1. Quantitative Thermochemistry

2. Chemistry Joke
Q: What weapon can you make from the elements potassium, nickel, and iron?
A: KNiFe!

3. Stoichiometry can be used with thermochemical equations.
Chemical Heat Changes
Stoichiometry can be used with thermochemical equations.
CH4 + 2O2 ® CO2 + 2H2O kJ
1 mole of CH4 releases kJ of energy.
When you make kJ you also make 2 moles of water.

4. Thermochemical Stoichiometry
CH4 + 2 O2 ® CO2 + 2 H2O kJ
If 10.3 grams of CH4 are burned, how much heat will be produced?
Use kJ in a stoichiometric ratio!
10. 3 g CH4
1 mol CH4 kJ
16.05 g CH4
1 mol CH4
= kJ

5. Thermochemical Stoichiometry
CH4 + 2 O2 ® CO2 + 2 H2O kJ
If you want to produce 500. kJ of heat, how many grams of CH4 need to be burned?
500. kJ
1 mol CH4
16.05 g CH4 kJ
1 mol CH4
= 10.0 g

6. Physical Heat Changes
Individual substances respond differently to heat changes.
Some substances, like water, can absorb a lot of heat before their temperature rises.
Other substances, like metals, only need a little bit of heat to cause a temperature change.

7. Physical Heat Change
The amount of heat it takes to raise the temperature of 1g of a substance 1oC is called the specific heat capacity of that substance.
“Specific” heat capacity is independent of mass.
Heat capacity is dependent on the mass of the object.

8. Specific Heat Capacity
Symbol is “C”
Units are J/g ·ºC or cal/g ·ºC
When considering physical heat changes, “q” is used to indicate heat.
C = q
m x T
Specific
Heat
heat
mass x temp change =
T = Tfinal - Tinitial

9. Specific Heat Problem
When 435 J of heat are added to 3.4 g of olive oil at 21oC, the temperature increases to 85oC. What is the specific heat of olive oil?
C = q / (m x T)
T = 85oC - 21oC = 64ºC
435 J / (3.4 g x 64ºC)
=  2.0 J/(g ·ºC)

10. Specific Heat Problem
It takes J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific heat of copper?
0.39 J/g·°C

11. Heat Problems
The equation for specific heat can be rearranged to solve for heat.
q = m C T
Note that heat, q, and specific heat, C, are two different things.
q can be positive or negative depending on the T.
(Remember, negative heat indicates an exothermic process, and positive heat indicates an endothermic process.)

12. Heat Problem
How much heat is required to raise the temperature of 250 g of mercury 52oC? The specific heat of Hg is 0.14 J/(g ·oC).
q = m x C x T
250 g x 0.14 J /(g ·ºC) x 52ºC
=1820  1800 J or 1.8 kJ

13. Heat Problem
How much energy must be absorbed by 20.0 g of water to increase its temperature from ºC to ºC? The specific heat of water is J/g ·oC.
1670 J 

14. Phase Changes
To find the heat required to melt a substance, q = m ∆Hfus.
To find the heat required to vaporize a substance, q = m ∆Hvap.
The mass units have to match.

15. Phase Changes Water: ∆Hfus = 333 J/g ∆HVap = 2250 J/g
How much heat is required to change 25 g of ice at 0 ºC to water at 0 ºC?
q = m ∆Hfus
25 g x 333 J/g
8300 J

16. Phase Changes Water: ∆Hfus = 333 J/g ∆HVap = 2250 J/g
How much heat in kJ is required to vaporize 2.0 moles of water?
81 kJ

17. Using a Heating Curve
How much energy would it take to change 25 g of water at 30 ºC to steam at 110 ºC?
Slanted line: q = m C ∆T
Flat lines: q = m ∆H

18. Using a Heating Curve
Step1: Draw a graph / label low and high temp from problem
Step 2: Determine any phase changes
Step 3: Draw slanted lines to and from phase changes and flat lines for phase changes
Step 4: Write one equation for each line segment. Use the correct C, ∆T, or ∆H.
Step 5: Add the energies together

19. 25 g of water at 30 ºC to steam at 110 ºC (Answer: 64,085 J)
Water: ∆Hfus = 333 J/g ∆HVap = 2250 J/g
Cwater = J/g ºC Cice = 2.15 J/g ºC Csteam = 2.05 J/g ºC
25 g of water at 30 ºC to steam at 110 ºC
(Answer: 64,085 J)

20. Chemistry Joke
Oh no! All the good chemistry jokes argon!
I don’t “zinc” so!