substance undergoing a chemical change

21. Don’t copy (Can be found on pg 132 ans 143 of textbook)

22. Calorimeter with Styrofoam
Two Styrofoam cups
One thermometer
- Insulated container that contains a thermometer

23. Q= m x c x ΔT Thermal Energy
Q = Quantity of heat energy, expressed in joules (J)
m= mass of the water in the calorimeter in g
c= the specific heat of water or what ever is acting as the environment (J/(g °C))
Specific heat: the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius (pg 134)
c for water = 4.19 J/(g °C)
ΔT = Temperature change (Tf – Ti) expressed in °C
PAGE

24. Calculate the quantity of thermal energy absorbed by a 5 kg block of concrete to raise its temperature from 17.1°C to 35.5°C.
m=5kg = 5000g
c= 2.10 J/(g °C) specific heat
Ti = 17.1 °C
Tf = 35.5 °C
ΔT = ?
Q = ?

25. A) ΔT = ?
ΔT = Tf - Ti
ΔT = 35.5 °C °C
ΔT = °C B)
Q= m x C x ΔT
= (5000 g)x (2.10 J/(g °C)) x (18.4 °C )
= J

26. Heat Transfer between two systems
Heat travels from hot  cold
-Q1 = Q Qlost = Qgained
Therefore
-(m1c1ΔT1)= m2c2ΔT2
ΔT1 = temp. change in system 1
ΔT2 = temp. change in system 2
Nov 3) !!!!

27. Calculate the mass of cold water at 10°C needed to cool to 30°C a 10-g piece of glass at 95°C.
c1 = 0.84 J/(g °C)
c2=4.19 J/(g °C)
Data
m1= 10 g
Ti1 = 95 °C
Tf1= 30 °C
m2= ?
Pg 138

28. 1) Calculate T1
ΔT1 = Tf1 – Ti1
= 30 °C - 95 °C
= - 65 °C
2) Calculate T2
ΔT2 = Tf2 – Ti2
= 30 °C - 10 °C
= 20 °C 3)
-(m1c1ΔT1)= m2c2ΔT2
m2= -(m1c1ΔT1)= -(10 x 0.84 x -65 ) = 6.5g of water
c2ΔT x 20

29. Temperature of the two systems
Tf = m2c2Ti2 + m1c1Ti1 m1c1 + m2c2
Tf = Final temperature of the two systems in °C
Tf = m2Ti2 + m1Ti1
m1 + m2

30. Calculate the final temperature when 200g of water at 20°C is mixed with 400g of water at 60°C.
Heat loss = Heat gained -Q= Q -(m x c x ΔT) = m x c x ΔT -( 400x 4.19 x (χ-60)) = (200)x(4.19)x ( χ- 20) -(1676 (χ-60)) = 838 (x-20) -1676χ = 838χ – = 2514 χ χ = 46.7 °C

31. Tf = mHTH + mCTC Tf = m2Ti2 + m1Ti1 mH + mc m1 + m2
Tf = Final temp. in °C
mH= mass of hot water
mc= mass of cold water
TH = temp. of hot water
Tc= temp. of cold water
Tf = (400x 60) + (200 x 20)
= (24000 ) + (400)
= 46.7 °C

32. Molar Enthalpy and Thermal Energy
Q= m x c x ΔT
n= mass/Molar Mass
ΔH = -Q/n (kJ/mol)
Nov 8
ΔH= (+)  Endothermic
ΔH= (-)  Exothermic

33. Find the molar heat of LiOH when you dissolve 4. 0g of LiOH in 200
Find the molar heat of LiOH when you dissolve 4.0g of LiOH in g of water causing the water temperature to increase from 25.0°C to 31.5°C
LiOH
m= 4.0g
Water
m= g
Ti = 25 °C
Tf= 31.5 °C
c=4.19 J/(g °C)

34. Step 1: Q= m x c x ΔT
= 200x 4.19 x (31.5 – 25)
= 5447 J
= kJ
Step 2: n= mass/Molar Mass
= 4.0g / 24.0 g/mol
= mol
Step 3: ΔH = -Q/n
= / 0.167
= kJ/ mol

35. In a calorimetry experiment, 4
In a calorimetry experiment, 4.24g of LiCl(s) is dissolved in 100g of water at an initial temperature of 16.3 °C. The final temperature of the solution is 25.1 °C. Calculate the molar enthalpy ΔH, for LiCl.
LiCl
m= 4.24g
Water
m= g
Ti = 16.3 °C
Tf= 25.1 °C
c=4.19 J/(g °C)

36. Step 1: Q= m x c x ΔT
= 100x 4.19 x (25.1 – 16.3)
= J
= 3.69 kJ
Step 2: n= mass/Molar Mass
= 4.24g / g/mol
= 0.10 mol
Step 3: ΔH = -Q/n
= / 0.1
= kJ/ mol

37. Percent error
Percent error = (experimental value) - (theoretical value) x 100 theoretical value

38. Find the molar heat of LiOH when you dissolve 4. 0g of LiOH in 200
Find the molar heat of LiOH when you dissolve 4.0g of LiOH in ml of water causing the water temp to increase from 25.0°C to 31.5°C
Step 1: Q= m x c x ΔT
= 200x 4.19 x (31.5 – 25)
= 5447 J
= kJ
Step 2: n= mass/Molar Mass
= 4.0g / 24.0 g/mol
= mols
Step 3:
ΔH = -Q/n
= / 0.167
= kJ/ mol

39. Percent error
Percent error = (experimental value) - (theoretical value) x 100 theoretical value

40. Making and breaking bonds.Pg
But examples of bond energies

41. ΔH = H bonds broken - H bonds formed (reactants) (products)
Each type of chemical bond has a characteristic bond energy.
Breaking bonds requires energy. It is endothermic.
Making new bonds gives out energy. It is exothermic.
The energy associated for different bonds can be found on page 156 of your textbook.
To calculate ΔH use the formula:
ΔH = H bonds broken - H bonds formed
(reactants) (products)
Breaking bonds absorbs energy – It is endo
Forming bonds releases energy – it is exothemic

42. Can be found on page 156 of textbook (Don’t Copy )
Bond
Energy (kJ/mol)
H-H
436
H-O
460
H-Cl
432
C=C
607
0=0
358
N=O
631
N Ξ N
418

43. H2(g) + Br2(g) 2HBr(g) Calculate the DH for the above reaction.
DH = H bonds broken - H bonds formed
DH =( ) - ( )
DH = kJ/mol
Bonds Broken:
H-H kJ/mol
Br-Br 192 kJ/mol
Bonds Formed:
H-Br kJ/mol

44. Calculate the enthalpy change for the following reactions
H2 (g) + Cl2 (g)  2 HCl (g) CH4(g) + 2O2(g)  CO2 (g) + 2H2O(g)

45. CH3COOH(aq) + 5/2 O2(g)  2CO2(g) + 3H2O(g)
H C C H O H