1. Ch. 16 Thermochemistry Notes
Heat changes that occur
during a chemical reaction

2. What is Thermochemistry?
The study of heat changes in chemical reactions and physical states.
Law of Conservation of Energy
Calorimetry
Temperature
Heat
Heat Capacity
Specific Heat

3. Energy The capacity for doing work or supplying heat
Energy is weightless, colorless, and odorless
Energy that is stored in the bonds of chemical substances is called chemical potential energy. (Example: gasoline)

4. Forms of Energy Potential Kinetic Work Enthalpy stored energy
Energy of motion
Work
energy used in a process
Enthalpy
Heat energy lost or gained in a process

5. Law of Conservation of Energy
Energy is neither created or destroyed only transferred.

6. What is work and heat? Work = when a force is used to move an object
Heat = energy that is transferred from one object to another, because of a temperature difference. (q)
Heat flows from warm to cold until equilibrium is reached

7. Heat flow into a system is an endothermic process (+q)
When describing the flow of heat:
Heat flow into a system is an endothermic process (+q)
Heat flow out of a system is an exothermic process (-q)
(Calorimeter—an instrument that measures heat changes in a closed system)

8. What is a calorie?
The quantity of heat required to raise 1 gram of water 1 °C.
The SI unit of measurement for heat is the joule.
1 joule = calories
1 calorie = joules

9. Specific Heat Capacity (Specific Heat)
The amount of heat needed to raise 1 g of a substance 1 °C.
Table 1 pg 533
Heat capacity depends upon the mass and the composition of the object.

10. Definition: Specific Heat
The amount of heat it takes
to raise 1 gram of a substance
1 °C.

11. Examples
Water has the highest specific heat capacity. It absorbs a lot of heat before finally increasing temperature.
Metals have the lowest specific heat capacity. A small amount of absorbed energy immediately changes temperature

12. There is a calculation that can be performed to determine the specific heat of an object.
q = C x m x ∆T
C = Specific Heat (J/g°C or cal/g°C)
q = heat (J or cal)
m = mass (g)
∆T = T final – T initial (°C)

13. Sample Problems

14. When 435 J of heat is added to 3
When 435 J of heat is added to 3.4 grams of olive oil at 21°C, the temperature increases to 85°C. What is the specific heat of olive oil?
C = ?
q = 435 J
m = 3.4 g
∆T = Tfinal – Tinitial = 85°C – 21°C = 64 °C
q = C x m x ∆T

15. C = ?
q = 435 J
m = 3.4 g
∆T = Tfinal – Tinitial = 85°C – 21°C = 64 °C
q = C x m x ∆T
Solve for C
C = q / (m x ∆T)
C = 435 J / (3.4 g x 64 °C) = 2.0 J/g°C

16. How much heat is required to raise the temperature of 250
How much heat is required to raise the temperature of g of mercury 52 C°? (specific heat of mercury is 0.14 J/g°C)
C = 0.14 J/g • °C
q = ?
m = g
∆T = 52 °C
q = C x m x ∆T
q =(0.14 J/g°C)(250.0 g)(52 °C)= 1820 J
Is this endothermic or exothermic?

17. Remember… Heat flow out of a system is an exothermic process (-q)
Heat flow into a system is an endothermic process (+q)
q =(0.14 J/g°C)(250.0 g)(52 °C)= 1800 J
Positive = Endothermic

18. Endo vs. Exo?
If the temperature of a reaction increases then the reaction is endothermic= heat in!
This makes a (+q)
If the temperature of a reaction decreases the then reaction is exothermic= heat out!
This makes a (-q)

19. Now you solve the problems
Practice --Heat problems
1. Define each variable in the equation for specific heat: q = C x m x ∆T
2. How many Joules of heat energy are required to raise the temp of 45 g of water from 10.2 C to 26.8 C?
3. How many calories of heat are released when 275 g of water cool from 85.2 C to 38.4 C?
4. What will the change in temperature be if 80 g of water absorb 7220 J of heat energy?
5. What is the specific heat of lead if a 30 g sample undergoes a 250 C temp change while absorbing Joules of energy?

20. Now do the practice problem worksheet!!
20 minutes to do then we review and continue notes

21. C = Specific Heat (J/g°C or cal/g°C) q = heat (J or cal) m = mass (g)
1. Define each variable in the equation for specific heat: q = C x m x ∆T
C = Specific Heat (J/g°C or cal/g°C)
q = heat (J or cal)
m = mass (g)
∆T = T final – T initial (°C)

22. 2. How many Joules of heat energy are required to raise the temp of 45 g of water from 10.2 C to 26.8 C?
C = 4.18 J/g • °C
q = ?
m = 45 g
∆T = 26.8 °C °C =16.6 °C
q = C x m x ∆T
q =(4.18 J/g°C)(45 g)(16.6 °C)= 3123J

23. 3. How many calories of heat are released when 275 g of water cool from 85.2 C to 38.4 C?
C = 1.0 cal/g • °C
q = ?
m = 275 g
∆T = 38.4 °C °C = °C
q = C x m x ∆T
q =(1.0 cal/g•°C)(275g)(-46.8°C)= cal

24. 4. What will the change in temp
4. What will the change in temp. be if 80 g of water absorb 7220 J of heat energy?
C = 4.18 J/g • °C
q = 7220 J
m = 80 g
∆T = ?
∆T= . q .
C x m
∆T= ( J )
(4.18 J/g • °C )(80 g)= 21.6 °C

25. 5. What is the specific heat of lead if a 30 g sample undergoes a 250 C temp change while absorbing Joules of energy?
C = ?
q = J
m = 30 g
∆T = 250 °C
C = q .
m x ∆T
C = ( J )
(30 g)(250 °C)= J/g°C

26. Enthalpy q= ∆H = C x m x ∆T ∆H = enthalpy
H=Heat of substance at a constant pressure
q= ∆H = C x m x ∆T
∆H = enthalpy
C = Specific Heat (J/g°C or cal/g°C)
q = heat (J or cal)
m = mass (g)
∆T = T final – T initial (°C)

27. Enthalpy
+ ∆H = endothermic rxn
- ∆H = exothermic rxn

28. Enthalpy of Reaction
The quantity of energy transferred as heat during a chemical reaction.
ΔH = Hreactants- Hproducts

29. Thermochemical Reactions ∆ H = Hreactants- Hproducts
Heat of Reaction – the heat required or released in a reaction (coefficients are mole quantities)
Exothermic Reaction
CaO (s) + H2O (l) → Ca(OH)2 (aq) kJ ∆H= kJ (heat is released)
See graph page 536
Endothermic Reaction
NaHCO3(s) kJ → Na2CO3 (s) + H2O (g) + CO2(g)
∆H= kJ (heat absorbed)
See graph page 537

30. Practice 1) Calculate the ∆ H of the reaction.
2NaHCO3 (s) kJ→ Na2CO3(s) + H2O(g) + CO2(g)
1) Calculate the ∆ H of the reaction.
∆ H = Hreactants- Hproducts
∆H= 129 kJ – 0 kJ= kJ
2) Calculate the kJ of heat required to decompose 2.24 mol NaHCO3(s).
2.24 mol NaHCO3 x (+ 129 kJ) = 144 kJ
2 mol NaHCO3

31. More Practice
2NaHCO3 (s) kJ→ Na2CO3(s) + H2O(g) + CO2(g)
3) How much energy is absorbed when 25g of NaHCO3 decomposes?
25g NaHCO3 x 1 mol NaHCO3 x 129 kJ = 19.2 kJ
84g NaHCO mol NaHCO3